Given below are two statements : One is labeled as Assertion (A) and the other is labeled as Reason (R). (Magnetism) Assertion (A) : In an uniform magnetic field, speed and energy remains the same for a moving charged particle. Reason (R) : Moving charged particle experiences magnetic force perpendicular to its direction of motion.

Both (A) and (R) are true and (R) is the correct explanation of (A).

Both (A) and (R) are true but (R) is NOT the correct explanation of (A).

NCERT Class 12 Physics Chapter 3 Solutions – Free PDF for Current Electricity

Q: 3.1 The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is 0.4 Ω, what is the maximum current that can be drawn from the battery?

3.1 EMF of the battery, E = 12 V Internal resistance of the battery, r = 0.4 Ω Let the maximum current drawn = I According to OHM’s law, E = Ir So I = Er = 120.4 amp = 30 amp Therefore, the maximum current can be drawn is 30 ampere.

Q: 3.2 A battery of emf 10 V and internal resistance 3 Ω is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?

3.2 EMF of the battery = 10 V Internal resistance of the battery, r = 3 Ω Current in the circuit, I = 0.5 A Let the resistance of the resistor be R According to Ohm’s law I = E (R+r) R + r = EI or R = EI - r = 100.5 - 3 = 17 Ω Terminal voltage of the battery when the circuit is closed is given by V = IR = 0.5 ×17=8.5V Therefore the resistance is 17 Ω and the terminal voltage is 8.5 V

Q: 3.3 (a) Three resistors 1 Ω, 2 Ω, and 3 Ω are combined in series. What is the total resistance of the combination? (b) If the combination is connected to a battery of emf 12 V and negligible internal resistance, obtain the potential drop across each resistor.

3.3 (a) The equivalent resistance of the resistor in series is given by R = 1 + 2 + 3 = 6 Ω (b) From Ohm’s law, I = VR we get I = 126 = 2 A. Potential drop across 1 Ω resistor = I ×R = 2 ×1 = 2 V Potential drop across 2 Ω resistor = I ×R = 2 ×2 = 4 V Potential drop across 3 Ω resistor = I ×R = 2 ×3 = 6 V

Q: 3.4 (a) Three resistors 2 Ω, 4 Ω and 5 Ω are combined in parallel. What is the total resistance of the combination? (b) If the combination is connected to a battery of emf 20 V and negligible internal resistance, determine the current through each resistor, and the total current drawn from the battery.

3.4 (a) Let R1 = 2 Ω, R2 = 4 Ω, R3 = 5 Ω If the equivalent resistance is R, then 1R = 1R1 + 1R2 + 1R3 = 12 + 14 + 15 = 10+5+420 = 1920 R = 2019 = 1.05 Ω (b) The EMF of the battery = 20 V Current through R1, I1 = VR1, = 202 = 10 A Current through R2, I2 = VR2, = 204 = 5A Current through R3, I3 = VR3, = 205 = 4 A Total current I = I1 + I2 + I3 = 10 + 5 + 4 = 19 A

Q: 3.5 At room temperature (27.0 °C) the resistance of a heating element is 100 Ω. What is the temperature of the element if the resistance is found to be 117 Ω, given that the temperature coefficient of the material of the resistor is 1.70 × 10–4 °C–1.

3.5 Let T be the room temperature and R be the resistance at room temperature and T1 be the required temperature and R1 be the resistance at that temperature and α be the coefficient of resistor. We have: T = 27 ?, R = 100 Ω, T1 = ?, R1 = 117 Ω, α = 1.70 ×10-4 ° C-1 We know the relation of α can be given as α = R1-RR(T1-T) = 117-100100(T1-27) = 1.70 ×10-4 or ( T1 - 27) = 117-100100×1.70×10-4 = 1000 T1 = 1000 +27 = 1027 ?

Q: Is electric current class 12 Physics considered a tough chapter to understand?

No, in fact, it is one of the easiest chapter of class 12 Physics. Other chapters which are considered comparatively easy are Ray Optics and Electric Charges and Fields.

