CBSE Class 12 Physics Exemplar Chapter 3: Current Electricity NCERT Solutions

Reviewed by NCERT Help Editorial Team · Updated on 29 Jul 2026

NCERT Solutions PDF Class 12 PDF

This resource provides detailed NCERT Solutions for Class 12 Physics Exemplar, Chapter 3: Current Electricity. It covers essential topics including the nature of current in a conductor, the combination of batteries in parallel, accurate resistance measurement using a meter bridge, and the working principles of a potentiometer for comparing emfs. The solutions explain the underlying physics concepts and provide step-by-step derivations for the given multiple-choice questions. These solutions are designed to help students grasp the nuances of current electricity, improve their problem-solving skills, and prepare effectively for their board examinations by offering clear explanations and accurate answers.

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Board	CBSE
Class	Class 12
Subject	Physics Exemplar
Session	2026
Language	English
Type	NCERT Solutions
Chapter	Chapter 3

Chapter summary

Chapter 3 of the CBSE Class 12 Physics Exemplar focuses on Current Electricity. The NCERT Solutions provided here address key concepts such as current density, the behavior of current in a circular wire, and the equivalent emf of parallel battery combinations. It also delves into practical applications like using a meter bridge for resistance measurement and a potentiometer for comparing emfs, emphasizing accuracy and proper experimental setup. The solutions offer clear explanations for multiple-choice questions, aiding students in understanding these fundamental principles.

Learning outcomes

Understand the concept of current density and its relation to electric field.
Analyze the equivalent emf of batteries connected in parallel.
Determine the optimal conditions for accurate resistance measurement using a meter bridge.
Explain the principle and application of a potentiometer for comparing emfs.
Calculate resistance based on the geometry of a conductor.

Topics covered

Paper topics

Current Density
Electric Field in Conductors
Parallel Combination of Batteries
Equivalent EMF
Meter Bridge
Resistance Measurement
Accuracy in Measurement
Potentiometer
Comparison of EMFs
Resistance Calculation
Geometry and Resistance

Important topics

Parallel Combination of Batteries
Meter Bridge Accuracy
Potentiometer Principle
Current Density
Resistance Calculation based on Geometry

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Questions and Solutions

Multiple Choice Questions (MCQs) - 1

1. Consider a current carrying wire in the shape of a circle. The agent that is essentially responsible for the change in the direction of current density (j) along the wire, while the current (I) remains unaffected, is:

Solution: The current density $j$ is a vector quantity representing the current flow per unit area, directed along the electric field $E$ within the conductor. For a current-carrying wire, charges tend to accumulate on the surface, creating an electric field. This electric field is responsible for directing the charge carriers, thus defining the direction of current density $j$. While the current $I$ represents the total charge flow across the entire cross-section and remains constant along the wire, the direction of $j$ can change if the cross-sectional area's orientation changes relative to the wire's path (like in a circular wire). Therefore, the electric field produced by the charges on the surface of the wire is the agent responsible for the directional changes in current density. The relationship is given by $j = \sigma E$, where $\sigma$ is the conductivity.

The correct option is (b).

Multiple Choice Questions (MCQs) - 2

2. Two batteries of emf $\varepsilon_1$ and $\varepsilon_2$ (where $\varepsilon_2 > \varepsilon_1$) and internal resistances $r_1$ and $r_2$ respectively are connected in parallel. Which of the following statements is true regarding the equivalent emf $\varepsilon_{eq}$?

Solution: When two batteries are connected in parallel, the equivalent emf $\varepsilon_{eq}$ and equivalent internal resistance $r_{eq}$ are given by the formulas:

$\varepsilon_{\text{eq}} = \frac{\varepsilon_1 r_2 + \varepsilon_2 r_1}{r_1 + r_2}$

$r_{\text{eq}} = \frac{r_1 r_2}{r_1 + r_2}$

Given that $\varepsilon_2 > \varepsilon_1$, let's analyze the expression for $\varepsilon_{eq}$. We can rewrite it as:

$\varepsilon_{\text{eq}} = \varepsilon_1 \left( \frac{r_2}{r_1 + r_2} \right) + \varepsilon_2 \left( \frac{r_1}{r_1 + r_2} \right)$

Since $\frac{r_2}{r_1 + r_2}$ and $\frac{r_1}{r_1 + r_2}$ are both positive fractions that sum to 1, $\varepsilon_{eq}$ is a weighted average of $\varepsilon_1$ and $\varepsilon_2$. Because $\varepsilon_2 > \varepsilon_1$, the equivalent emf $\varepsilon_{eq}$ will always lie between $\varepsilon_1$ and $\varepsilon_2$. It cannot be equal to $\varepsilon_1$ or $\varepsilon_2$ unless one of the emfs is zero or one of the resistances is infinite (which is not the case here). Therefore, $\varepsilon_1 < \varepsilon_{eq} < \varepsilon_2$.

