CBSE Class 12 Physics Current Electricity Boards Questions Worksheet

Question. Two resistors of resistance R₁ and R₂ having R₁ > R₂ are connected in parallel. For equivalent resistance R, the correct statement is:
(a) R > R₁ + R₂
(b) R₁ < R₁ < R₂
(c) R₂ < R₁ < (R₁ + R₂)
(d) R < R₂ < R₁

Answer: D

Question. In a Wheatstone bridge, all the four arms have equal resistance R. If resistance of the galvanometer arm is also R, then equivalent resistance of the combination is
(a) R
(b) 2R
(c) R/2
(d) R/4

Answer: A

Question. A potentiometer is an accurate and versatile device to make electrical measurement of EMF because the method involves
(a) potential gradients
(b) a condition of no current flow through the galvanometer
(c) a combination of cells, galvanometer and resistance
(d) cells

Answer: B

Question. Consider a current carrying wire (current I ) in the shape of a circle. Note that as the current progresses along the wire, the direction of j (current density) changes in an exact manner, while the current I remain unaffected. The agent that is essentially responsible for is
(a) source of emf.
(b) electric field produced by charges accumulated on the surface of wire.
(c) the charges just behind a given segment of wire which push them just the right way by repulsion.
(d) the charges ahead.

Answer: B

Question. The drift velocity of the free electrons in a conducting wire carrying a current i is v. If in a wire of the same metal, but of double the radius, the current be 2I, then the drift velocity of the electrons will be
(a) v/4
(b) v/2
(c) v
(d) 4v

Answer: B

Question. A resistance R is to be measured using a meter bridge. Student chooses the standard resistance S to be 100 Ω. He finds the null point at l₁ = 2.9 cm. He is told to attempt to improve the accuracy. Which of the following is a useful way?
(a) He should measure l₁ more accurately.
(b) He should change S to 1000 Ω and repeat the experiment.
(c) He should change S to 3 Ω and repeat the experiment.
(d) He should give up hope of a more accurate measurement with a meter bridge.

Answer: C

Question. Two cells of emf ’s approximately 5 V and 10 V are to be accurately compared using a potentiometer of length 400 cm.
(a) The battery that runs the potentiometer should have voltage of 8V.
(b) The battery of potentiometer can have a voltage of 15 V and R adjusted so that the potential drop across the wire slightly exceeds 10 V.
(c) The first portion of 50 cm of wire itself should have a potential drop of 10 V.
(d) Potentiometer is usually used for comparing resistances and not voltages.

Answer: B

Question. The resistivity of iron is 1 × 10^–7 ohm-meter. The resistance of the given wire of a particular thickness and length is 1 ohm. If the diameter and length of the wire both are doubled the resistivity will be (in ohm-meter)
(a) 1 × 10^–7
(b) 2 × 10^–7
(c) 4 × 10^–7
(d) 8 × 10^–7

Answer: A

Question. A student connects 10 dry cells each of emf E and internal resistance r in series, but by mistake the one cell gets wrongly connected. Then net emf and net internal resistance of the combination will be
(a) 8E, 8r
(b) 8E, 10r
(c) 10E, 10r
(d) 8E, r/10

Answer: B

Question. A metal rod of length 10 cm and a rectangular cross-section of 1cm × (1/2) cm is connected to a battery across opposite faces. The resistance will be
(a) maximum when the battery is connected across 1 cm × 1/2) cm faces.
(b) maximum when the battery is connected across 10 cm × 1 cm faces.
(c) maximum when the battery is connected across 10 cm × (1/2) cm faces.
(d) same irrespective of the three faces.

Answer: A

Question. Which of the following characteristics of electrons determines the current in a conductor?
(a) Drift velocity alone
(b) Thermal velocity alone
(c) Both drift velocity and thermal velocity
(d) Neither drift nor thermal velocity.

Answer: A

Question. Temperature dependence of resistivity r(T) of semiconductors insulators and metals is significantly based on the following factors.
(a) Number of charge carriers can change with temperature T.
(b) Time interval between two successive collision can depend on T.
(c) Length of material can be a function of T.
(d) Mass of carriers is a function of T.

Answer: A, B

Question. Kirchhoff ’s junction rule is a reflection of
(a) conservation of current density vector.
(b) conservation of charge.
(c) the fact that the momentum with which a charged particle approaches a junction is unchanged (as a vector) as the charged particle leaves the junction.
(d) the fact that there is no accumulation of charged at a junction.

Answer: B, D

Question. Two filaments of same length are connected first in series then in parallel. For the same amount of main current flowing, the ratio of the heat produced is:   
(a) 1 : 2
(b) 4 : 1
(c) 1 : 4
(d) 2 : 1
Answer  D

Question. Given a current carrying wire of non-uniform cross-section. Which one of the following is constant throughout the length of wire ?   
(a) current only
(b) current and drift speed
(c) drift speed only
(d) current, electric field and drift speed
Answer  B

Question. Same length of two identical wires are first connected is series and then in parallel, then the amount of heat produced in both the conditions are in the ratio :  
(a) 1 : 4
(b) 4 : 1
(c) 3 : 1
(d) 1 : 2
Answer  B

Question. An electric bulb is rated 60 W, 220 V. The resistance of its filament is   
(a) 870 W
(b) 780 W
(c) 708 W
(d) 807 W
Answer  D

Question. Two bulbs are of (40 W, 200 V), and (100 W, 200 V).   
Then correct relation for their resistances is
(a) R40 < R100
(b) R40 > R100
(c) R40 = R100
(d) no relation can be predicted. 
Answer   B

Question. A 5°C rise in temperature is observed in a conductor by passing a current. When the current is doubled the rise in temperature will be approximately   
(a) 20°C
(b) 16°C
(c) 10°C
(d) 12°C 
Answer  A 

Question. The charge flowing through a resistance R varies with time t as Q = at – bt2, where a and b are positive constants. The total heat produced in R is   
(a) a Rb32
(b) a R b 3
(c) a R b 3 6
(d) a R b 3 3
Answer  C

Question. Two cities are 150 km apart. Electric power is sent from one city to another city through copper wires. The fall of potential per km is 8 volt and the average resistance per km is 0.5 W. The power loss in the wire is   
(a) 19.2 W
(b) 19.2 kW
(c) 19.2 J
(d) 12.2 kW
Answer  B

Question 201. Define current density. Write its S.I. unit. Is it a scalar or vector quantity ? 
Answer: Current density (\( J \)) at a point in a conductor is defined as the amount of electric current flowing perpendicularly per unit cross-sectional area around that point. \[ J = \frac{I}{A} \] * **SI Unit:** Ampere per square meter (\( \text{A/m}^2 \)). * **Type of quantity:** It is a **vector** quantity, pointing in the direction of the flow of positive charges.
In simple words: Current density measures how crowded the flow of electric current is through a wire's cross-section. It is a vector quantity measured in Amperes per square meter.

Exam Tip: State both the formula and the vector nature clearly, as these are highly valued in the grading key.

Question 202. (a) Define resistance of a conductor. Write its S.I. unit.
(b) What are the factors on which the resistance of a conductor depends ? 
Answer: The parameters are defined below:
(a) **Resistance:** The electrical resistance (\( R \)) of a conductor is defined as the ratio of the potential difference \( V \) applied across its ends to the current \( I \) flowing through it: \[ R = \frac{V}{I} \] * **SI Unit:** **Ohm** (\( \Omega \)).
(b) **Factors affecting resistance:** The resistance of a conductor depends on:
(i) **Length of the conductor (\( L \)):** Directly proportional (\( R \propto L \)).
(ii) **Area of cross-section (\( A \Delta \)):** Inversely proportional (\( R \propto \frac{1}{A} \)).
(iii) **Nature of the material:** Dependent on free electron density.
(iv) **Temperature:** Resistance typically increases with temperature for metallic conductors.
In simple words: (a) Resistance is how much a material opposes the flow of electric current, measured in Ohms. (b) It increases if the wire is longer, decreases if the wire is thicker, and depends on the wire's material and temperature.

Exam Tip: List all four factors explicitly with their proportional relations (\( R \propto L \) and \( R \propto 1/A \)) to ensure full marks.

Question 203. (a) Define resistivity of a conductor. Write its S.I. unit.
(b) On what factors does the resistivity of a conductor depend ? 
Answer: The details are:
(a) **Resistivity:** The electrical resistivity (\( \rho \)) of a material is defined as the resistance of a conductor of that material having a unit length and unit cross-sectional area: \[ \rho = \frac{R A}{L} \] * **SI Unit:** **Ohm-meter** (\( \Omega\cdot\text{m} \)).
(b) **Factors affecting resistivity:** Unlike resistance, resistivity is an intrinsic property of the material and does not depend on the dimensions of the conductor. It depends on:
(i) **Nature of the material** (specifically, the number density of free electrons, \( n \)).
(ii) **Temperature of the conductor** (which alters the average relaxation time, \( \tau \)).
In simple words: (a) Resistivity is the specific resistance of a standard-sized block of a material, measured in Ohm-meters. (b) It is a constant for a given material and only changes if you change the material itself or its temperature.

Exam Tip: Clearly state that resistivity is independent of the length and area of the conductor to avoid a common conceptual trap.

Question 204. Draw a graph showing the variation of resistance of a metal wire as a function of its diameter keeping its length and material constant. 
Answer: The resistance \( R \) of a wire is given by: \[ R = \rho \frac{L}{A} = \rho \frac{L}{\pi r^2} = \rho \frac{4L}{\pi D^2} \] Since the length \( L \) and material resistivity \( \rho \) are constant: \[ R \propto \frac{1}{D^2} \] where \( D \) is the diameter of the wire. The graph showing this inverse square relationship is a hyperbola:

In simple words: Since resistance is inversely proportional to the square of the wire's diameter, a thicker wire has much less resistance. This relation plots as a smooth downward curve.

Exam Tip: Write the proportionality \( R \propto \frac{1}{D^2} \) on the graph page to justify the hyperbolic shape of the curve.

Question 205. Two wires, one of copper and the other of manganin, have same resistance and equal thickness. Which wire is longer? Justify your answer. 
Answer: The **copper wire** will be longer.
**Justification:**
The resistance is defined as: \[ R = \rho \frac{L}{A} \implies L = \frac{R A}{\rho} \] Since both wires have the same resistance \( R \) and thickness (which means equal cross-sectional area \( A \)): \[ L \propto \frac{1}{\rho} \] The resistivity of copper (\( \rho_c \)) is significantly lower than that of manganin (\( \rho_m \)): \[ \rho_c < \rho_m \] Because of this inverse relationship, the copper wire must be longer to have the same resistance as the manganin wire: \[ L_c > L_m \]
In simple words: Copper conducts electricity much better than manganin (it has lower resistivity). Therefore, for a copper wire to have the exact same resistance as a manganin wire of the same thickness, it has to be much longer.

Exam Tip: Write down the proportional relationship \( L \propto \frac{1}{\rho} \) to support your qualitative conclusion.

Question 206. Two wires of equal length, one of copper and the other of manganin have the same resistance. Which wire is thicker ? 
Answer: The **manganin wire** will be thicker.
**Justification:**
The resistance equation is: \[ R = \rho \frac{L}{A} \implies A = \rho \frac{L}{R} \] Since both wires have the same length \( L \) and resistance \( R \): \[ A \propto \rho \] The resistivity of manganin (\( \rho_m \)) is much higher than that of copper (\( \rho_c \)): \[ \rho_m > \rho_c \] Thus, the cross-sectional area of the manganin wire must be larger: \[ A_m > A_c \] This means the manganin wire is thicker.
In simple words: Manganin opposes current much more than copper does. To keep the total resistance equal for the same length of wire, the manganin wire needs a much larger, thicker pathway for the current to flow.

Exam Tip: Relate the thickness to the area relation \( A \propto \rho \) to provide a mathematically sound justification.

Question 207. Nichrome and copper wires of same length and same radius are connected in series. Current I is passed through them. Which wire gets heated up more ? Justify your answer.
Answer: The **nichrome wire** will heat up more.
**Justification:**
When two wires are connected in series, the same current \( I \) flows through both. The heat generated \( H \) in a conductor is given by Joule's law: \[ H = I^2 R t \] Since \( I \) and \( t \) are identical for both: \[ H \propto R \] Resistance \( R = \rho \frac{L}{A} \). Since the length \( L \) and radius (and hence area \( A \)) are the same: \[ R \propto \rho \] Since the resistivity of nichrome (\( \rho_{\text{nichrome}} \)) is much higher than that of copper (\( \rho_{\text{copper}} \)): \[ R_{\text{nichrome}} > R_{\text{copper}} \implies H_{\text{nichrome}} > H_{\text{copper}} \]
In simple words: In a series connection, both wires carry the same current. Since nichrome has a much higher electrical resistance than copper, it opposes the current more, converting much more electrical energy into heat.

Exam Tip: Clearly state that "in series, current remains constant" as the starting point of your logical deduction.