Q: Define current electricity in class 12 Physics.

In simple words, current electricity can be defined as the electric charge continuously moving from one place to another along a pathway. It is measured in amperes (A). Electric current is needed for electrical devices to work.

Q: According to current and electricity class 12, what is the SI unit of current?

The current's SI unit is ampere. The symbol for ampere is A. The term ampere is named after the French physicist André-Marie Ampère.

Q: How many types of electricity in current electricity for class 12?

There are two types of electricity - Static and Current electricity. The electric charges buildup on a material's surface is called the static electricity. The continuous flow of electric charge is termed as the current electricity. Current electricity is of two types - Alternating Current (AC) and Direct Current (DC). In AC, the charge direction reverses periodically and in DC, charge flows in one direction.

Q: What is a galvanometer according to current and electricity class 12 Physics chapter?

According to this chapter, a galvanometer is used to find and measure the small electric currents in a circuit. The principle that works in a galvanometer is the electromagnetic induction.

Updated on Sep 19, 2025 12:27 IST

By Pallavi Pathak, Assistant Manager Content

CBSE Class 12 Physics Chapter 3 is all about Electric Current and its behaviour. This chapter explains the behaviour of current when passed through the circuit under various conditions like resistance, resistivity, temperature, circular loops, etc. Knowing these topics is essential for both theoretical and practical knowledge. This chapter also covers topics like the role of power and energy in circuits and tools to analyze electrical circuits. Therefore, practising these topics through NCERT Class 12 Physics solutions will help to solve the problems and boost confidence.

Physics Class 12 Current Electricity has at least 8-10% weightage in the CBSE Class 12 exam. Topics like Ohm’s law, Drift velocity, resistivity vs temperature, combination of cells, Kirchhoff's rules, and Wheatstone bridge are very important from the viewpoint of the board exam as well as entrance tests like JEE Main, MHT CET, WJEE, etc.

The subject matter experts at Shiksha have taken utmost care to prepare the NCERT Class 12 Physics Ch 3 solutions. In Class 12 Physics Current Electricity NCERT solution by Shiksha, students will find solutions for Class 12 Physics textbook exercise problems, multiple choice questions, and JEE-level problems. At Shiksha, we aim to provide the best study material to the students so that they can excel at any competitive exam.

Also Refer: NCERT Exemplar Solutions for Class 12th Chapter 3

Table of content

Class 12 Current Electricity Chapter NCERT Solution PDF: Download PDF for Free
Class 12 Physics Current Electricity NCERT Exercise Solutions
NCERT Class 12 Physics Chapter 3 Solutions - Important Topics
CBSE Class 12 Physics Chapter 3 Current Electricity Important Formulas & Concepts
Why Refer to NCERT Physics Solutions PDF?
Why are Shiksha NCERT Solutions Reliable?
NCERT Physics Chapter 3 Current Electricity – FAQs

Class 12 Current Electricity Chapter NCERT Solution PDF: Download PDF for Free

Find below the link to download the free PDF.

Class 12 Current Electricity NCERT Solution PDF: Download Free PDF

Class 12 Physics Current Electricity NCERT Exercise Solutions

Q.3.1 The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is 0.4 Ω, what is the maximum current that can be drawn from the battery?

Ans.3.1 EMF of the battery, E = 12 V

Internal resistance of the battery, r = 0.4 Ω

Let the maximum current drawn = I

According to OHM’s law, E = Ir

So I = $\frac{E}{r}$ = $\frac{12}{0.4}$ amp = 30 amp

Therefore, the maximum current can be drawn is 30 ampere.

Q.3.2 A battery of emf 10 V and internal resistance 3 Ω is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?

