Class 12 Physics · NCERT Chapter 3

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Current Electricity Class 12 Physics Notes

A complete, exam-ready pass on current electricity — electric current, current density and drift velocity, Ohm's law, resistance and resistivity and their temperature dependence, series–parallel combinations, Kirchhoff's junction and loop rules, the Wheatstone bridge and the metre bridge, emf versus terminal voltage and internal resistance, ammeter–voltmeter–galvanometer conversion, and the potentiometer with its two classic measurements (emf comparison and internal resistance). Every topic ends with mark-wise question banks matching the CBSE pattern, followed by the chapter's full NCERT exercises (3.1–3.23) and two free CBSE-pattern test PDFs. No signup required.

Chapter:Current ElectricityClass:Class 12Subject:PhysicsBoard:CBSEExams:CBSE · JEE · NEET

15 Key Formulas6 Solved Examples40 Practice MCQs

DWritten byDeep Narayan

UpdatedSeptember 2026

Key Concept Summary

What is electric current in one line?

Electric current is the rate of flow of electric charge through a conductor, measured in amperes (A), defined as I = q/t for steady current or I = dq/dt in general.

Electric Current and Drift Velocity

Definition

Electric current

I = \frac{d q}{d t}

The rate of flow of charge. Conventional current direction is opposite to the electron flow direction. The SI unit is the ampere (1 A = 1 C/s).

Definition

Drift velocity

v_{d} = \frac{e E}{m} τ

The average velocity acquired by electrons in a conductor under an applied electric field. $τ$ is the average relaxation time between collisions. Relating it to current: $I = n A e v_{d}$ , where $n$ is electron density, $A$ the cross-section.

Current in terms of drift velocity

I = n e A v_{d}, v_{d} = \frac{I}{n e A}

MCQ Questions

Electric current through a conductor is the

The drift velocity of electrons in a conductor is proportional to the

If n is the number of free electrons per unit volume, the current density in a conductor is

Conventional current direction is taken as

1 Mark Questions

1
Define electric current. Write its SI unit.
2
What is current density? Write its SI unit.
3
Define drift velocity of electrons in a conductor.
4
Write the relation between current, drift velocity and the free-electron number density.

2 Mark Questions

1
A current of 5 A flows through a wire for 2 minutes. How much charge crosses any section of the wire?
2
A steady current of 3.0 A flows through a copper wire of area of cross-section 2.0 × 10⁻⁶ m². Find the drift velocity of the free electrons if their number density is 8.5 × 10²⁸ m⁻³.
3
Distinguish between the drift speed and the thermal speed of free electrons in a conductor. Which is the larger at room temperature?
4
How does the drift velocity of electrons in a metallic conductor change when (i) the temperature is increased and (ii) the potential difference across the conductor is doubled?

3 Mark Questions

1
The number density of free electrons in a copper conductor 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 is 2.0 × 10⁻⁶ m² and it carries a current of 3.0 A.
2
Show that the current density in a conductor is j = n e v_d, and hence derive the relation I = n A e v_d.
3
A wire of cross-sectional area 1.0 × 10⁻⁶ m² carries a current of 2.0 A. If the free-electron number density is 8.5 × 10²⁸ m⁻³, find the time taken by an electron to drift through a 1.0 cm length of the wire.

Ohm's Law and Resistance

Definition

Ohm's law

V = I R

At constant temperature, the current through a conductor is directly proportional to the potential difference across it, where R is the resistance in ohms (Ω). Not all materials obey it — semiconductors and electrolytes do not.

$Resistivity:$
$ρ = \frac{R A}{L}$
— a material property independent of dimensions.
$Conductivity:$
$σ = \frac{1}{ρ}$
.
Resistance of metals increases with temperature; that of semiconductors decreases.
Resistors in series:
$R = R_{1} + R_{2} + \dots$
.
Resistors in parallel:
$\frac{1}{R} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + \dots$
.

Resistance and conductivity

R = ρ \frac{L}{A}, σ = \frac{1}{ρ}

MCQ Questions

Ohm's law asserts that

The resistance of a conductor of resistivity ρ, length L and area of cross-section A is

When the temperature of a metallic conductor is raised, its resistance

Which element does NOT obey Ohm's law?

1 Mark Questions

1
State Ohm's law.
2
Write the formula for resistivity of a material and state its SI unit.
3
How does the resistance of a metal change when its temperature rises?
4
Define one ohm.