Question 208. Define the term conductivity of a conductor. Write its S.I. unit. On what factors does it depend ? 
Answer: The parameters are defined as:
The electrical conductivity (\( \sigma \)) of a conductor is defined as the reciprocal of its electrical resistivity (\( \rho \)): \[ \sigma = \frac{1}{\rho} \] Alternatively, it is the ratio of current density \( J \) to the applied electric field \( E \): \[ \sigma = \frac{J}{E} \] * **SI Unit:** **Siemens per meter** (\( \text{S/m} \)) or Ohm-inverse meter-inverse (\( \Omega^{-1}\cdot\text{m}^{-1} \)).
**Factors affecting conductivity:** Conductivity depends on:
1. **Nature of the material** (the number density of free electrons, \( n \)).
2. **Temperature of the conductor** (specifically affecting the average relaxation time, \( \tau \)).
In simple words: Conductivity is a measure of how easily a material allows electric current to flow through it. It is the exact opposite of resistivity and depends on what the material is made of and its temperature.

Exam Tip: Give both definition paths (reciprocal of resistivity and \( J/E \) ratio) to show comprehensive knowledge.

Question 209. Resistance of a conductor increases with the rise in temperature. Why ?
Answer: When the temperature of a metallic conductor rises, the thermal kinetic energy of the metal ions/atoms increases, causing them to vibrate more violently. This significantly increases the frequency of collisions of the drifting free electrons with these vibrating ions. As a result, the average relaxation time (\( \tau \)) decreases, which increases the resistance: \[ R = \frac{m L}{n e^2 \tau A} \implies R \propto \frac{1}{\tau} \]
In simple words: Heating up a metal makes its atoms vibrate faster. This makes it much harder for flowing electrons to navigate through, as they crash into these vibrating atoms more frequently, raising the resistance.

Exam Tip: Use the key term "relaxation time (\( \tau \)) decreases" to explain the rise in resistance scientifically.

Question 210 . If a wire is stretched to double its original length without loss of mass, what will be its new- (a) resistivity (b) resistance ? 
Answer: The changes are described below:
(a) **Resistivity:** The resistivity remains **completely unchanged** because it is an intrinsic material property and does not depend on the physical dimensions (length or area) of the wire.
(b) **Resistance:** When a wire of initial resistance \( R = \rho \frac{L}{A} \) is stretched to double its length (\( L' = 2L \)) without any loss of mass, its volume (\( V = A \cdot L \)) must remain constant. This means the area is halved (\( A' = \frac{A}{2} \)). The new resistance \( R' \) is: \[ R' = \rho \frac{L'}{A'} = \rho \frac{2L}{A/2} = 4 \left( \rho \frac{L}{A} \right) = 4R \] Thus, the resistance becomes **4 times** the original value.
In simple words: (a) Resistivity is a constant for the metal, so stretching it doesn't change it. (b) Stretching a wire to double its length also makes it twice as thin. This combined effect makes the new resistance exactly four times larger.

Exam Tip: Memorize the shortcut for stretching: \( R' = n^2 R \), where \( n \) is the stretching factor. Here, \( n = 2 \), so \( R' = 2^2 R = 4R \).

Question 211. Two materials, Si and Cu, are cooled from 300K to 60K. What will be the effect on their resistivity ?
Answer: The cooling effect on the resistivity of the two materials is:
1. **For Silicon (Si, a semiconductor):** The resistivity will **increase**.
* **Reason:** Semiconductors have a negative temperature coefficient of resistivity. Lowering the temperature reduces the thermal energy available to break covalent bonds, which drastically decreases the free charge carrier density (\( n \)).
2. **For Copper (Cu, a metallic conductor):** The resistivity will **decrease**.
* **Reason:** Metallic conductors have a positive temperature coefficient of resistivity. Lowering the temperature reduces the thermal vibrations of the lattice ions, which increases the relaxation time (\( \tau \)).
In simple words: Cooling Silicon increases its resistance because fewer free electrons can break loose to carry current. Cooling Copper decreases its resistance because its atoms vibrate less, making it easier for electrons to flow through smoothly.

Exam Tip: Contrast the two material types explicitly: Si is a semiconductor, whereas Cu is a metallic conductor.

Question 212. Explain, why allows like constantan and manganin are used for making standard resistors ? 
Answer: These alloys are preferred for manufacturing standard resistors because:
1. They possess a high electrical resistivity.
2. They have an exceptionally small, negligible temperature coefficient of resistivity, meaning their resistance remains almost constant even if their temperature rises during operation.
In simple words: We use these alloys because they have high resistance and, more importantly, their resistance doesn't change when they heat up while carrying current.

Exam Tip: Mention "very small temperature coefficient of resistance" as it is the most critical characteristic for standard resistors.

Question 213. The I-V graph for a metallic wire at two different temperatures T1 and T2 is as shown in the figure. Which of the two temperatures is higher and why ? 
Answer: The temperature **\( T_1 \)** is higher.
**Reason:**
From Ohm's law, the reciprocal of the slope of the \( I\text{-}V \) graph represents the electrical resistance: \[ R = \frac{V}{I} = \frac{1}{\text{Slope}} \] From the plotted graph, for any fixed current \( I \), the corresponding voltage for \( T_1 \) is greater than that for \( T_2 \) (\( V_1 > V_2 \)). This indicates: \[ R_1 > R_2 \] Since the resistance of a metallic conductor is directly proportional to its temperature (\( R \propto T \)), the higher resistance \( R_1 \) implies a higher temperature: \[ T_1 > T_2 \]

In simple words: The steeper the voltage climb on the graph for a given current, the higher the resistance. Since the line for \( T_1 \) has a higher resistance, it must represent a hotter, higher temperature.

Exam Tip: Explicitly state that the slope of an \( I\text{-}V \) graph represents \( 1/R \) (or conductance), not \( R \).

Question 214. The I-V graph for two identical conductors of different materials A and B is shown in figure. Which one of the two has higher resistivity and why ? 
Answer: Conductor **B** has the higher resistivity.
**Reason:**
The resistance is inversely proportional to the slope of the \( I\text{-}V \) graph: \[ R = \frac{1}{\text{Slope}} \] The line for B has a lower slope than A, indicating B has a higher resistance (\( R_B > R_A \)). Since both conductors are identical in length \( L \) and cross-sectional area \( A \): \[ R \propto \rho \] Because B has a larger resistance, it must possess a higher resistivity: \[ \rho_B > \rho_A \]
In simple words: Conductor B has a flatter slope, meaning it conducts less current for the same voltage and has higher resistance. Since both wires are identical in size, B's higher resistance means it has a higher resistivity.

Exam Tip: Connect the graphical slope directly to resistance first, and then link resistance to resistivity using the identical dimensions.

Question 215. Two metallic resistors are connected first in series and then in parallel across a d.c. supply. Plot of I-V graph is shown for the two cases. Which one represents a parallel combination of the resistors and why ?
Answer: The line **A** represents the parallel combination of the resistors.
**Reason:**
The equivalent resistance in a parallel combination (\( R_P \)) is always smaller than the individual resistances, whereas the equivalent resistance in a series combination (\( R_S \)) is larger: \[ R_P < R_S \] From the \( I\text{-}V \) graph, the slope of line A is steeper than that of line B. Since resistance is inversely proportional to the slope of the \( I\text{-}V \) curve: \[ R = \frac{1}{\text{Slope}} \] The steeper slope of A indicates a lower resistance value: \[ R_A < R_B \] Therefore, line A represents the parallel combination, and line B represents the series combination.
In simple words: A parallel connection has a lower overall resistance than a series connection. On an I-V graph, a steeper slope represents lower resistance, so line A must be the parallel connection.

Exam Tip: Explicitly state the relation \( R_P < R_S \) to build a logically complete argument.

Question 216. Figure shows a plot of current flowing through the cross section of a wire versus the time . Use the plot to find the charge flowing in 10 s through the wire.
Answer: The total electric charge \( q \) flowing through the cross-section is equal to the area under the current-time (\( I\text{-}t \)) curve from \( t = 0 \) to \( t = 10\text{ s} \).
The area can be calculated by splitting the shape into a triangle (from 0 to 5 s) and a rectangle (from 5 to 10 s): \[ q = \text{Area of Triangle} + \text{Area of Rectangle} \] \[ q = \left[ \frac{1}{2} \times \text{base} \times \text{height} \right] + [ \text{length} \times \text{breadth} ] \] \[ q = \left[ \frac{1}{2} \times 5 \times 5 \right] + [ (10 - 5) \times 5 ] \] \[ q = 12.5 + 25 = 37.5\text{ C} \] Thus, the total charge flowing through the wire is \( 37.5\text{ Coulombs} \).

In simple words: The total charge is just the area under the current-time graph. Splitting the graph's shape into a triangle and a rectangle and adding their areas gives a total charge of 37.5 Coulombs.

Exam Tip: Always specify the units of your final answer clearly as Coulombs (\( \text{C} \)) to secure full marks.

Question 217. Show that the current density is related to the applied electric field by the relation \vec{J} = \sigma \vec{E} Where \sigma defines the conductivity of the material. 
Answer: The relationship between current \( I \) and drift velocity \( v_d \) is: \[ I = n e A v_d \] Dividing both sides by the cross-sectional area \( A \) to obtain current density \( J \): \[ J = \frac{I}{A} = n e v_d \] The expression for drift velocity \( v_d \) in terms of electric field \( E \) and relaxation time \( \tau \) is: \[ v_d = \frac{e E \tau}{m} \] Substituting \( v_d \) into the current density equation: \[ J = n e \left( \frac{e E \tau}{m} \right) = \left( \frac{n e^2 \tau}{m} \right) E \] Since the electrical conductivity of the material is defined as \( \sigma = \frac{n e^2 \tau}{m} \), we substitute it in: \[ J = \sigma E \] In vector form, this is expressed as: \[ \vec{J} = \sigma \vec{E} \]
In simple words: By combining the current density formula with the drift velocity equation, we show that current density is directly proportional to the electric field. The constant linking them is the material's electrical conductivity.

Exam Tip: Clearly define the conductivity constant \( \sigma = \frac{ne^2\tau}{m} \) during your derivation.

Question 218. Define the term (a) Emf of a cell (b) Terminal voltage of a cell. 
Answer: (a) **Electromotive Force (emf):** The emf (\( E \)) of a cell represents the total work done or energy supplied by the source in driving a unit positive charge once around the entire closed electrical circuit (which includes the internal path through the cell itself): \[ E = \frac{W}{q} \]
(b) **Terminal Voltage:** The terminal voltage (\( V \)) of a cell is the potential difference measured between its two electrodes (terminals) when the cell is active and delivering current to an external closed circuit: \[ V = E - I r \] where \( I \) is the current and \( r \) is the internal resistance.
In simple words: (a) Emf is the maximum chemical push or voltage a cell can provide when it is not delivering any current. (b) Terminal voltage is the actual usable voltage available at the cell's terminals when it is connected to a working circuit and current is flowing.

Exam Tip: Draw a clear distinction: emf is an open-circuit characteristic of a cell, whereas terminal voltage is measured in a closed circuit.

Question 219. Define internal resistance of a cell. Write any two factors on which it depends. 
Answer: The internal resistance (\( r \)) of a cell is the inherent opposition offered by the electrolyte and electrodes inside the cell to the passage of electric current flowing through it.

**Factors affecting internal resistance:** The internal resistance of a cell depends on:
(i) **Distance between electrodes:** Directly proportional to the separation between the plates.
(ii) **Concentration of electrolyte:** Increases with an increase in the concentration of the electrolyte.
(iii) **Area of electrodes:** Inversely proportional to the area of the electrodes submerged in the electrolyte.
(iv) **Temperature:** Decreases as the temperature of the electrolyte increases.
In simple words: Internal resistance is the internal friction or obstacle that the chemical liquid inside a battery presents to the flowing current. It gets higher if the liquid is concentrated or if the battery gets cold.

Exam Tip: State any two factors clearly, such as "separation between electrodes" and "concentration of electrolyte," to easily get full marks.

Question 220. The emf of a cell is always greater than its terminal voltage. Give reason
Answer: When a cell is connected in a closed circuit and current \( I \) begins to flow, a small potential drop equal to \( I r \) occurs internally across its own internal resistance \( r \). This internal drop subtracts from the electromotive force \( E \), leaving the terminal voltage \( V \) always smaller than the emf: \[ V = E - I r \implies V < E \]
In simple words: When a battery starts powering a circuit, some of its energy is wasted pushing current through its own internal chemicals. This internal loss reduces the available voltage at its terminals.

Exam Tip: Use the terminal voltage equation \( V = E - I r \) to mathematically back up your explanation.

Question 221. Can the value of terminal potential difference be greater than the emf of a cell ? 
Answer: Yes, the terminal potential difference can exceed the cell's emf. This occurs specifically during the charging process of the cell. Under these conditions, the external charger drives current into the positive terminal of the cell, reversing the internal drop and yielding: \[ V = E + I r \implies V > E \]
In simple words: Yes, this happens when you are actively charging a battery. Because you are forcing current backward into the battery, the charger has to push with a voltage higher than the battery's natural emf.

Exam Tip: Show the charging equation \( V = E + I r \) to demonstrate why the voltage becomes larger than the emf during charging.