Ans.3.2 EMF of the battery = 10 V

Internal resistance of the battery, r = 3 Ω

Current in the circuit, I = 0.5 A

Let the resistance of the resistor be R

According to Ohm’s law

I = $\frac{E}{(R + r)}$

R + r = $\frac{E}{I}$ or R = $\frac{E}{I}$ - r = $\frac{10}{0.5}$ - 3 = 17 Ω

Terminal voltage of the battery when the circuit is closed is given by

V = IR = 0.5 $\times 17 = 8.5 V$

Therefore the resistance is 17 Ω and the terminal voltage is 8.5 V

Q.3.3 (a) Three resistors 1 Ω, 2 Ω, and 3 Ω are combined in series. What is the total resistance of the combination?

(b) If the combination is connected to a battery of emf 12 V and negligible internal resistance, obtain the potential drop across each resistor.

Ans.3.3 c

Q.3.4 (a) Three resistors 2 Ω, 4 Ω and 5 Ω are combined in parallel. What is the total resistance of the combination?

(b) If the combination is connected to a battery of emf 20 V and negligible internal resistance, determine the current through each resistor, and the total current drawn from the battery.

Ans.3.4 Let $R_{1}$ = 2 Ω, $R_{2}$ = 4 Ω, $R_{3}$ = 5 Ω

If the equivalent resistance is R, then $\frac{1}{R}$ = $\frac{1}{R_{1}}$ + $\frac{1}{R_{2}}$ + $\frac{1}{R_{3}}$ = $\frac{1}{2}$ + $\frac{1}{4}$ + $\frac{1}{5}$ = $\frac{10 + 5 + 4}{20}$ = $\frac{19}{20}$

R = $\frac{20}{19}$ = 1.05 Ω

The EMF of the battery = 20 V

Current through $R_{1,}$ $I_{1}$ = $\frac{V}{R_{1,}}$ = $\frac{20}{2}$ = 10 A

Current through $R_{2,}$ $I_{2}$ = $\frac{V}{R_{2,}}$ = $\frac{20}{4}$ = 5A

Current through $R_{3,}$ $I_{3}$ = $\frac{V}{R_{3,}}$ = $\frac{20}{5}$ = 4 A

Total current I = $I_{1}$ + $I_{2}$ + $I_{3}$ = 10 + 5 + 4 = 19 A

Commonly asked questions

3.1 The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is 0.4 Ω, what is the maximum current that can be drawn from the battery?

3.2 A battery of emf 10 V and internal resistance 3 Ω is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?

3.3 (a) Three resistors 1 Ω, 2 Ω, and 3 Ω are combined in series. What is the total resistance of the combination?

(b) If the combination is connected to a battery of emf 12 V and negligible internal resistance, obtain the potential drop across each resistor.

3.4 (a) Three resistors 2 Ω, 4 Ω and 5 Ω are combined in parallel. What is the total resistance of the combination?

(b) If the combination is connected to a battery of emf 20 V and negligible internal resistance, determine the current through each resistor, and the total current drawn from the battery.

3.5 At room temperature (27.0 °C) the resistance of a heating element is 100 Ω. What is the temperature of the element if the resistance is found to be 117 Ω, given that the temperature coefficient of the material of the resistor is 1.70 × 10^–4 °C^–1.

3.6 A negligibly small current is passed through a wire of length 15 m and uniform cross-section 6.0 × 10^–7 m², and its resistance is measured to be 5.0 Ω. What is the resistivity of the material at the temperature of the experiment?

3.7 A silver wire has a resistance of 2.1 Ω at 27.5 °C, and a resistance of 2.7 Ω at 100 °C. Determine the temperature coefficient of resistivity of silver.

3.8 A heating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds to a steady value of 2.8 A. What is the steady temperature of the heating element if the room temperature is 27.0 °C? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is 1.70 × 10^–4 °C^–1.

3.9 Determine the current in each branch of the network shown in Fig. 3.30:

3.10 (a) In a meter bridge [Fig. 3.27], the balance point is found to be at 39.5 cm from the end A, when the resistor S is of 12.5 Ω. Determine the resistance of R. Why are the connections between resistors in a Wheatstone or meter bridge made of thick copper strips?

(b) Determine the balance point of the bridge above if R and S are interchanged.