2 Mark Questions

1
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. Find the resistivity of the material.
2
At room temperature (27.0 °C) the resistance of a heating element is 100 Ω. Find the temperature of the element when its resistance becomes 117 Ω, given the temperature coefficient of the material is 1.70 × 10⁻⁴ °C⁻¹.
3
A silver wire has a resistance of 2.1 Ω at 27.5 °C and 2.7 Ω at 100 °C. Determine the temperature coefficient of resistivity of silver.
4
Distinguish between resistance and resistivity of a conductor.

3 Mark Questions

1
Derive the relation R = ρL/A for a conducting wire in terms of the microscopic quantities m, n, e and τ, and state the physical assumptions involved.
2
A wire of resistance 5 Ω is stretched so that its length becomes double. Find its new resistance, assuming the volume and resistivity remain unchanged.
3
Explain why the resistance of a metal increases while that of a semiconductor (e.g. carbon) decreases with temperature. Support your answer with the concept of relaxation time τ and drift velocity.

Combination of Resistances

Definition

Resistors in series

R_{e q} = R_{1} + R_{2} + R_{3}

The same current flows through every resistor in a series combination; the total potential drop is the sum of the individual drops V = V₁ + V₂ + V₃. The equivalent resistance is therefore larger than the largest individual resistance and equals the algebraic sum of all the resistances.

Definition

Resistors in parallel

\frac{1}{R _{e q}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + \frac{1}{R _{3}}

Every resistor in a parallel combination has the same potential difference across it; the total current is the sum of the branch currents. The reciprocal of the equivalent resistance is the sum of the reciprocals, and the equivalent resistance is smaller than the smallest individual resistance.

For n equal resistors of value R: series gives nR, parallel gives R/n. The maximum resistance (all in series) and the minimum (all in parallel) are therefore in the ratio n² : 1 — a favourite one-liner in CBSE and JEE papers.

Definition

Combination formulas

series R = R_{1} + R_{2}, parallel \frac{1}{R} = \frac{1}{R _{1}} + \frac{1}{R _{2}}

Two resistors in series carry the same current and divide the voltage proportionally to their resistances; two in parallel share the same voltage and divide the current inversely to their resistances. Identical resistors in a network can often be reduced by symmetry before applying the series–parallel laws.

MCQ Questions

Resistors R₁, R₂, R₃ connected in series give an equivalent resistance

Two resistors of 4 Ω and 6 Ω connected in parallel give an equivalent resistance

To obtain the maximum resistance from n identical resistors each of resistance R, they must be connected

The ratio of the maximum to the minimum resistance obtainable from n identical resistors each of resistance R is

1 Mark Questions

1
Write the formula for the equivalent resistance of resistors connected (i) in series and (ii) in parallel.
2
Why is the equivalent resistance of a parallel combination smaller than the smallest individual resistance?
3
State the ratio of maximum to minimum resistance obtainable with n identical resistors of resistance R.
4
What is the effective resistance of two identical resistors R connected in parallel?

2 Mark Questions

1
Three resistors 1 Ω, 2 Ω and 3 Ω are combined in series and connected to a battery of emf 12 V (negligible internal resistance). Find the total resistance and the potential drop across each resistor.
2
Three resistors 2 Ω, 4 Ω and 5 Ω are combined in parallel and connected to a battery of emf 20 V (negligible internal resistance). Determine the current through each resistor and the total current drawn.
3
Given resistors 1 Ω, 2 Ω and 3 Ω, combine them to obtain an equivalent resistance of (i) 11/3 Ω, (ii) 11/5 Ω, (iii) 6 Ω and (iv) 6/11 Ω.
4
A ladder network (as in Fig. 3.31(i)) has four identical loops: each loop is a 1 Ω + 1 Ω series combination in parallel with a 2 Ω + 2 Ω series combination. Find the total end-to-end resistance.

3 Mark Questions

1
Given n resistors each of resistance R, how will you combine them to get (i) the maximum and (ii) the minimum effective resistance? What is the ratio of the maximum to the minimum resistance?
2
Determine the equivalent resistance of the networks shown in Fig. 3.31: (i) the ladder of four loops built from 1 Ω and 2 Ω resistors, and (ii) five resistors each of resistance R in series.
3
A wire of resistance 12 Ω is bent to form a closed square. Find the resistance between two diagonally opposite corners of the square.

Kirchhoff's Laws

Definition

Kirchhoff's junction rule

The algebraic sum of currents meeting at a junction is zero. Equivalent to conservation of charge: current entering a junction equals current leaving it.

Definition

Kirchhoff's loop rule

The algebraic sum of the potential differences around any closed loop is zero. Equivalent to conservation of energy.