Question 222. The figure shows a plot of terminal voltage ‘V’ versus the current ‘i’ of a given cell. Calculate from the graph (a) emf of the cell and (b) internal resistance of the cell. 
Answer: The relationship between terminal voltage \( V \), current \( I \), and internal resistance \( r \) is described by: \[ V = E - I r \]
**(a) Calculating EMF (\( E \)):** The electromotive force \( E \) is equal to the terminal voltage when the circuit is open (current \( I = 0 \)). From the given graph, the y-intercept (where \( I = 0 \)) corresponds to a voltage of \( 6\text{ V} \). \[ E = 6\text{ V} \]
**(b) Calculating Internal Resistance (\( r \)):** From the graph, when the current \( I = 1.0\text{ A} \), the terminal voltage \( V \) drops to \( 4\text{ V} \). Substituting these coordinates into the voltage relation: \[ 4 = 6 - (1.0 \times r) \] \[ r = 6 - 4 = 2\text{ }\Omega \] Thus, the internal resistance is \( 2\text{ }\Omega \).

In simple words: (a) The graph's starting point on the vertical axis shows the maximum voltage of 6 Volts when no current flows, which is the emf. (b) As current increases to 1 Ampere, the voltage drops to 4 Volts. The rate of this drop shows the internal resistance is 2 ohms.

Exam Tip: Be sure to state the coordinates \( (0\text{ A}, 6\text{ V}) \) and \( (1\text{ A}, 4\text{ V}) \) that you extracted from the graph to make your steps clear.

Question 223. Find the resistance of the following carbon resistors.
Answer: Using the standard color code sequence for carbon resistors (B B R O Y G B V G W):
**(i) Yellow, Violet, Brown, Gold:**
* Yellow = 4
* Violet = 7
* Brown = \( 10^1 \) (multiplier)
* Gold = \( \pm 5\% \) (tolerance)
The resistance value is: \[ R = 47 \times 10^1\text{ }\Omega \pm 5\% = 470\text{ }\Omega \pm 5\% \]
**(ii) Red, Red, Red, Silver:**
* Red = 2
* Red = 2
* Red = \( 10^2 \) (multiplier)
* Silver = \( \pm 10\% \) (tolerance)
The resistance value is: \[ R = 22 \times 10^2\text{ }\Omega \pm 10\% = 2200\text{ }\Omega \pm 10\% \]
In simple words: (i) Yellow (4), Violet (7), and Brown (multiply by 10) gives a base resistance of 470 ohms. The gold band means the margin of error is within 5%. (ii) Red (2), Red (2), and Red (multiply by 100) gives 2200 ohms with a silver margin of error of 10%.

Exam Tip: Keep the color code mnemonic sentence ("B B ROY of Great Britain had a Very Good Wife") in mind to quickly recall values during exam stress.

Question 224. State Ohm’s law. 
Answer: Ohm's law states that the electric current \( I \) flowing through a conductor is directly proportional to the potential difference \( V \) applied across its ends, provided that physical conditions like temperature, mechanical strain, and pressure remain completely constant: \[ V \propto I \implies V = I R \] where \( R \) is the constant electrical resistance of the given conductor.
In simple words: Ohm's law says that if you don't change a wire's temperature or stretch it, the current running through it grows in exact proportion to the voltage you apply.

Exam Tip: Be sure to emphasize the condition "provided physical conditions remain constant," as omitting this phrase will lose you valuable marks.

Question 225. Graph showing the variation of current versus voltage for a material GaAs as shown in figure. Identify the region of (i) negative resistance  (ii) where Ohm’s law is obeyed.
Answer: Based on the provided current-voltage graph for Gallium Arsenide (GaAs):
(i) **Negative Resistance Region:** This corresponds to the **region DE**. In this section, as the applied potential difference increases, the resulting current decreases, giving a negative slope (\( R = \frac{\Delta V}{\Delta I} < 0 \)).
(ii) **Ohm's Law Obeyed Region:** This corresponds to the **region AB**. Here, the current rises linearly and proportionally with the increasing potential difference.

In simple words: (i) In region DE, raising the voltage actually slows down the current, which acts as a "negative" resistance. (ii) In region AB, current grows in a straight line with voltage, obeying Ohm's law perfectly.

Exam Tip: Label the regions directly on your sketch to show examiners where the linear (Ohmic) and negative-resistance zones lie.

Question 226. Two identical cells each of emf E,having negligible internal resistance, are connected in parallel with each other across an external resistance R. What is the current through the resistance ? 
Answer: When two identical cells with electromotive force \( E \) and zero internal resistance (\( r \approx 0 \)) are connected in parallel, the net equivalent emf of the parallel combination remains equal to the emf of a single cell: \[ E_{\text{eq}} = E \] Since the internal resistance is negligible, the total resistance in the circuit is simply the external load resistor \( R \). Applying Ohm's law, the current \( I \) flowing through the resistor is: \[ I = \frac{E_{\text{eq}}}{R + r_{\text{eq}}} = \frac{E}{R} \]
In simple words: Connecting batteries of equal voltage in parallel doesn't double the voltage; the net push remains equal to one battery. Since there is no internal loss, the current is simply this voltage divided by the external resistance.

Exam Tip: State clearly that "the equivalent emf of identical cells in parallel is equal to the emf of a single cell" to justify your starting step.

Question 227. A 10 V battery of negligible internal resistance is connected across a 200 V battery and a resistance of 38 \Omega as shown. Find the value of the current in the circuit. 
Answer: The two batteries are connected in opposition (positive terminal connected to positive terminal). Therefore, the net equivalent electromotive force \( V_{\text{net}} \) in the circuit is the difference between their individual values: \[ V_{\text{net}} = 200\text{ V} - 10\text{ V} = 190\text{ V} \] Since the internal resistance is negligible, the total resistance is \( R = 38\text{ }\Omega \). Applying Ohm's law, the current \( I \) in the circuit is: \[ I = \frac{V_{\text{net}}}{R} = \frac{190}{38} = 5\text{ A} \]

In simple words: The 10V battery is connected backward against the 200V battery, fighting its push. The net pushing voltage is 190 Volts. Dividing this by the 38-ohm resistor gives a current of 5 Amperes.

Exam Tip: Pay close attention to how the terminals are wired to verify if the voltages add up (series-support) or subtract (series-opposition).

Question 228. Define the term drift velocity of charge carriers in a conductor and write its relation with the current flowing through it. 
Answer: The average velocity with which free electrons drift inside a conductor under the influence of an applied external electric field is called drift velocity (\( v_d \)). The direction of this drift is opposite to the direction of the applied electric field.
The relationship between drift velocity and the electric current \( I \) flowing through the conductor is: \[ I = n e A v_d \] where:
\( n \) is the number density of free electrons,
\( e \) is the charge of an electron,
\( A \) is the cross-sectional area of the conductor, and
\( v_d \) is the drift velocity.
In simple words: Drift velocity is the slow, net speed at which electrons are nudged along a wire when you turn on a voltage. The total current is directly proportional to this drift speed.

Exam Tip: Define each variable in the current-drift velocity relation \( I = n e A v_d \) to score full marks.

Question 229. How does the random motion of free electrons in a conductor gets affected when a potential difference is applied across its ends. 
Answer: When an external potential difference is established across the ends of a conductor, an electric field is created inside. This field exerts an electrostatic force on the free electrons, causing their highly random thermal motions to be partially directed towards the positive terminal of the conductor, resulting in a net directional drift.
In simple words: Before applying a voltage, electrons bounce around randomly in all directions. Applying a voltage adds a steady, gentle push that makes them all slowly drift together towards the positive end, while still bouncing randomly.

Exam Tip: Use the phrase "partially directed towards the positive end" to describe the combined thermal and drift motion.

Question 230. When electrons drift in a metal from lower to higher potential, does it mean that all the ‘free’ electrons of the metal are moving in the same direction?
Answer: No, this does not imply that all free electrons are moving in the exact same direction. At any instant, the electrons still possess large random thermal velocities in all directions. The applied electric field merely superimposes a tiny, uniform drift velocity on top of this highly chaotic random motion, biasing the overall flow towards the higher potential terminal.
In simple words: No. It is like a swarm of bees blowing slowly in the wind; each bee is flying frantically in its own random direction, but the entire swarm has a slow, net movement in the direction of the wind.

Exam Tip: Emphasize that "drift velocity is superposed on top of large random thermal velocities."

Question 231. The electron drift speed is estimated to be only a few mm s–1 for currents in the range of a few amperes ? How then is current established almost the instant a circuit is closed ? 
Answer: When the switch is closed, an electromagnetic wave travels along the wires at nearly the speed of light. This establishes an electric field throughout the entire circuit almost instantaneously, causing electrons at every single point in the conductor to begin drifting immediately. Consequently, current starts flowing instantly, even though individual electrons travel very slowly.
In simple words: It is like a pipe completely packed with tennis balls. When you push one ball into the pipe, another ball pops out of the far end almost instantly, even though no single ball traveled all the way through the pipe yet.

Exam Tip: Explain that "the electric field is set up instantly with the speed of electromagnetic waves" to provide a precise scientific answer.

Question 232. If the electron drift speed is so small, and the electron’s charge is small, how can we still obtain large amounts of current in a conductor ? 
Answer: Although the drift velocity \( v_d \) and the charge \( e \) of an individual electron are extremely small, metals possess an incredibly high number density \( n \) of free electrons (on the order of \( 10^{29}\text{ m}^{-3} \)). Since the current is proportional to this density (\( I = n e A v_d \)), this massive pool of charge carriers produces a very large overall current.
In simple words: Even though each electron moves slowly and carries very little charge, there are billions of trillions of them inside the wire. When they all move together, their combined flow adds up to a large electric current.

Exam Tip: Quote the numerical magnitude of free electron density in metals (\( \approx 10^{29}\text{ m}^{-3} \)) to show detailed knowledge.

Question 233. The electron drift arises due to the force experienced by electrons in the electric field inside the conductor. But force should cause acceleration. Why then do the electrons acquire a steady average drift speed ? 
Answer: Individual free electrons do accelerate between collisions under the influence of the electric field. However, they frequently collide with the heavy lattice ions and atoms of the conductor, which randomizes their velocities. This continuous cycle of acceleration followed by collisional slowing limits their speed, giving them a constant, steady average speed called drift speed.
In simple words: It is like walking through a very crowded market. You try to run (accelerate) in the empty spaces, but you keep bumping into people (colliding), which keeps your overall walking speed down to a steady, slow average.

Exam Tip: Highlight the "frequent collisions with lattice ions" as the mechanism that converts acceleration into a steady average velocity.

Question 234. How does the drift velocity of electrons in a metallic conductor vary with increase in temperature ? 
Answer: The drift velocity of free electrons inside a metallic conductor **decreases** when the temperature is increased.
**Reason:** The formula for drift velocity is: \[ v_d = \frac{e E \tau}{m} \] As temperature rises, the thermal vibrations of the metal atoms increase, causing more frequent collisions with the drifting electrons. This decreases the average relaxation time (\( \tau \)), which directly reduces the drift velocity (\( v_d \propto \tau \)).
In simple words: Heating the wire makes the metal atoms jiggle violently. This causes electrons to collide more frequently, reducing their average relaxation time and slowing down their drift speed.

Exam Tip: Write the mathematical relation \( v_d \propto \tau \) to support your qualitative explanation.

Question 235. If a potential difference V applied across a conductor is increased to 2V, how will the drift velocity of electrons change ? 
Answer: The drift velocity of the free electrons will be **doubled**.
**Reason:** The drift velocity \( v_d \) is related to the applied potential difference \( V \) by: \[ v_d = \frac{e E \tau}{m} = \frac{e V \tau}{m L} \implies v_d \propto V \] Since drift velocity is directly proportional to the applied potential difference, doubling the voltage to \( 2V \) will double the drift velocity of the electrons.
In simple words: The electrical force pushing the electrons is directly determined by the voltage. Doubling the voltage doubles the push, which doubles the speed at which the electrons drift.

Exam Tip: Clearly state the linear proportionality \( v_d \propto V \) to make your answer mathematically solid.

Question 236. Define the term ‘relaxation time’ in a conductor.
Answer: Relaxation time (\( \tau \)) in a conductor is defined as the average time interval that elapses between two consecutive collisions of a drifting free electron with the lattice ions or atoms of the conductor.
In simple words: Relaxation time is the average "free time" an electron enjoys between bumping into one metal atom and the next.

Exam Tip: Use the phrase "successive collisions" to define this term precisely.

Question 237. If the temperature of a good conductor increases, how does the relaxation time of electrons in the conductor change ? 
Answer: The relaxation time of the free electrons will **decrease**.
**Reason:** Raising the temperature causes the metal lattice ions to vibrate with greater amplitude. This increases the probability and frequency of electron-ion collisions, which shortens the average free time (relaxation time) between consecutive collisions.
In simple words: When the conductor gets hotter, the atoms vibrate more wildly. Electrons crash into them much more quickly, reducing the average time they spend flying freely between collisions.

Exam Tip: State clearly that "temperature rise increases collision frequency, thereby reducing relaxation time."