The correct option is (a).

Multiple Choice Questions (MCQs) - 3

3. A resistance R is to be measured using a meter bridge. A student chooses the standard resistance S to be $100 \Omega$ and finds the null point at $l_1 = 2.9$ cm. To improve the accuracy of the measurement, which of the following actions would be most useful?

Solution: In a meter bridge experiment, the resistance R is related to the standard resistance S and the lengths $l_1$ and $l_2$ (where $l_2 = 100 - l_1$) by the formula:

$\frac{R}{S} = \frac{l_1}{l_2} = \frac{l_1}{100 - l_1}$

The percentage error in the measurement of R is approximately given by $\frac{\Delta R}{R} \times 100 \approx \left( \frac{\Delta l_1}{l_1} + \frac{\Delta l_2}{l_2} + \frac{\Delta S}{S} \right) \times 100$, where $\Delta l_1$, $\Delta l_2$, and $\Delta S$ are the errors in measuring lengths and standard resistance. For maximum accuracy, the balance point $l_1$ should be as close to the middle of the bridge (50 cm) as possible. This minimizes the error contribution from the length measurements. In this case, $l_1 = 2.9$ cm, so $l_2 = 100 - 2.9 = 97.1$ cm. The ratio $\frac{R}{S} = \frac{2.9}{97.1} \approx \frac{1}{33.5}$. This means R is much smaller than S. To bring the balance point closer to 50 cm, we need to adjust S. If we want the ratio $\frac{R}{S}$ to be closer to 1, we need to decrease S significantly. If we choose S = $3 \Omega$, the ratio $\frac{R}{S}$ would be approximately $\frac{R}{3}$. If R is around $100 \Omega$ (as initially assumed with S=100), then $\frac{100}{3} \approx 33.3$, which would shift the balance point to $l_1 \approx \frac{33.3}{1+33.3} \times 100 \approx 97$ cm. This is also not ideal. However, the question implies we want to improve accuracy for measuring R. The current setup gives R ≈ $\frac{2.9}{97.1} \times 100 \approx 3 \Omega$. If we change S to $3 \Omega$, then $\frac{R}{S} = \frac{R}{3} = \frac{2.9}{97.1}$ (using the original R value derived from the first measurement). This implies R ≈ $3 \times \frac{2.9}{97.1} \approx 0.09 \Omega$. This is not correct. The goal is to make $l_1$ near 50 cm. The current ratio $R/S = 2.9/97.1$. To make $l_1$ near 50, we need $R/S \approx 1$. Since $R \approx 3 \Omega$ (from $R = S \times \frac{l_1}{100-l_1} = 100 \times \frac{2.9}{97.1}$), we should choose S to be close to R. Thus, changing S to $3 \Omega$ would bring the ratio $R/S$ closer to 1, leading to a balance point nearer to 50 cm, thus improving accuracy.

The correct option is (c).

Multiple Choice Questions (MCQs) - 4

4. Two cells with approximate emfs of 5 V and 10 V are to be accurately compared using a potentiometer of length 400 cm. Which of the following statements is correct regarding the setup?

Solution: A potentiometer is used to measure or compare emfs of cells. For accurate measurement, the potential drop across the entire length of the potentiometer wire must be greater than the maximum emf of the cell being measured. This ensures that a balance point can be found along the wire for any cell within that range. In this case, the maximum emf to be measured is approximately 10 V. Therefore, the potential drop across the 400 cm potentiometer wire must be slightly greater than 10 V. This can be achieved by adjusting the rheostat (variable resistance R) in the primary circuit. If the battery voltage is 15 V, we can adjust R such that the potential drop across the wire is, for example, 12 V. This would allow us to measure emfs up to 12 V. If the battery voltage were only 8 V, it would be insufficient to provide a potential drop greater than 10 V across the wire, making it impossible to measure the 10 V cell accurately.

The correct option is (b).

Multiple Choice Questions (MCQs) - 5

5. A metal rod of length 10 cm and a rectangular cross-section of 1 cm × 0.5 cm is connected to a battery across opposite faces. In which orientation will the resistance of the rod be maximum?

Solution: The resistance of a conductor is given by the formula $R = \rho \frac{L}{A}$, where $\rho$ is the resistivity of the material, L is the length of the conductor in the direction of current flow, and A is the cross-sectional area perpendicular to the current flow. The rod has dimensions: length = 10 cm, width = 1 cm, and thickness = 0.5 cm. We need to consider three possible orientations for connecting the battery across opposite faces: 1. Connected across the faces with dimensions 1 cm × 0.5 cm: In this case, the length of the rod in the direction of current flow is L = 10 cm, and the cross-sectional area is A = 1 cm × 0.5 cm = 0.5 cm².