(c) What happens if the galvanometer and cell are interchanged at the balance point of the bridge? Would the galvanometer show any current?

3.11 A storage battery of emf 8.0 V and internal resistance 0.5 Ω is being charged by a 120 V dc supply using a series resistor of 15.5 Ω. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit?

3.12 In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35.0 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63.0 cm, what is the emf of the second cell?

3.13 The number density of free electrons in a copper conductor estimated in Example 3.1 is 8.5 × 10²⁸ m^–3. How long does an electron take to drift from one end of a wire 3.0 m long to its other end? The area of cross-section of the wire is 2.0 × 10^–6 m² and it is carrying a current of 3.0 A.

3.14 The earth’s surface has a negative surface charge density of 10^–9 C m^–2. The potential difference of 400 kV between the top of the atmosphere and the surface results (due to the low conductivity of the lower atmosphere) in a current of only 1800 A over the entire globe. If there were no mechanism of sustaining atmospheric electric field, how much time (roughly) would be required to neutralize the earth’s surface? (This never happens in practice because there is a mechanism to replenish electric charges, namely the continual thunderstorms and lightning in different parts of the globe). (Radius of earth = 6.37 × 10⁶ m.)

3.15 (a) Six lead-acid type of secondary cells each of emf 2.0 V and internal resistance 0.015 Ω are joined in series to provide a supply to a resistance of 8.5 Ω. What are the current drawn from the supply and its terminal voltage?

(b) A secondary cell after long use has an emf of 1.9 V and a large internal resistance of 380 Ω. What maximum current can be drawn from the cell? Could the cell drive the starting motor of a car?

3.16 Two wires of equal length, one of aluminium and the other of copper have the same resistance. Which of the two wires is lighter? Hence explain why aluminium wires are preferred for overhead power cables. ( $ρ_{A l}$ = 2.63 × 10^–8 Ωm, $ρ_{C u}$ = 1.72 × 10^–8 Ω m, Relative density of Al = 2.7, of Cu = 8.9.)

3.17 What conclusion can you draw from the following observations on a resistor made of alloy manganin?

Current (A)	Voltage (V)	Current (A)	Voltage (V)
0.2	3.94	3.0	59.2
0.4	7.87	4.0	78.8
0.6	11.8	5.0	98.6
0.8	15.7	6.0	118.5
1.0	19.7	7.0	138.2
2.0	39.4	8.0	158.0

Q.3.18 Answer the following questions:

(a) A steady current flows in a metallic conductor of non-uniform cross-section. Which of these quantities is constant along the conductor: current, current density, electric field, drift speed?

(b) Is Ohm’s law universally applicable for all conducting elements? If not, give examples of elements which do not obey Ohm’s law.

(c) A low voltage supply from which one needs high currents must have very low internal resistance. Why?

(d) A high tension (HT) supply of, say, 6 kV must have a very large internal resistance. Why?

Q.3.19 Choose the correct alternative:

(a) Alloys of metals usually have (greater/less) resistivity than that of their constituent metals.

(b) Alloys usually have much (lower/higher) temperature coefficients of resistance than pure metals.

(c) The resistivity of the alloy manganin is nearly independent of/ increases rapidly with increase of temperature.

(d) The resistivity of a typical insulator (e.g., amber) is greater than that of a metal by a factor of the order of (10²²/10²³).

3.20 (a) Given n resistors each of resistance R, how will you combine them to get the (i) maximum (ii) minimum effective resistance? What is the ratio of the maximum to minimum resistance?

(b) Given the resistances of 1 Ω, 2 Ω, 3 Ω, how will be combine them to get an equivalent resistance of (i) (11/3) Ω (ii) (11/5) Ω, (iii) 6 Ω, (iv) (6/11) Ω?

(c) Determine the equivalent resistance of networks shown in Fig. 3.31.

3.21 Determine the current drawn from a 12V supply with internal resistance 0.5 Ω by the infinite network shown in Fig. 3.32. Each resistor has 1Ω resistance.