Question 238. (i) How is the relaxation time related to the drift velocity of free electrons ?
(ii) Obtain an expression for the current density in terms of relaxation time. 
Answer: The relationships are developed below:
(i) **Relationship:** The drift velocity \( v_d \) of free electrons is directly proportional to the relaxation time \( \tau \): \[ v_d = \frac{e E}{m} \tau \] where \( e \) is the electronic charge, \( m \) is the electron mass, and \( E \) is the applied electric field.
(ii) **Expression for Current Density:** The current density \( j \) is given by the relation: \[ j = n e v_d \] Substituting the value of drift velocity \( v_d = \frac{e E \tau}{m} \): \[ j = n e \left( \frac{e E \tau}{m} \right) = \left( \frac{n e^2 \tau}{m} \right) E \]
In simple words: (i) Drift speed is the electron's charge-to-mass ratio multiplied by the electric field and the relaxation time. (ii) Current density is this drift speed times the charge density, which simplifies to the product of electron density, charge squared, relaxation time, and electric field, all divided by mass.

Exam Tip: Keep track of your subscripts and constants so that your final equation clearly defines the conductivity term \( \sigma = \frac{ne^2\tau}{m} \).

Question 239. (i) Define mobility of a charge carrier. Write its S.I. unit.
(ii) What is its relation with relaxation time ? 
(iii) How does the electron mobility change if (a) temperature is increased , (b) potential difference in doubled ?
Answer: The parameters are described below:
(i) **Mobility (\( \mu \)):** The mobility of a charge carrier is defined as the magnitude of its drift velocity per unit applied electric field: \[ \mu = \frac{v_d}{E} \] * **SI Unit:** square meter per Volt-second (\( \text{m}^2\cdot\text{V}^{-1}\cdot\text{s}^{-1} \)).
(ii) **Relationship with Relaxation Time:** Substituting \( v_d = \frac{e E \tau}{m} \) into the mobility equation: \[ \mu = \frac{\left( \frac{e E \tau}{m} \right)}{E} = \frac{e \tau}{m} \]
(iii) **Effects of changes:**
(a) **If temperature is increased:** The mobility will **decrease** because a temperature rise decreases the relaxation time \( \tau \) due to more frequent collisions (\( \mu \propto \tau \)).
(b) **If potential difference is doubled:** The mobility remains **unchanged**. Since mobility \( \mu = \frac{e\tau}{m} \) depends only on the material properties and temperature, it is independent of the applied potential difference.
In simple words: (i) Mobility is how easily a charge carrier drifts through a material when pushed by an electric field. (ii) It is directly proportional to the relaxation time. (iii) (a) Heating the wire decreases mobility because the jiggling atoms slow down the electrons. (b) Changing the voltage has no effect on this native ease of movement.

Exam Tip: Be careful to state the units correctly as \( \text{m}^2\cdot\text{V}^{-1}\cdot\text{s}^{-1} \) and not confuse them with conductivity units.

Question 240. What happens if the galvanometer and cell are interchanged at the balanced point of the Wheatstone bridge? Would the galvanometer show any current ?
Answer: If the cell and the galvanometer are interchanged at the null (balanced) point, the balance condition of the Wheatstone bridge remains completely unaffected. Therefore, the galvanometer will still show zero deflection and no current will flow through it.
In simple words: Swapping the battery and the galvanometer when the bridge is balanced doesn't change anything; the bridge stays perfectly balanced and the meter still reads zero.

Exam Tip: State clearly that "the balance condition remains satisfied" as this is the core technical justification.

Question 241. What is a meter bridge ? Write the principle of working meter bridge. 
Answer: **Meter Bridge:** It is an electrical apparatus designed as the simplest practical setup of a Wheatstone bridge, primarily utilized to measure unknown electrical resistances.
**Working Principle:** It operates on the principle of a balanced Wheatstone bridge. When the bridge is balanced, no current flows through the central galvanometer branch, satisfying the ratio: \[ \frac{P}{Q} = \frac{R}{S} \]

In simple words: A meter bridge is a practical lab tool used to find an unknown resistance. It works on the principle of a Wheatstone bridge, balancing four resistances so that no current flows through the center meter.

Exam Tip: Draw a small Wheatstone bridge diamond diagram alongside the meter bridge description to show how the physical wire lengths correspond to the resistors \( P \) and \( Q \).

Question 242. Why are the connections between the resistors in a meter bridge made of thick metal (copper) strips ? 
Answer: The connections are made of thick copper strips because their large cross-sectional area \( A \) and low native resistivity \( \rho \) make their electrical resistance (\( R = \rho \frac{L}{A} \)) virtually negligible. This ensures that the resistance of the junction connections does not alter the measured resistances of the coils, allowing for an exceptionally accurate null point.
In simple words: Thick copper plates have almost zero resistance. Using them to connect parts ensures we don't add accidental "extra" resistance at the joints, making our measurements highly accurate.

Exam Tip: Mention both "low resistivity" and "large cross-sectional area" to mathematically explain why the connection resistance is negligible.

Question 243. Why is it generally preferred to obtain the balance point in the middle of the meter bridge wire? 
Answer: The sensitivity of a Wheatstone bridge is highest when all four arm resistances are of comparable magnitudes (\( P \approx Q \approx R \approx S \)). This condition is met when the null point falls near the middle of the meter bridge wire (around \( 50\text{ cm} \)), thereby minimizing potential measurement errors.
In simple words: A bridge circuit is most sensitive and accurate when all its resisting sides are roughly equal. Having the balance point near the middle of the wire (\( 50\text{ cm} \)) keeps the two sides of the wire equal, making the reading highly precise.

Exam Tip: Connect the "middle of the wire" directly to the concept of "maximum bridge sensitivity" to show deep understanding.

Question 244. State the principle of potentiometer. 
Answer: The principle of a potentiometer states that when a steady, constant electric current flows through a wire of uniform cross-sectional area and composition, the potential drop \( V \) across any segment of the wire is directly proportional to the physical length \( l \) of that segment: \[ V \propto l \implies V = K l \] where \( K \) is the constant potential gradient along the wire.
In simple words: If you run a steady current through a perfectly uniform wire, the voltage drops at a constant rate along its length. This means the voltage across any section is directly proportional to how long that section is.

Exam Tip: State both the proportional relation \( V \propto l \) and the required constant conditions (constant current, uniform area) to get full marks.

Question 245. Of which material a metre bridge/potentiometer wire normally made and why ?
Answer: The wire is typically manufactured from alloys such as **constantan, manganin, or nichrome**.
**Reason:** These alloys exhibit a high specific electrical resistivity along with an exceptionally low temperature coefficient of resistance. Consequently, their resistance remains stable and does not fluctuate even if the wire heats up due to the continuous flow of electric current.
In simple words: The wire is made of alloys like manganin or constantan because their resistance stays completely constant even when they heat up from the flowing current.

Exam Tip: Use the phrase "very small temperature coefficient of resistance" as it is the critical keyword.

Question 246. Why should the potentiometer wire be of uniform cross section and composition ? 
Answer: A perfectly uniform cross-sectional area and material composition ensure that the electrical resistance per unit length of the wire remains constant throughout. Only under this condition will the potential drop per unit length (potential gradient \( K \)) remain uniform, which is the foundational requirement for the potentiometer principle (\( V \propto l \)).
In simple words: If the wire has thick and thin spots, the resistance per centimeter will vary. To ensure the voltage drops at a perfectly steady rate along the wire, its thickness and material must be identical from end to end.

Exam Tip: Relate the uniformity of cross-section directly to having a "constant potential gradient (\( K \))" along the wire.

Question 247. Why do we prefer a potentiometer with a longer wire ? 
Answer: We prefer a longer potentiometer wire because the sensitivity of a potentiometer is inversely proportional to its potential gradient \( K \): \[ \text{Sensitivity} \propto \frac{1}{K} \] Since \( K = \frac{V}{L} \), using a longer wire (larger \( L \)) decreases the potential gradient \( K \) for a given total potential difference \( V \). This smaller potential gradient allows us to measure much smaller voltage differences with a much larger, more readable balancing length.
In simple words: A longer wire spreads the voltage over a greater distance, making the voltage drop per centimeter much smaller. This makes the tool much more sensitive and able to measure tiny voltages accurately.

Exam Tip: Write out the relation \( K = V/L \) and show how a larger \( L \) leads to a smaller \( K \), enhancing sensitivity.

Question 248.. What is meant by sensitivity of a potentiometer ? 
Answer: The sensitivity of a potentiometer refers to its ability to measure extremely small electrical potential differences.
Specifically, a highly sensitive potentiometer:
1. Is capable of accurately resolving very minute potential differences.
2. Exhibits a remarkably large and easily readable shift in its balancing length \( l \) for any minor variation in the potential difference under measurement.
In simple words: Sensitivity is how good a potentiometer is at measuring tiny voltages. A highly sensitive one will show a big, noticeable change in the balancing spot even for a tiny change in voltage.

Exam Tip: List the two aspects of sensitivity (resolving tiny potentials, showing large changes in balancing length) to write a complete answer.

Question 249. How can a given potentiometer be made more sensitive ?
Answer: We can enhance the sensitivity of a potentiometer by reducing its potential gradient \( K \). This can be achieved through:
1. Increasing the physical length of the potentiometer wire.
2. Lowering the current flowing in the primary circuit (for example, by inserting an adjustable series rheostat).
3. Placing a high resistance in series with the potentiometer wire in the primary circuit.
In simple words: To make it more sensitive, we need to lower the voltage drop per centimeter. We can do this by using a longer wire, or by adding a resistor in series to throttle the current running through the main wire.

Exam Tip: Mentioning "increasing wire length" and "reducing primary current" are the two most common and highly-graded methods.

Question 250. The emf of the driving cell used in the main circuit of the potentiometer should be more than the potential Difference to be measured. Why ? 
Answer: The emf of the primary driving cell must exceed the potential difference being measured because the maximum potential drop across the potentiometer wire is limited by the driving cell's voltage. If the voltage to be measured is larger than this maximum drop, we will never find a null balance point on the wire.
In simple words: The main battery has to push harder than the battery we are testing. If the voltage we want to measure is higher than the total voltage spread along the wire, we will run out of wire before finding the balance point.

Exam Tip: Clearly state that "otherwise, the null point will lie outside the wire length" to earn full marks.

Question 251. The variation of potential difference V with length l in case of two potentiometer wires P and Q is as shown. (a) Which potentiometer is more sensitive ?
(b) Which of these will you prefer for comparing emfs of two primary cells and why ?
Answer: The configurations are:
(a) **Potentiometer Q** is more sensitive.
**Reason:** The sensitivity is inversely proportional to the potential gradient \( K \) (where \( K = \frac{V}{l} = \text{slope} \)): \[ \text{Sensitivity} \propto \frac{1}{K} \] From the graph, wire Q has a shallower slope than wire P, indicating Q has a smaller potential gradient \( K_Q < K_P \). Because of this smaller gradient, Q is much more sensitive.
(b) We would prefer **Potentiometer Q** because of its higher sensitivity, which allows us to measure the balance point with a much larger and more precise balancing length.

In simple words: (a) Potentiometer Q is more sensitive because its line rises more slowly on the graph, meaning it has a smaller potential gradient. (b) We prefer Q because its higher sensitivity makes our voltage measurements much more accurate and easy to read.

Exam Tip: Use the slope of the \( V\text{-}l \) graph to explicitly define the potential gradient \( K \) to support your answer.

Question 252. When a metallic conductor is subjected to a certain potential V across its ends, discuss briefly how the phenomenon of drift occurs. 
Answer: When we apply a potential difference across a conductor, an internal electric field is established. This electric field exerts a force on the free electrons, causing them to accelerate in the opposite direction. However, as they move, these electrons frequently collide with the positive ions of the conductor's lattice. Each collision resets their velocity, making their path random. Despite these constant interruptions, they manage to slowly drift along the length of the conductor. The average speed they achieve in this direction is called the drift velocity, which remains constant over time.
In simple words: When a voltage is connected to a metal wire, it pushes the free electrons inside. Because they keep bumping into the metal's atoms, they can only slowly drift forward with a steady average speed.

Exam Tip: Define both the cause (electric field force) and the limiting factor (lattice collisions resetting velocity) to secure full marks for drift velocity explanations.