Sign convention

Traverse a loop in a chosen direction: a rise in potential is positive, a drop is negative. A battery gives +EMF if traversed from negative to positive terminal.

MCQ Questions

Kirchhoff's junction rule is based on the conservation of

Kirchhoff's loop rule is based on the conservation of

At a junction, currents I₁ and I₂ enter while I₃ leaves. The junction rule gives

A network with b branches and n junctions has this many independent loop equations

1 Mark Questions

1
State Kirchhoff's junction rule.
2
State Kirchhoff's loop rule.
3
Name the conservation law underlying each of Kirchhoff's two rules.
4
State the sign convention for applying the loop rule to resistors and batteries while traversing a loop.

2 Mark Questions

1
Using Kirchhoff's junction and loop rules, write three independent equations for the five branch currents of the bridge network of Fig. 3.30 (AB = 10 Ω, BC = 5 Ω, CD = 10 Ω, DA = 5 Ω, BD = 5 Ω, source 10 V in series with 10 Ω across A–C).
2
At a junction, a current of 5 A enters and the currents leaving are I₁ and I₂. If I₂ = 2 A, find I₁.
3
Distinguish between Kirchhoff's junction rule and loop rule, and state the conservation principle behind each.
4
Explain why Kirchhoff's junction rule remains valid even when the connecting wires at a junction are bent or reoriented.

3 Mark Questions

1
State Kirchhoff's two laws and justify each with a conservation principle. Under what conditions are they valid in a circuit?
2
A cell of emf 4 V and internal resistance 2 Ω is connected in parallel with a cell of emf 2 V and internal resistance 2 Ω, and a 6 Ω resistor is connected across the common terminals. Using Kirchhoff's laws find the current in the 6 Ω resistor.
3
A circuit contains two loops sharing a common branch: loop 1 has a 6 V battery (r = 1 Ω) and loop 2 a 3 V battery (r = 1 Ω), with a 3 Ω resistor in the shared branch. Use Kirchhoff's laws to find the current through the shared 3 Ω resistor.

Wheatstone Bridge and Meter Bridge

Definition

Wheatstone bridge condition

\frac{P}{Q} = \frac{R}{S}

When the four resistances P, Q, R, S are arranged in a bridge and the galvanometer shows zero deflection (balanced), the ratio condition holds — no current flows through the galvanometer.

Meter bridge

A meter bridge is a practical Wheatstone bridge. In balance:

\frac{R}{S} = \frac{l}{100 - l}

, where

l

is the balancing length in cm. The closer the null point to the middle, the more accurate the measurement.

MCQ Questions

A Wheatstone bridge is balanced when

At balance, the galvanometer in a Wheatstone bridge

In a metre bridge, if the balance point is l cm from end A, then

The connections between resistors in a Wheatstone or metre bridge are made of thick copper strips to

1 Mark Questions

1
State the balance condition of a Wheatstone bridge.
2
What is the function of the galvanometer in a metre bridge experiment?
3
Why is a metre bridge also called a slide-wire bridge?
4
Why are the connecting strips of a metre bridge made of thick copper?

2 Mark Questions

1
In a metre bridge, the balance point is found to be at 39.5 cm from the end A when the resistor S in the right gap is 12.5 Ω. Determine the resistance R. Why are the connections between resistors in a Wheatstone or metre bridge made of thick copper strips?
2
Determine the balance point of the metre bridge of the question above if the resistors R and S are interchanged.
3
What happens if the galvanometer and the cell are interchanged at the balance point of a metre bridge? Would the galvanometer show any current?
4
The four arms of a Wheatstone bridge are PQ = 2 Ω, QR = 4 Ω, RS = 3 Ω and SP = 6 Ω. Is the bridge balanced? A 3 Ω resistor is now placed in parallel with QR — is the bridge still balanced?

3 Mark Questions

1
Describe, with a labelled diagram, how an unknown resistance is determined experimentally using a metre bridge.
2
Derive the balance condition P/Q = R/S of a Wheatstone bridge using Kirchhoff's laws.
3
In a metre bridge experiment the balance point shifts from 39.5 cm to 60.5 cm when R and S are interchanged. If S = 12.5 Ω, find R and show that the balance condition is unchanged when the galvanometer and cell are interchanged.

EMF, Internal Resistance and Cells

Definition

EMF and terminal voltage

V = ε - I r

The terminal voltage of a cell is the EMF minus the drop across its internal resistance $r$ . $I$ is the current drawn. For a battery of cells, connect in series to increase EMF and in parallel to reduce internal resistance.