$R_1 = \rho \frac{10 \text{ cm}}{0.5 \text{ cm}^2}$

2. Connected across the faces with dimensions 10 cm × 1 cm: In this case, the length of the rod in the direction of current flow is L = 0.5 cm, and the cross-sectional area is A = 10 cm × 1 cm = 10 cm².

$R_2 = \rho \frac{0.5 \text{ cm}}{10 \text{ cm}^2}$

3. Connected across the faces with dimensions 10 cm × 0.5 cm: In this case, the length of the rod in the direction of current flow is L = 1 cm, and the cross-sectional area is A = 10 cm × 0.5 cm = 5 cm².

$R_3 = \rho \frac{1 \text{ cm}}{5 \text{ cm}^2}$

To find the maximum resistance, we compare the values of $\frac{L}{A}$ for each case: 1. $\frac{L}{A} = \frac{10}{0.5} = 20$ cm⁻¹ 2. $\frac{L}{A} = \frac{0.5}{10} = 0.05$ cm⁻¹ 3. $\frac{L}{A} = \frac{1}{5} = 0.2$ cm⁻¹ The largest value of $\frac{L}{A}$ is 20 cm⁻¹, which occurs when the current flows along the 10 cm length and the cross-sectional area is 1 cm × 0.5 cm. This corresponds to connecting the battery across the 1 cm × 0.5 cm faces.

The correct option is (a).

Common mistakes

Choosing an inappropriate standard resistance (S) for meter bridge leading to inaccurate results.
Incorrectly setting up the potentiometer circuit, where the potential drop across the wire is less than the emf to be measured.
Misinterpreting the direction of current density versus current in a conductor.
Assuming simple addition for parallel battery emfs without considering internal resistances.

Revision tips

Focus on understanding the formula for equivalent emf when batteries are in parallel.
Review the conditions for accurate measurement using a meter bridge, especially the role of S and the balance point.
Practice problems involving potentiometers to ensure the potential drop is always greater than the emf being measured.
Visualize the direction of current density and its relation to the electric field in different conductor shapes.

Practice MCQs

Q1. What is the primary agent responsible for the change in the direction of current density (j) along a current-carrying wire, while the current (I) remains unaffected?

Q2. When two batteries with emfs $ $\varepsilon$ _1$ and $ $\varepsilon$ _2$ (where $ $\varepsilon$ _2 > $\varepsilon$ _1$) and internal resistances $ $r_1$ $ and $ $r_2$ $ are connected in parallel, what is the range of the equivalent emf $ $\varepsilon$ _{eq}$?

Q3. A student measures a resistance R using a meter bridge with a standard resistance $S = 100$ $ $\Omega$ $ and finds the null point at $l = 2.9 cm. To improve accuracy$ , what is the most useful change?

Q4. For accurately comparing emfs of two cells (approx. 5 V and 10 V) using a 400 cm potentiometer, what condition must the potentiometer's battery voltage satisfy?

Q5. A metal rod (10 cm length, 1 cm x 0.5 cm cross-section) is connected to a battery across opposite faces. When will the resistance be maximum?

Frequently asked questions

What is current density and how does it relate to current?

Current density (j) is the current per unit area perpendicular to the current flow. It's a vector quantity directed along the electric field (E) within the conductor, following the relation $j = \sigma E$, where $\sigma$ is conductivity. While current (I) is the total flow, current density describes the flow rate through a specific cross-section.

How are batteries connected in parallel for calculating equivalent emf?

When batteries with emfs $\varepsilon_1, \varepsilon_2$ and internal resistances $r_1, r_2$ are connected in parallel, the equivalent emf $\varepsilon_{eq}$ is given by $\varepsilon_{eq} = \frac{\varepsilon_2 r_1 + \varepsilon_1 r_2}{r_1 + r_2}$. This value lies between the individual emfs.

What is the key to improving accuracy when measuring resistance with a meter bridge?

To improve accuracy, the standard resistance (S) should be chosen such that the null point (balance point) occurs near the middle of the meter bridge wire (around 50 cm). This minimizes the percentage error in the measured resistance (R).

Why must the potential drop across a potentiometer wire be greater than the emf of the cell being measured?

A potentiometer measures emf by balancing it against the potential drop along the wire. If the potential drop is less than the cell's emf, no balance point can be found, making measurement impossible. The potential drop must exceed the emf to allow for a measurable balance length.

How does the geometry of a conductor affect its resistance?

Resistance (R) is directly proportional to the length (L) and inversely proportional to the cross-sectional area (A), given by $R = \rho \frac{L}{A}$, where $\rho$ is resistivity. Therefore, a longer, thinner conductor has higher resistance than a shorter, thicker one made of the same material.

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