Question 253. Derive an expression for drift velocity of free electrons in a conductor in terms of relaxation time of electrons. 
Answer: Suppose we apply a voltage \( V \) across a metallic conductor. This sets up an electric field \( \vec{E} \) inside it. Each free electron of mass \( m \) and charge \( -e \) experiences an electrostatic force:
\( \vec{F} = -e\vec{E} \)
According to Newton's second law, this force produces an acceleration:
\( \vec{a} = \frac{\vec{F}}{m} \)
\( \implies \vec{a} = -\frac{e\vec{E}}{m} \)
Let \( \vec{u}_1, \vec{u}_2, \dots, \vec{u}_N \) be the random thermal velocities of the \( N \) free electrons before applying the field. Their average thermal velocity is zero:
\[ \frac{\vec{u}_1 + \vec{u}_2 + \dots + \vec{u}_N}{N} = 0 \]
Upon applying the field, the velocity of the \( i \)-th electron after a time interval \( \tau_i \) (since its last collision) is given by:
\( \vec{v}_i = \vec{u}_i + \vec{a}\tau_i \)
The average of all these post-collision velocities is defined as the drift velocity \( \vec{v}_d \):
\[ \vec{v}_d = \frac{\vec{v}_1 + \vec{v}_2 + \dots + \vec{v}_N}{N} \]
\[ \vec{v}_d = \frac{(\vec{u}_1 + \vec{a}\tau_1) + (\vec{u}_2 + \vec{a}\tau_2) + \dots + (\vec{u}_N + \vec{a}\tau_N)}{N} \]
\[ \vec{v}_d = \left(\frac{\vec{u}_1 + \vec{u}_2 + \dots + \vec{u}_N}{N}\right) + \vec{a}\left(\frac{\tau_1 + \tau_2 + \dots + \tau_N}{N}\right) \]
Since the first term is zero, and we define the average relaxation time as \( \tau = \frac{\tau_1 + \tau_2 + \dots + \tau_N}{N} \), we have:
\( \vec{v}_d = 0 + \vec{a}\tau \)
\( \implies \vec{v}_d = -\frac{e\vec{E}}{m}\tau \)
Taking only the magnitude of drift velocity:
\( v_d = \frac{eE}{m}\tau \)
In simple words: The electric field accelerates electrons, but collisions slow them down. By adding up all their individual velocities and taking the average, we find that the average speed depends directly on the electric field and the average time between collisions.

Exam Tip: Remember to explicitly state that the average of the initial thermal velocities of all free electrons is zero; missing this step will result in a loss of marks.

Question 254. Deduce the relation between current flowing through a conductor and drift velocity of free electrons. 
Answer: Consider a conductor of uniform cross-sectional area \( A \) connected across a potential difference \( V \). Let the number density of free electrons (number of free electrons per unit volume) be \( n \).
The drift velocity of the electrons is \( v_d \). In a small time interval \( \Delta t \), the distance traveled by the electrons is:
\( \Delta x = v_d \Delta t \)
The volume of the conductor of this length is:
\( \text{Volume} = A \Delta x = A v_d \Delta t \)
Thus, the total number of free electrons in this volume element is given by:
\( N = n A v_d \Delta t \)
The total charge \( \Delta Q \) that crosses this cross-section in time \( \Delta t \) is:
\( \Delta Q = N e = n e A v_d \Delta t \)
Using the definition of electric current \( I \):
\( I = \frac{\Delta Q}{\Delta t} \)
\( \implies I = \frac{n e A v_d \Delta t}{\Delta t} \)
\( \implies I = n e A v_d \)
Additionally, the current density \( j \) is defined as the current per unit area:
\( j = \frac{I}{A} \)
\( \implies j = n e v_d \)
In simple words: The current through a wire is equal to the number of electrons per unit volume multiplied by their charge, the wire's thickness, and their average drift speed. More electrons or faster movement means a stronger current.

Exam Tip: Be sure to define each term (\( n \), \( A \), and \( v_d \)) clearly at the start of your derivation to guarantee full credit.

Question 255. Deduce Ohm’s law using the concept of drift velocity. 
OR
On the basis of electron drift, derive an expression for resistivity of a conductor in terms of number density of free electrons and relaxation time. 
Answer: Let a potential difference \( V \) be applied across a conductor of length \( l \) and cross-sectional area \( A \). The uniform electric field \( E \) generated inside the conductor is:
\( E = \frac{V}{l} \)
The drift velocity \( v_d \) of free electrons under this electric field is:
\( v_d = \frac{e E}{m} \tau \)
Substituting the value of \( E \) into the drift velocity equation:
\( v_d = \frac{e V}{m l} \tau \)
The current \( I \) flowing through the conductor is related to the drift velocity by:
\( I = n e A v_d \)
Substituting the value of \( v_d \) into this relation:
\( I = n e A \left( \frac{e V}{m l} \tau \right) \)
\( \implies I = \left( \frac{n e^2 \tau A}{m l} \right) V \)
Rearranging this expression to solve for the ratio \( \frac{V}{I} \):
\( \frac{V}{I} = \left( \frac{m}{n e^2 \tau} \right) \left( \frac{l}{A} \right) \) ----(1)
If the physical state of the conductor, such as its temperature, is kept constant, all the terms inside the parentheses (mass \( m \), charge \( e \), number density \( n \), and relaxation time \( \tau \)) remain constant. Therefore, we can write:
\( \left( \frac{m}{n e^2 \tau} \right) \left( \frac{l}{A} \right) = \text{constant} = R \) ----(2)
where \( R \) is the electrical resistance of the conductor.
Using equation (2) in equation (1), we get:
\( \frac{V}{I} = R \)
\( \implies V = I R \)
This is Ohm's law.
Furthermore, the electrical resistance is also given by the formula:
\( R = \rho \frac{l}{A} \)
Comparing this standard relation with the expression for \( R \) in equation (2), we get the formula for resistivity \( \rho \):
\( \rho = \frac{m}{n e^2 \tau} \)
In simple words: By combining the formulas for current and electron drift speed, we show that voltage divided by current is equal to a constant value. This constant is the resistance, which depends on the metal's properties and shape. From this, we also find the formula for resistivity.

Exam Tip: Be prepared to answer both parts of this combined question; always state clearly that physical conditions like temperature must remain constant to justify Ohm's law.

Question 256. (i) Plot a graph showing the variation of resistivity with temperature in the case of a conductor.
(ii) How does one explain such behaviour, using the mathematical expression of the resistivity.
Answer:
(i) The graph depicting how the resistivity \( \rho \) of a metallic conductor varies with temperature \( T \) is shown below:

(ii) The mathematical expression for electrical resistivity is given by:
\( \rho = \frac{m}{n e^2 \tau} \)
In a metallic conductor, the number density \( n \) of free electrons is virtually independent of temperature. However, as the temperature of the conductor rises, the thermal motion of both the free electrons and the metal ions increases. This causes the electrons to collide more frequently with the vibrating lattice ions, reducing the average time between consecutive collisions (the relaxation time \( \tau \)). Since resistivity \( \rho \) is inversely proportional to \( \tau \), a decrease in \( \tau \) results in an increase in the resistivity of the metal.
Over a moderate temperature range, this relation can be written as:
\( \rho_T = \rho_0 [1 + \alpha (T - T_0)] \)
where \( \alpha \) is the temperature coefficient of resistivity.
In simple words: In metals, heating makes the metal atoms vibrate faster, which gets in the way of the moving electrons. This makes collisions happen more often, slowing down the current and increasing the resistivity.

Exam Tip: Be sure to state that the electron number density \( n \) in metals remains constant with temperature, as this is the key difference from semiconductors.

Question 257. (i) Plot a graph showing the variation of resistivity with temperature in the case of a semiconductor.
(ii) How does one explain such behaviour, using the mathematical expression of the resistivity.
Answer:
(i) The graph depicting how the resistivity of a semiconductor decreases with temperature is shown below:

(ii) The resistivity of a material is given by:
\( \rho = \frac{m}{n e^2 \tau} \)
\( \implies \rho \propto \frac{1}{n \tau} \)
In semiconductors, the band gap \( E_g \) between the valence and conduction bands is relatively small. As the temperature rises, more covalent bonds break, freeing a large number of charge carriers. The carrier concentration \( n \) increases exponentially with temperature according to:
\( n(T) = n_0 e^{-\frac{E_g}{k_B T}} \)
This rapid exponential increase in the carrier density \( n \) completely dominates the minor decrease in relaxation time \( \tau \) caused by lattice collisions. Consequently, the resistivity \( \rho \) of semiconductors drops exponentially as temperature increases, following:
\( \rho(T) = \rho_0 e^{\frac{E_g}{k_B T}} \)
In simple words: Heating a semiconductor breaks chemical bonds, which releases a massive number of free electrons. Even though these electrons collide with the lattice, the sheer increase in their number makes it much easier for current to flow, so resistivity drops quickly.

Exam Tip: Clearly state that the exponential increase in carrier density \( n \) dominates over the decrease in relaxation time \( \tau \) to get full marks on semiconductor temperature dependence questions.

Question 258. Explain by plotting a graph, variation of resistivity with temperature for an allow such as Nichrome (Constantan or manganin). 
Answer: Alloys like Nichrome, Constantan, and Manganin possess a very high resistivity even at absolute zero temperature. Their temperature coefficient of resistivity \( \alpha \) is extremely small. Consequently, their electrical resistivity exhibits a very weak, nearly linear dependency on temperature, as illustrated in the graph below:

Since \( \rho_T = \rho_0 [1 + \alpha(T - T_0)] \), and \( \alpha \) is negligible for these alloys, heating them causes almost no change in their overall resistance. This makes them highly suitable for fabricating standard resistance coils and standard resistors.
In simple words: Standard alloys like Nichrome barely change their resistance when heated. This stable behavior is why they are used to make accurate electrical instruments and standard resistors.

Exam Tip: Mentioning the extremely low temperature coefficient of resistivity (\( \alpha \)) is crucial for obtaining full marks when discussing why alloys are used for standard resistors.

Question 259. Plot a graph showing the variation of conductivity with temperature for a metallic conductor. How does one explain such behaviour, using the mathematical expression of the conductivity of a material. 
Answer:
(i) The graph below illustrates how the electrical conductivity \( \sigma \) of a metallic conductor decreases as the temperature rises:

(ii) Electrical conductivity \( \sigma \) is defined as the reciprocal of resistivity \( \rho \):
\( \sigma = \frac{1}{\rho} = \frac{n e^2 \tau}{m} \)
For a metallic conductor, the free electron density \( n \) remains constant with temperature. However, as temperature increases, the thermal speed of the electrons and the amplitude of the lattice vibrations both increase. This leads to more frequent collisions, which directly decreases the average relaxation time \( \tau \). Since conductivity \( \sigma \) is directly proportional to \( \tau \), a shorter relaxation time results in a corresponding reduction in the conductivity of the metal.
In simple words: When metals are heated, their atoms vibrate more, causing moving electrons to bump into them more frequently. These constant collisions reduce the average time between impacts, making the metal a poorer conductor of electricity.

Exam Tip: Be sure to write the reciprocal relation \( \sigma = \frac{1}{\rho} \) first, as it links the physical explanation of conductivity directly to the resistivity formulas you already know.

Question 260. A wire whose cross sectional area is increasing linearly from it one end to another, is connected across a battery Of volts. Which of following quantities remain constant in the wire ?
(a) drift speed (b) current density (c) electric current (d) electric field. Justify your answer.
Answer: The quantity that remains constant throughout the wire is (c) electric current.

Justification:
When a steady potential difference is applied across a conductor of varying cross-section, the electric current \( I \) flowing through any cross-section must remain constant. This is a direct consequence of the principle of conservation of charge, as charge cannot accumulate at any point along the conductor under steady-state conditions.
On the other hand, the other physical quantities depend on the cross-sectional area \( A \):
1. Current density: \( j = \frac{I}{A} \implies j \propto \frac{1}{A} \)
2. Drift speed: \( v_d = \frac{I}{n e A} \implies v_d \propto \frac{1}{A} \)
3. Electric field: \( E = \frac{j}{\sigma} = \frac{I}{\sigma A} \implies E \propto \frac{1}{A} \)
Since the cross-sectional area \( A \) varies along the wire, the drift speed, current density, and electric field will all change from one point to another.
In simple words: Electric current is like water flowing through a pipe; the same amount of water must enter and leave each second, regardless of how wide or narrow the pipe gets. However, the speed of flow (drift speed) and crowding (current density) will change in wider sections.

Exam Tip: Always cite the conservation of charge as the fundamental reason why steady-state current is independent of the cross-sectional area of a conductor.

Question 261. Two conducting wires X and Y of same diameter but different materials are joined in series across a battery. If the number density of electrons in X is twice that in Y, find the ratio of drift velocities of electrons in the two wires. 
Answer: Since the two wires \( X \) and \( Y \) are connected in series, the same electric current \( I \) flows through both of them:
\( I_X = I_Y \)
Also, because they have the same diameter, their cross-sectional areas are equal:
\( A_X = A_Y = A \)
The relationship between current and drift velocity is given by:
\( I = n e A v_d \)
Let the number density of electrons in wire \( Y \) be \( n_Y = n \). Then, the number density in wire \( X \) is given as:
\( n_X = 2n \)
Writing the current equations for both wires:
\( n_X e A v_X = n_Y e A v_Y \)
Substituting the values of \( n_X \) and \( n_Y \):
\( (2n) e A v_X = n e A v_Y \)
Canceling the common terms \( n, e, A \) from both sides:
\( 2 v_X = v_Y \)
\( \implies \frac{v_X}{v_Y} = \frac{1}{2} \)
Thus, the ratio of the drift velocities of electrons in the two wires is \( v_X : v_Y = 1 : 2 \).
In simple words: Since the wires are in series, they carry the same current. Wire X has twice as many free electrons as wire Y, so those electrons only need to move at half the speed of the electrons in wire Y to carry that same amount of current.