3.22 Figure 3.33 shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Ω maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V (for very moderate currents up to a few mA) gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of 600 kΩ is put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf $?$ and the balance point found similarly, turns out to be at 82.3 cm length of the wire.

(a) What is the value e ?

(b) What purpose does the high resistance of 600 kΩ have?

(c) Is the balance point affected by this high resistance?

(d) Would the method work in the above situation if the driver cell of the potentiometer had an emf of 1.0V instead of 2.0V?

(e) Would the circuit work well for determining an extremely small emf, say of the order of a few mV (such as the typical emf of a thermo-couple)? If not, how will you modify the circuit?

3.23 Figure 3.34 shows a 2.0 V potentiometer used for the determination of internal resistance of a 1.5 V cell. The balance point of the cell in open circuit is 76.3 cm. When a resistor of 9.5 Ω is used in the external circuit of the cell, the balance point shifts to 64.8 cm length of the potentiometer wire. Determine the internal resistance of the cell.

CBSE Class 12 Physics Chapter 3 Current Electricity Important Formulas & Concepts

Refer to the table below for important formulas and concepts of Chapter 3 Current Electricity:

Concept	Formula / Key Point
Electric Current (I)	$I = \frac{q}{t}$
Ohm's Law	$V = I R$
Resistance of a conductor	$R = ρ \frac{l}{A}$
Electrical conductivity	$σ = \frac{1}{ρ}$
Electrical Power	$P = I V = I^{2} R = \frac{V^{2}}{R}$
Work / Energy	$W = I V t$
Power loss in transmission line	$P_{c} = I^{2} R_{c} = \frac{P^{2} R_{c}}{V^{2}}$
Cell equation	$V = ε - I r$
Current in the circuit	$I = \frac{ε}{R + r}$
Cell in series	$Equivalent emf: ε_{eq} = ε_{1} + ε_{2} + \dots$ $Equivalent resistance: r_{eq} = r_{1} + r_{2} + \dots$
Cell in parallel	$\frac{1}{r_{eq}} = \frac{1}{r_{1}} + \frac{1}{r_{2}} + \dots$ $\frac{ε_{eq}}{r_{eq}} = \frac{ε_{1}}{r_{1}} + \frac{ε_{2}}{r_{2}} + \dots$
Drift velocity	$v_{d} = \frac{I}{n e A} = \frac{e E τ}{m}$
Current density	$j = \frac{I}{A} = σ E = n q v_{d}$
Mobility	$μ = \frac{v_{d}}{E} = \frac{e τ}{m}$
Kirchhoff's Law	$Junction Rule: \sum I_{in} = \sum I_{out}$ $Loop Rule: \sum Δ V = 0 in a closed loop$
Wheatstone Rule	Balance Condition: $\frac{R_{1}}{R_{2}} = \frac{R_{3}}{R_{4}}$

Why Refer to NCERT Physics Solutions PDF?

Online NCERT solutions, a dependable academic reference to prepare for the CBSE board examination. The solutions PDF is accessible anytime, anywhere. The answers provided to the NCERT problem are well-structured and provide conceptual clarity.

Why are Shiksha NCERT Solutions Reliable?

The best subject matter experts in the industry prepare Shiksha NCERT Solutions pdf. Each experts have sound knowledge of the subject. While preparing the class 12 physics chapter 3 ncert solutions, the experts adhere to the CBSE standard and explain all the problems in a structured format. Also, here at Shiksha, we provide detailed notes of current electricity class 12.

NCERT Physics Chapter 3 Current Electricity – FAQs

The following are the frequently asked questions on Class 12 Physics Chapter 3 NCERT Solutions - Current Electricity:

Commonly asked questions

Is electric current class 12 Physics considered a tough chapter to understand?

Define current electricity in class 12 Physics.

According to current and electricity class 12, what is the SI unit of current?

How many types of electricity in current electricity for class 12?

What is a galvanometer according to current and electricity class 12 Physics chapter?

Physics Ncert Solutions Class 12th Exam

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