Exam Tip: Under series connection, the currents are always equal (\( I_X = I_Y \)). Using this key equality is the correct starting point for this calculation.

Question 262. Explain giving reasons, how the internal resistance of a cell changes in the following cases : 
(i) When concentration of the electrolyte is increased
(ii) When area of the anode is decreased
(iii) When temperature of the electrolyte is increased
Answer:
(i) Concentration is increased: The internal resistance of the cell increases. This is because a higher solute concentration brings ions closer together, which enhances inter-ionic attractions. These strong attractions make it harder for the ions to migrate freely through the liquid.

(ii) Anode area is decreased: The internal resistance of the cell increases. A smaller surface area for the anode reduces the physical contact region with the electrolyte, thereby limiting its capacity to collect and pull the oppositely charged ions from the solution.

(iii) Temperature of the electrolyte is increased: The internal resistance of the cell decreases. At elevated temperatures, the thermal energy of the system increases, which weakens both the inter-ionic electrostatic forces and the viscous drag of the solvent. This allows the ions to move with greater mobility.
In simple words: (i) A thicker electrolyte makes it crowded and hard for ions to swim. (ii) A smaller electrode gives less space for ions to land, increasing resistance. (iii) Heating the fluid makes it less sticky and lets the ions swim faster, lowering resistance.

Exam Tip: When explaining internal resistance changes, always use the terminology of ion mobility, viscous drag, and electrode surface area for precise, high-scoring answers.

Question 263. Derive an expression for the equivalent resistance of combination of cells in series. 
Answer: Let us consider two cells connected in series. The first cell has an electromotive force \( E_1 \) and an internal resistance \( r_1 \), while the second cell has an electromotive force \( E_2 \) and an internal resistance \( r_2 \). They are joined at point \( B \) as shown in the diagram below:
₁ r₁ B E₂ r₂ C
Let \( I \) be the current flowing through the combination. The potential difference across the first cell (between points \( A \) and \( B \)) is:
\( V_{AB} = V_A - V_B = E_1 - I r_1 \)
Similarly, the potential difference across the second cell (between points \( B \) and \( C \)) is:
\( V_{BC} = V_B - V_C = E_2 - I r_2 \)
The total potential difference across the entire combination (between points \( A \) and \( C \)) is given by:
\( V_{AC} = V_A - V_C = (V_A - V_B) + (V_B - V_C) \)
Substituting the expressions for \( V_{AB} \) and \( V_{BC} \):
\( V_{AC} = (E_1 - I r_1) + (E_2 - I r_2) \)
\( \implies V_{AC} = (E_1 + E_2) - I(r_1 + r_2) \) ----(1)
If we replace this series combination of cells with a single equivalent cell between \( A \) and \( C \) having an equivalent electromotive force \( E_{eq} \) and an equivalent internal resistance \( r_{eq} \), the terminal potential difference is:
\( V_{AC} = E_{eq} - I r_{eq} \) ----(2)
Comparing equation (1) and equation (2), we get:
\( E_{eq} = E_1 + E_2 \)
and
\( r_{eq} = r_1 + r_2 \)
Thus, the equivalent internal resistance of the series combination of cells is the sum of their individual internal resistances.
In simple words: When cells are connected back-to-back in a row, their voltages add up to give a stronger total voltage, and their internal resistances also add up to make a higher total internal resistance.

Exam Tip: Remember to use the node potentials (\( V_A - V_B \) and \( V_B - V_C \)) to set up the derivation; this logical step makes the proof clean and easy to follow.

Question 264. Two cells of emfs \( E_1 \) and \( E_2 \) and internal resistances \( r_1 \) and \( r_2 \) are connected in parallel as shown in the figure. Deduce an expression for the 
(i) equivalent emf of the combination
(ii) equivalent internal resistance of the combination
(iii) potential difference between the points A and C
Answer: Let two cells of emfs \( E_1, E_2 \) and internal resistances \( r_1, r_2 \) be connected in parallel between the junctions \( B_1 \) and \( B_2 \). Let \( I_1 \) and \( I_2 \) be the currents leaving the positive terminals of the two cells, and \( I \) be the total current in the main circuit:
₁ E₁ r₁ E₂ r₂ B₂ C I₁ I₂ I
By applying Kirchhoff's junction rule at \( B_1 \):
\( I = I_1 + I_2 \)
Let \( V \) be the common potential difference across the parallel combination (between points \( B_1 \) and \( B_2 \)). For the upper branch containing the first cell:
\( V = E_1 - I_1 r_1 \)
\( \implies I_1 = \frac{E_1 - V}{r_1} \)
For the lower branch containing the second cell:
\( V = E_2 - I_2 r_2 \)
\( \implies I_2 = \frac{E_2 - V}{r_2} \)
Substitute the expressions for \( I_1 \) and \( I_2 \) into the total current equation:
\( I = \frac{E_1 - V}{r_1} + \frac{E_2 - V}{r_2} \)
\( \implies I = \left( \frac{E_1}{r_1} + \frac{E_2}{r_2} \right) - V \left( \frac{1}{r_1} + \frac{1}{r_2} \right) \)
\( \implies I = \left( \frac{E_1 r_2 + E_2 r_1}{r_1 r_2} \right) - V \left( \frac{r_1 + r_2}{r_1 r_2} \right) \)
Rearranging this equation to solve for \( V \):
\( V \left( \frac{r_1 + r_2}{r_1 r_2} \right) = \left( \frac{E_1 r_2 + E_2 r_1}{r_1 r_2} \right) - I \)
Multiplying both sides by \( \frac{r_1 r_2}{r_1 + r_2} \):
\( V = \left( \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2} \right) - I \left( \frac{r_1 r_2}{r_1 + r_2} \right) \) ----(1)
If we replace this combination with a single equivalent cell between \( A \) and \( C \), the potential difference across it is:
\( V = E_{eq} - I r_{eq} \) ----(2)
Comparing equations (1) and (2):

(i) Equivalent emf (\( E_{eq} \)):
\( E_{eq} = \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2} \)

(ii) Equivalent internal resistance (\( r_{eq} \)):
\( r_{eq} = \frac{r_1 r_2}{r_1 + r_2} \) (or \( \frac{1}{r_{eq}} = \frac{1}{r_1} + \frac{1}{r_2} \))

(iii) Potential difference between \( A \) and \( C \):
The potential difference across the terminal points \( A \) and \( C \) is the same as the potential difference across the parallel junctions:
\( V_{AC} = V = E_{eq} - I r_{eq} = \left( \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2} \right) - I \left( \frac{r_1 r_2}{r_1 + r_2} \right) \)
In simple words: For cells connected in parallel, the total current is divided between them. By using the terminal voltage equation for each branch and combining them, we find that the equivalent voltage is a weighted average of their individual voltages, and the reciprocal of the total internal resistance is the sum of the reciprocals of each resistance.

Exam Tip: In parallel combinations, ensure you state that the potential difference \( V \) across both branches remains identical; this is the key constraint that makes the math work.

Question 265. State Kirchhoff’s rules in electrostatics and explain on what basis they are justified ? 
Answer: Kirchhoff's rules for electric circuits are stated as follows:

(i) Junction Rule (First Law):
The algebraic sum of all currents meeting at any electrical junction in a closed circuit is zero.
\( \sum I = 0 \)
In other words, the total current entering a junction must equal the total current leaving that junction:
\( I_{\text{in}} = I_{\text{out}} \)
\( \implies i_1 + i_2 = i_3 + i_4 + i_5 \)
Justification: This rule is based on the law of conservation of charge. Since charge cannot accumulate or be depleted at any point under steady-state conditions, the rate at which charge enters a point must equal the rate at which it leaves.

(ii) Loop Rule (Second Law):
In any closed loop or mesh of an electric circuit, the algebraic sum of the potential differences (which is the sum of the products of current and resistance, \( IR \), in each segment) is equal to the algebraic sum of the electromotive forces (\( E \)) acting in that loop.
\( \sum I R = \sum E \)
Justification: This rule is based on the law of conservation of energy. Since the electrostatic field is conservative, the total work done in moving a unit charge around any closed path must be zero.
In simple words: The first rule says that whatever current goes into a junction must come out, because charge cannot disappear. The second rule says that if you travel in a complete loop around a circuit, the total voltage drops must equal the total voltage gains, because energy is conserved.

Exam Tip: Clearly state the physical conservation law associated with each rule (charge for the Junction Rule and energy for the Loop Rule), as these are highly valued by evaluators.

Question 266. What is Wheatstone bridge ? When is the bridge said to be balanced ? Use Kirchhoff’s rules to obtain conditions for the balanced condition in a Wheatstone bridge. 
Answer:
Wheatstone Bridge: It is an electrical network consisting of four resistors arranged in a closed loop (or bridge shape). It is primarily used to determine the value of an unknown resistance in terms of three other known resistances.

Balanced Condition: The bridge is said to be in a balanced state when the resistances are adjusted such that no electric current flows through the central galvanometer branch (\( I_g = 0 \)). In this balanced state, the ratio of the adjacent resistances is equal:
\( \frac{P}{Q} = \frac{R}{S} \)


Derivation of Balanced Condition using Kirchhoff's Rules:
Let \( I_1 \) be the current flowing through resistor \( P \) and \( I_2 \) be the current flowing through resistor \( R \). At balance, the current through the galvanometer is zero (\( I_g = 0 \)). Therefore, by Kirchhoff's junction rule, the current through \( Q \) is also \( I_1 \), and the current through \( S \) is also \( I_2 \).

Applying Kirchhoff's loop rule to the closed loop \( ABDA \):
\( I_1 P + I_g G - I_2 R = 0 \)
Since the bridge is balanced, \( I_g = 0 \):
\( I_1 P + 0 - I_2 R = 0 \)
\( \implies I_1 P = I_2 R \) ----(1)

Now, applying Kirchhoff's loop rule to the closed loop \( BCDB \):
\( I_1 Q - I_2 S - I_g G = 0 \)
Since \( I_g = 0 \):
\( I_1 Q - I_2 S - 0 = 0 \)
\( \implies I_1 Q = I_2 S \) ----(2)

Dividing equation (1) by equation (2):
\( \frac{I_1 P}{I_1 Q} = \frac{I_2 R}{I_2 S} \)
\( \implies \frac{P}{Q} = \frac{R}{S} \)
This is the balanced condition for a Wheatstone bridge.
In simple words: A Wheatstone bridge uses a diamond of four resistors. When the opposite sides are balanced, no current flows down the middle path. We use this to calculate an unknown resistance using the simple ratio formula.

Exam Tip: Draw the loop direction arrows clearly in your diagram and show the intermediate step setting \( I_g = 0 \) to prevent any loss of presentation marks.

Question 267. How a metre bridge is used to determine the unknown resistance of a given wire ? Write the necessary precautions to minimize the error in the result. 
Answer: A meter bridge works on the principle of a balanced Wheatstone bridge. The experimental setup is represented below:

Let a standard known resistance \( R \) be connected in the left gap of the meter bridge, and the unknown resistance \( S \) be connected in the right gap. A uniform resistance wire \( AC \) of length \( 100\text{ cm} \) is stretched on a wooden board.
A jockey is moved along the wire until a point \( D \) is reached where the galvanometer shows zero deflection (null point). Let the balancing length from end \( A \) be \( l\text{ cm} \). The length of the remaining wire is \( (100 - l)\text{ cm} \).
Since the wire is uniform, the resistance of the segment \( AD \) (representing resistor \( P \)) and the segment \( DC \) (representing resistor \( Q \)) are proportional to their respective lengths:
\( P \propto l \)
and
\( Q \propto (100 - l) \)
Under the balanced bridge condition:
\( \frac{P}{Q} = \frac{R}{S} \)
\( \implies \frac{l}{100 - l} = \frac{R}{S} \)
Solving for the unknown resistance \( S \):
\( S = R \left( \frac{100 - l}{l} \right) = R \left( \frac{100}{l} - 1 \right) \)
By changing the value of the resistance \( R \), we can repeat the experiment to get multiple readings, and the average of these calculated values gives a highly accurate measurement of the unknown resistance \( S \).

Precautions to Minimize Error:
1. All electrical connections must be kept clean, tightly bound, and free from any insulating oxide layers.
2. The value of the standard resistance \( R \) should be chosen such that the null point \( D \) lies close to the middle of the wire (between \( 40\text{ cm} \) and \( 60\text{ cm} \)), which reduces percentage measurement errors.
In simple words: A meter bridge uses a 1-meter wire to balance two resistors. By finding the spot on the wire where the current is zero, we can measure the length of both sides and use their ratio to find the unknown resistance.

Exam Tip: Remember to specify the two major precautions - tight/clean connections and choosing \( R \) to get a mid-scale null point - as these are frequently asked in laboratory-based theory questions.

Question 268. (i) With the help of a circuit diagram, explain how a potentiometer is used to compare the emf’s of two primary cells. obtain the required expression used for comparing the emfs. 
(ii) Write two possible causes for one sided deflection in a potentiometer experiment. CBSE (D)-2013
Answer:
(i) Comparing EMFs of Two Cells:
The experimental circuit diagram for comparing the electromotive forces of two primary cells using a potentiometer is shown below:
₁ E₂ G J
Let a steady current be maintained through the uniform potentiometer wire \( AB \) by a driver cell, establishing a constant potential gradient \( K \) (potential drop per unit length) along the wire.
Let \( E_1 \) and \( E_2 \) be the emfs of the two cells to be compared. First, we connect the cell \( E_1 \) in the circuit. Let \( l_1 \) be the balancing length obtained on the wire where the galvanometer shows zero deflection. According to the principle of a potentiometer:
\( E_1 = K l_1 \) ----(1)
Next, we disconnect the first cell and insert the second cell \( E_2 \) into the circuit. Let \( l_2 \) be the new balancing length where zero deflection is achieved. Thus:
\( E_2 = K l_2 \) ----(2)
Dividing equation (1) by equation (2), we get the ratio of their emfs:
\( \frac{E_1}{E_2} = \frac{l_1}{l_2} \)

(ii) Two Possible Causes for One-Sided Deflection:
1. The electromotive force of the main driver cell in the auxiliary/primary circuit is smaller than the emf (\( E_1 \) or \( E_2 \)) of the cells being measured.
2. The positive terminals of the driver cell and the test cells are not connected to the same initial zero end (\( A \)) of the potentiometer wire.
In simple words: (i) A potentiometer drops voltage steadily along its wire. By finding the lengths where two different cells balance this voltage drop, the ratio of those lengths gives the exact ratio of their voltages. (ii) If the galvanometer only swings one way, it usually means your driver battery is too weak or you hooked up the positive wires to the wrong ends.

Exam Tip: Be sure to explicitly define the potential gradient \( K \) in your derivation, as this links the physical lengths directly to the electromotive forces.

Question 269. With the help of a circuit diagram, explain how a potentiometer is used to determine the internal resistance of a cell. tain the required expression used. 
Answer:
The circuit diagram for determining the internal resistance of a cell using a potentiometer is shown below:

Question 269. With the help of a circuit diagram, explain how a potentiometer is used to determine the internal resistance of a cell. tain the required expression used.
Answer:
The experimental setup used to find the internal resistance of a cell is illustrated in the schematic below: Let \( K \) represent the potential gradient along the potentiometer wire \( AB \).

First Step: Keep the key \( K_2 \) in the secondary circuit open so that no current passes through the external resistance box \( R \). Find the balancing length \( l_1 \) on the wire where the galvanometer reads zero. Here, the potentiometer balances the open-circuit electromotive force (\( \varepsilon \)) of the cell:
\( \varepsilon = K l_1 \) ----(1)

Second Step: Close key \( K_2 \), allowing current to flow through the shunt resistance \( R \). The balanced voltage is now the terminal potential difference \( V \) across the cell. Let \( l_2 \) be the new balancing length:
\( V = K l_2 \) ----(2)

Dividing equation (1) by equation (2):
\( \frac{\varepsilon}{V} = \frac{K l_1}{K l_2} = \frac{l_1}{l_2} \) ----(3)

We know that \( \varepsilon = I(R + r) \) and \( V = IR \), where \( r \) is the internal resistance of the cell. Substituting these values into equation (3):
\( \frac{I(R + r)}{IR} = \frac{l_1}{l_2} \)

\( \implies \frac{R + r}{R} = \frac{l_1}{l_2} \)

\( \implies 1 + \frac{r}{R} = \frac{l_1}{l_2} \)

\( \implies \frac{r}{R} = \frac{l_1}{l_2} - 1 \)

\( \implies r = R \left( \frac{l_1}{l_2} - 1 \right) \)
This is the expression used to calculate the internal resistance.
In simple words: First, balance the cell's open-circuit voltage, then shunt it with a resistor to balance its terminal voltage. The ratio of these two lengths gives the internal resistance.

Exam Tip: Ensure that the primary cell used as the driver has an EMF larger than the experimental cell, or you won't get a balance point on the wire.

Question 270. Why is potentiometer preferred over a voltmeter for comparison of emf. of cells ? 
Answer:
At the balanced null point, the potentiometer draws zero current from the cell under evaluation. This enables it to calculate the true open-circuit terminal voltage, which is the actual electromotive force (\( \varepsilon \)) of the cell. Conversely, a voltmeter must draw a small current to show a reading, which causes an internal voltage drop inside the cell. Thus, a voltmeter measures only the terminal potential difference (\( V = \varepsilon - Ir \)), which is always slightly lower than the true EMF.

In simple words: A voltmeter must draw current to work, which lowers the voltage it is trying to measure. A potentiometer measures at a null point where zero current is drawn, giving the exact true voltage.

Exam Tip: The core reason to highlight is that the potentiometer behaves as an ideal voltmeter with infinite resistance because it works on a null-deflection principle.

Question 271. (i) Define potential gradient. Write its S.I. unit. Obtain an expression for potential gradient in terms of the resistivity of the potentiometer wire. 
(ii) In a potentiometer experiment, if the area of cross section of the wire increases uniformly from one end to another, draw a graph showing how potential gradient would vary as the length of the wire increases from one end ? 
Answer:
(i) Potential Gradient:
The potential drop per unit length across a current-carrying wire is defined as the potential gradient (\( K \)). Its S.I. unit is **V/m** (volts per meter).

Expression in terms of resistivity:
Let a steady current \( I \) flow through a potentiometer wire of length \( L \), uniform cross-sectional area \( A \), and material resistivity \( \rho \). The resistance of the wire is:
\( R = \rho \frac{L}{A} \)
The total voltage drop \( V \) across the wire is given by Ohm's law:
\( V = I R = I \left( \rho \frac{L}{A} \right) \)
The potential gradient \( K \) is defined as:
\( K = \frac{V}{L} \)

\( \implies K = \frac{I \rho \frac{L}{A}}{L} \)

\( \implies K = \frac{I \rho}{A} \)

(ii) If the cross-sectional area of the wire increases uniformly as a function of length \( x \) from one end, the area at any point \( A(x) \) increases. Since \( K = \frac{I \rho}{A} \), the potential gradient \( K \) is inversely proportional to the area of cross-section (
\( K \propto \frac{1}{A} \)). Therefore, as length \( x \) increases and the wire becomes thicker, the potential gradient decreases. The graph of this variation is shown below: In simple words: (i) Potential gradient is the voltage drop over each meter of the wire. (ii) If the wire gets thicker as you go along, its local resistance drops, so the voltage gradient decreases.

Exam Tip: Always remember the relation \( K = \frac{I \rho}{A} \). When writing the unit, both V/m and V/cm are acceptable, but V/m is the official SI unit.

Question 272. A conductor of length 'l' is connected to a d.c. source of potential 'V'. If the length of the conductor is tripled by gradually stretching it, keeping 'V' constant, how will (i) drift speed of electrons and (ii) resistance of the conductor be affected ? Justify your answer. 
Answer:
(i) Effect on Drift Speed:
The drift speed \( v_d \) of free electrons is given by:
\( v_d = \frac{e V}{m l} \tau \)
Since the potential difference \( V \), electron charge \( e \), mass \( m \), and relaxation time \( \tau \) remain constant, we have:
\( v_d \propto \frac{1}{l} \)
When the length is tripled (
\( l' = 3l \)), the new drift speed \( v_d' \) becomes:

\( \implies v_d' = \frac{v_d}{3} \)
Thus, the drift velocity reduces to **one-third** of its initial value.

(ii) Effect on Resistance:
When a wire is stretched, its volume remains constant. Therefore, if the length becomes three times (
\( l' = 3l \)), the cross-sectional area \( A' \) must decrease to one-third (
\( A' = \frac{A}{3} \)) to keep the product of area and length constant.
The initial resistance is:
\( R = \rho \frac{l}{A} \)
The new resistance \( R' \) is:
\( R' = \rho \frac{l'}{A'} = \rho \frac{3l}{A / 3} = 9 \left( \rho \frac{l}{A} \right) \)

\( \implies R' = 9 R \)
Hence, the resistance increases to **nine times** the original value.
In simple words: Tripling the length under the same voltage drops the electric field strength, so electrons drift at 1/3 the speed. Because stretching also makes the wire thinner, the resistance jumps to 9 times the original.

Exam Tip: Ensure you clearly state that the volume is conserved during stretching so that the change in area is accounted for alongside the change in length.

Question 273. Two wires X and Y have the same resistivity but their cross sectional areas are in the ratio 2:3 and lengths in the ratio 1:2. They are first connected in series and then in parallel to a d.c. source. Find out the ratio of the drift speeds of the electrons in the two wires for the two cases. 
Answer:
Let the ratio of cross-sectional areas be \( \frac{A_X}{A_Y} = \frac{2}{3} \) and the ratio of lengths be \( \frac{l_X}{l_Y} = \frac{1}{2} \). Since they have the same resistivity, they are made of the same material, meaning the number density of free electrons \( n \) is the same for both.

Case (i): When connected in series:
In a series combination, the same electric current \( I \) flows through both wires (\( I_X = I_Y \)).
Since \( I = n e A v_d \), we can write:
\( n e A_X v_{dX} = n e A_Y v_{dY} \)

\( \implies A_X v_{dX} = A_Y v_{dY} \)

\( \implies \frac{v_{dX}}{v_{dY}} = \frac{A_Y}{A_X} = \frac{3}{2} \)
Therefore, the ratio of drift speeds in series is **3 : 2**.

Case (ii): When connected in parallel:
In a parallel combination, both wires are connected across the same potential difference \( V \).
The drift speed is given by:
\( v_d = \frac{e V}{m l} \tau \)
Since the potential \( V \) is constant and \( e, m, \tau \) are properties of the material which are identical for both wires, we get:

\( \implies v_d \propto \frac{1}{l} \)

\( \implies \frac{v_{dX}}{v_{dY}} = \frac{l_Y}{l_X} = \frac{2}{1} \)
Therefore, the ratio of drift speeds in parallel is **2 : 1**.
In simple words: In series, the current is identical, so drift speed depends only on the area ratio. In parallel, the voltage is identical, so drift speed depends only on the length ratio.

Exam Tip: For parallel connections, avoid using \( I = neAv_d \) directly without first relating the currents using \( V = IR \). Using \( v_d \propto \frac{V}{l} \) is much cleaner and less prone to calculation errors.

Question 274. A potential difference V is applied across a conductor of length L and diameter D. How is the drift velocity \( v_d \), of charge carriers in the conductor is affected when (i) V is halved (ii) L is doubled and (iii) D is halved ? Justify your answer in each case. 
Answer:
The magnitude of the drift velocity of free electrons is given by:
\( v_d = \frac{e V}{m L} \tau \)

(i) When \( V \) is halved:
Since \( v_d \propto V \), if the applied voltage is reduced to half, the drift velocity will also be **halved**.

(ii) When \( L \) is doubled:
Since \( v_d \propto \frac{1}{L} \), if the length of the conductor is doubled while keeping \( V \) constant, the drift velocity will become **halved**.

(iii) When \( D \) is halved:
The expression for drift velocity does not contain any diameter \( D \) term. Thus, drift speed is independent of the thickness or diameter of the wire. Halving \( D \) will have **no effect** on the drift velocity.
In simple words: (i) Halving voltage cuts drift speed in half. (ii) Doubling length cuts drift speed in half. (iii) Changing diameter has no effect because drift velocity does not depend on the wire's width.

Exam Tip: State the formula \( v_d = \frac{e V \tau}{m L} \) at the very beginning to justify each relationship clearly and earn full marks.

Question 275. Two cells of emf 1.5 V and 2.0 V having internal resistances 0.2\Omega and 0.3\Omega respectively are connected in parallel. Calculate the emf and internal resistance of the equivalent cell. 
Answer:
We are given:
\( E_1 = 1.5\text{ V}, \quad r_1 = 0.2\,\Omega \)
\( E_2 = 2.0\text{ V}, \quad r_2 = 0.3\,\Omega \)

1. Equivalent Electromotive Force (
\( E_{eq} \)):
For two cells connected in parallel, the equivalent emf is given by:
\( E_{eq} = \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2} \)
Substituting the given values:
\( E_{eq} = \frac{(1.5 \times 0.3) + (2.0 \times 0.2)}{0.2 + 0.3} \)

\( \implies E_{eq} = \frac{0.45 + 0.40}{0.5} \)

\( \implies E_{eq} = \frac{0.85}{0.5} = 1.7\text{ V} \)

2. Equivalent Internal Resistance (
\( r_{eq} \)):
The formula for equivalent internal resistance of a parallel combination is:
\( r_{eq} = \frac{r_1 r_2}{r_1 + r_2} \)
Substituting the given values:
\( r_{eq} = \frac{0.2 \times 0.3}{0.2 + 0.3} \)

\( \implies r_{eq} = \frac{0.06}{0.5} = 0.12\,\Omega \)

The equivalent cell has an EMF of **1.7 V** and an internal resistance of **\( 0.12\,\Omega \)**.
In simple words: In parallel, the combined voltage of the cells becomes a weighted average of 1.7 V, and their parallel internal resistance works out to 0.12 ohms.

Exam Tip: Always perform the calculation steps with decimals carefully, as minor division errors are highly common in parallel cell problems.

Question 276. Calculate the current drawn from the battery by the network of the resistors shown in figure.
Answer:
Let the given circuit be viewed as a standard diamond network with points \( A, B, C, D \) as represented in the layout below: The ratio of the resistances in the opposite arms is:

\( \frac{R_{AB}}{R_{AD}} = \frac{1\,\Omega}{2\,\Omega} = \frac{1}{2} \)

\( \frac{R_{BC}}{R_{CD}} = \frac{2\,\Omega}{4\,\Omega} = \frac{1}{2} \)
Since the ratios are equal, the network is a balanced Wheatstone bridge:

\( \frac{R_{AB}}{R_{AD}} = \frac{R_{BC}}{R_{CD}} \)
Consequently, the electric potential at point \( B \) is equal to the potential at point \( D \). No current flows through the central branch \( BD \) which has the \( 5\,\Omega \) resistor. This central resistor can be omitted from our calculations.

Now, the circuit simplifies into two parallel branches:
1. The upper series branch \( ABC \):
\( R_{\text{upper}} = 1\,\Omega + 2\,\Omega = 3\,\Omega \)
2. The lower series branch \( ADC \):
\( R_{\text{lower}} = 2\,\Omega + 4\,\Omega = 6\,\Omega \)

The total equivalent resistance \( R \) of these two branches in parallel is:
\( \frac{1}{R} = \frac{1}{R_{\text{upper}}} + \frac{1}{R_{\text{lower}}} \)

\( \implies \frac{1}{R} = \frac{1}{3} + \frac{1}{6} = \frac{2 + 1}{6} = \frac{3}{6} = \frac{1}{2} \)

\( \implies R = 2\,\Omega \)

The current \( I \) drawn from the \( 4\text{ V} \) battery is:

\( I = \frac{V}{R} = \frac{4\text{ V}}{2\,\Omega} = 2\text{ A} \).
In simple words: The ratios of opposite resistors are equal, so the bridge is balanced and no current goes through the central 5-ohm branch. The rest simplifies to 2 ohms of total resistance connected to 4 volts, drawing 2 amps.

Exam Tip: Always show the ratio test explicitly (\( \frac{1}{2} = \frac{2}{4} \)) in your written answer to demonstrate that you identified the balanced bridge correctly.

Question 277. In a meter bridge with R and S in the gaps, the null point is found at 40 cm from A. If the resistance of 30 \Omega is connected in parallel with S, the null point occurs at 50 cm from A. Determine the value of R and S. 
Answer:
First Case:
Let \( l_1 = 40\text{ cm} \) be the initial balancing length from end \( A \). The formula for the meter bridge is:
\( \frac{R}{S} = \frac{l_1}{100 - l_1} \)

\( \implies \frac{R}{S} = \frac{40}{100 - 40} = \frac{40}{60} = \frac{2}{3} \)

\( \implies R = \frac{2}{3} S \) ----(1)

Second Case:
When a resistor of \( 30\,\Omega \) is shunted in parallel with \( S \), the new equivalent resistance in the right gap becomes:

\( S' = \frac{30 S}{30 + S} \)
The new null point occurs at \( l_2 = 50\text{ cm} \). Thus:
\( \frac{R}{S'} = \frac{50}{100 - 50} = \frac{50}{50} = 1 \)

\( \implies R = S' \) ----(2)

Equating the values of \( R \) from equations (1) and (2):
\( \frac{2}{3} S = \frac{30 S}{30 + S} \)
Since \( S \neq 0 \), we can cancel \( S \) from both sides:
\( \frac{2}{3} = \frac{30}{30 + S} \)

\( \implies 2(30 + S) = 90 \)

\( \implies 60 + 2S = 90 \)

\( \implies 2S = 30 \)

\( \implies S = 15\,\Omega \)

Substituting this back into equation (1):
\( R = \frac{2}{3} \times 15 = 10\,\Omega \).
Thus, the values are \( R = 10\,\Omega \) and \( S = 15\,\Omega \).
In simple words: The initial setup tells us R is 2/3 of S. Shunting S with 30 ohms brings the balance point to the center (making R equal to the shunted resistance). Solving these gives S = 15 ohms and R = 10 ohms.

Exam Tip: In calculations, cancel the common variable \( S \) on both sides carefully instead of expanding into a quadratic equation, which saves time.

Question 278. In a meter bridge, the null point is found at a distance of \( l_1 \) cm from A. If now a resistance of X is connected in parallel with S, the null point occurs at \( l_2 \) cm from A. Obtain a formula for X in terms of \( l_1 \), \( l_2 \) and S. 
Answer:
In the first case:
Let \( l_1 \) be the initial balancing length. The balance condition of the bridge is:
\( \frac{R}{S} = \frac{l_1}{100 - l_1} \) ----(1)

In the second case:
When resistor \( X \) is connected in parallel with \( S \), the effective resistance in the right gap is:
\( S' = \frac{X S}{X + S} \)
The new null point is found at distance \( l_2 \). The balance condition is:
\( \frac{R}{S'} = \frac{l_2}{100 - l_2} \) ----(2)

Dividing equation (1) by equation (2):

\( \implies \frac{S'}{S} = \frac{l_1 (100 - l_2)}{l_2 (100 - l_1)} \)
Substituting the expression for \( S' \):
\( \frac{X S}{S(X + S)} = \frac{l_1 (100 - l_2)}{l_2 (100 - l_1)} \)

\( \implies \frac{X}{X + S} = \frac{l_1 (100 - l_2)}{l_2 (100 - l_1)} \)
Taking the reciprocal of both sides:

\( \implies 1 + \frac{S}{X} = \frac{l_2 (100 - l_1)}{l_1 (100 - l_2)} \)

\( \implies \frac{S}{X} = \frac{l_2 (100 - l_1)}{l_1 (100 - l_2)} - 1 \)

\( \implies X = \frac{S}{\frac{l_2 (100 - l_1)}{l_1 (100 - l_2)} - 1} \)
The parallel combination in the gap is illustrated in the diagram below: In simple words: We set up equations for both balancing states and divide them to remove R. Taking the reciprocal gives a clean expression for X.

Exam Tip: Simplifying this with algebraic reciprocals avoids complex division steps and is highly recommended during exams.

Question 279. A resistance of R \Omega draws current from a potentiometer. The potentiometer wire AB, has a total resistance of \( R_0 \, \Omega \). A voltage V is supplied to the potentiometer. Derive an expression for the voltage across R when the sliding contact is in the middle of the potentiometer wire. 
Answer:
Since the sliding contact is in the middle of the wire, the wire of total resistance \( R_0 \) is split into two equal parts, each of resistance \( \frac{R_0}{2} \).
The resistance of the segment \( AC \) (which is \( \frac{R_0}{2} \)) is in parallel with the external resistor \( R \). Let the equivalent resistance of this parallel section be \( R_1 \):
\( R_1 = \frac{R \left( \frac{R_0}{2} \right)}{R + \frac{R_0}{2}} = \frac{R R_0}{2R + R_0} \)

The total resistance \( R_{eq} \) of the whole circuit (parallel section \( R_1 \) in series with the remaining half-wire \( \frac{R_0}{2} \)) is:
\( R_{eq} = R_1 + \frac{R_0}{2} = \frac{R R_0}{2R + R_0} + \frac{R_0}{2} \)

\( \implies R_{eq} = \frac{2 R R_0 + R_0(2R + R_0)}{2(2R + R_0)} = \frac{4 R R_0 + R_0^2}{2(2R + R_0)} \)

The current \( I \) flowing through the circuit from the source voltage \( V \) is:
\( I = \frac{V}{R_{eq}} \)

\( \implies I = \frac{2 V (2R + R_0)}{R_0(4R + R_0)} \)

The potential difference \( V_1 \) across the resistor \( R \) is the voltage drop across the parallel segment \( R_1 \):
\( V_1 = I R_1 \)

\( \implies V_1 = \left( \frac{2 V (2R + R_0)}{R_0(4R + R_0)} \right) \left( \frac{R R_0}{2R + R_0} \right) \)

\( \implies V_1 = \frac{2 V R}{4R + R_0} \)
The circuit diagram representing this middle-contact potentiometer is shown below: In simple words: When the slider is in the middle, the wire is cut into two equal halves. Finding the parallel resistance of R with the first half, and adding it to the second, yields the overall circuit current and the voltage drop across R.

Exam Tip: The key step is combining the parallel parts first, then calculating total circuit resistance to find the main current.

Question 280. In the circuit diagram given below, AB is a uniform wire of resistance 15 \Omega and length 1 m. It is connected to a cell \( E_1 \) of emf 2V and negligible internal resistance and a resistance R. The balance point with another cell \( E_2 \) of emf 75 mV is found at 30 cm from end A. Calculate the value of R. [Ans 105 \Omega] 
Answer:
We are given:
Resistance of potentiometer wire, \( R_{AB} = 15\,\Omega \)
Length of wire, \( L = 1\text{ m} = 100\text{ cm} \)
Primary source EMF, \( E_1 = 2\text{ V} \)
Balanced cell EMF, \( E_2 = 75\text{ mV} = 75 \times 10^{-3}\text{ V} \)
Balancing length, \( l = 30\text{ cm} \)

The current \( I \) flowing through the main potentiometer circuit is:
\( I = \frac{E_1}{R + R_{AB}} = \frac{2}{R + 15} \)

The voltage drop \( V_{AB} \) across the wire \( AB \) is:
\( V_{AB} = I \times R_{AB} = \left( \frac{2}{R + 15} \right) \times 15 = \frac{30}{R + 15}\text{ V} \)

The potential gradient \( K \) per unit length along the wire is:
\( K = \frac{V_{AB}}{L} = \frac{30}{100(R + 15)} = \frac{0.3}{R + 15}\text{ V/cm} \)

Using the balancing condition of the potentiometer:
\( E_2 = K l \)

\( \implies 75 \times 10^{-3} = \left( \frac{0.3}{R + 15} \right) \times 30 \)

\( \implies 75 \times 10^{-3} = \frac{9}{R + 15} \)

\( \implies R + 15 = \frac{9}{75 \times 10^{-3}} = \frac{9000}{75} = 120 \)

\( \implies R = 120 - 15 = 105\,\Omega \)
The circuit diagram for this setup is shown below: In simple words: The primary battery sets up a voltage drop across the wire. Finding the gradient for a 30 cm balance length of 75 mV shows that the total wire drop is 250 mV, leading to a calculated series resistance R of 105 ohms.

Exam Tip: Remember to convert the balanced EMF from mV to V (\( 75\text{ mV} = 0.075\text{ V} \)) to avoid missing decimal units in the final equation.

Question 281. In the following potentiometer circuit AB is a uniform wire of length 1 m and resistance 10 \Omega. Calculate the (i) potential gradient along the wire, and (ii) balance length AO (= l ).
Answer:
We are given:
Wire length, \( L = 1\text{ m} \)
Wire resistance, \( R_{AB} = 10\,\Omega \)
Primary cell EMF, \( E_1 = 2\text{ V} \)
Series resistance, \( R = 15\,\Omega \)

(i) Potential gradient along the wire (\( K \)):
The total resistance of the primary circuit is:
\( R_{\text{total}} = 15\,\Omega + 10\,\Omega = 25\,\Omega \)
The current \( I \) in the primary circuit is:
\( I = \frac{E_1}{R_{\text{total}}} = \frac{2\text{ V}}{25\,\Omega} = 0.08\text{ A} \)
The potential difference across the potentiometer wire \( AB \) is:
\( V_{AB} = I \times R_{AB} = 0.08\text{ A} \times 10\,\Omega = 0.8\text{ V} \)
The potential gradient \( K \) is:
\( K = \frac{V_{AB}}{L} = \frac{0.8\text{ V}}{1\text{ m}} = 0.8\text{ V/m} \)

(ii) Balancing length \( AO = l \):
In the secondary circuit, the cell has \( E_2 = 1.5\text{ V} \), internal resistance \( r_2 = 1.2\,\Omega \), and is shunted by \( R = 0.3\,\Omega \).
The current in this closed secondary loop is:
\( I_{\text{sec}} = \frac{E_2}{R + r_2} = \frac{1.5}{0.3 + 1.2} = \frac{1.5}{1.5} = 1\text{ A} \)
The terminal potential difference \( V \) across this cell is:
\( V = I_{\text{sec}} \times R = 1\text{ A} \times 0.3\,\Omega = 0.3\text{ V} \)
According to the potentiometer principle:
\( V = K l \)

\( \implies 0.3 = 0.8 \times l \)

\( \implies l = \frac{0.3}{0.8} = 0.375\text{ m} = 37.5\text{ cm} \)
The potentiometer circuit is shown below: In simple words: (i) The potential gradient is found by dividing the total wire drop (0.8 V) by the length (1 m) to get 0.8 V/m. (ii) Since the experimental cell is closed with a resistor, its terminal voltage is 0.3 V. This balances at 37.5 cm on the wire.

Exam Tip: Always compute the terminal voltage of the cell when it is connected to a shunt resistor, as the potentiometer balances the terminal voltage \( V \), not the EMF \( E \).