Cell equations

V = ε - I r, ε = I (R + r)

MCQ Questions

The terminal voltage of a cell of emf ε and internal resistance r when it delivers current I is

The maximum current that can be drawn from a cell of emf ε and internal resistance r is

n identical cells each of emf ε joined in series give a total emf of

While a secondary cell is being charged, its terminal voltage is

1 Mark Questions

1
Define the emf of a cell.
2
Write the relation between emf ε, terminal voltage V, current I and internal resistance r.
3
When is the terminal voltage of a cell equal to its emf?
4
What is the difference between the emf and the terminal voltage of a cell?

2 Mark Questions

1
A battery of emf 10 V and internal resistance 3 Ω is connected to a resistor and delivers a current of 0.5 A. Find the resistance of the resistor and the terminal voltage of the battery.
2
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. Find the terminal voltage during charging and state the purpose of the series resistor.
3
Six lead-acid secondary cells each of emf 2.0 V and internal resistance 0.015 Ω are joined in series to supply a resistance of 8.5 Ω. Find the current drawn from the supply and the terminal voltage.
4
The storage battery of a car has an emf of 12 V and an internal resistance of 0.4 Ω. What is the maximum current that can be drawn from the battery?

3 Mark Questions

1
Derive the expression for the power delivered by a cell of emf ε and internal resistance r to an external resistance R. For what value of R is the power maximum?
2
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
Explain why a low-voltage supply that must give high currents needs a very low internal resistance, and why a high-tension (HT) supply of 6 kV must have a very large internal resistance.

Ammeter, Voltmeter and Galvanometer

Definition

Galvanometer

τ = N I A B

A moving-coil galvanometer is a sensitive device that detects tiny currents (of the order of 10⁻³ A or less) and shows both the magnitude and direction of current. A radial magnetic field keeps the deflecting torque proportional to the current, so the scale is uniform. Its resistance G is typically a few tens of ohms.

Definition

Ammeter (galvanometer + low shunt)

S = \frac{I _{g} G}{I - I _{g}}

A galvanometer shunted by a small resistance S becomes an ammeter. Only a fraction I_g of the current I flows through the galvanometer; the rest bypasses it through the shunt. An ideal ammeter has zero resistance and is always connected in series with the branch whose current is being measured.

Definition

Voltmeter (galvanometer + high series resistance)

R = \frac{V}{I _{g}} - G

A galvanometer in series with a large resistance R becomes a voltmeter. The high resistance keeps the current through the galvanometer below I_g and limits the current drawn from the circuit, so the voltmeter reads the true potential difference. An ideal voltmeter has infinite resistance and is connected in parallel.

Exam favouite: an ammeter must have very low resistance (so it does not disturb the current) and is connected in series; a voltmeter must have very high resistance (so it draws negligible current) and is connected in parallel. Reverse them and the ammeter acts as a shunt while the voltmeter opens the circuit.

MCQ Questions

A galvanometer is used to

To convert a galvanometer into an ammeter, a

To convert a galvanometer into a voltmeter, a

An ammeter is connected in the circuit in series and a voltmeter in

1 Mark Questions

1
State the function of an ammeter and how it is connected in a circuit.
2
State the function of a voltmeter and how it is connected in a circuit.
3
Why must an ammeter have a very low resistance and a voltmeter a very high resistance?
4
Why does a shunt used with a galvanometer have a very small resistance?

2 Mark Questions

1
A galvanometer of resistance 20 Ω gives full-scale deflection for a current of 10 mA. Find the shunt resistance that converts it into an ammeter of range 0–5 A.
2
The same galvanometer (20 Ω, 10 mA) is to be converted into a voltmeter of range 0–50 V. Find the resistance that must be connected in series.
3
Explain why a galvanometer is converted into an ammeter with a low shunt in parallel rather than with a series resistance, and why a voltmeter uses a large series resistance rather than a shunt.
4
A galvanometer of resistance G gives full-scale deflection for a current I_g. It is shunted with S. Give the formula for the range of the resulting ammeter.

3 Mark Questions

1
Describe how a galvanometer is converted into an ammeter, and derive the expression for the shunt resistance S = I_g G/(I − I_g).
2
Describe how a galvanometer is converted into a voltmeter, and derive the expression for the required series resistance R = V/I_g − G.
3
A galvanometer of resistance 30 Ω gives full-scale deflection for a current of 2 mA. Find (i) the shunt needed to measure currents up to 2 A and (ii) the series resistance needed to measure voltages up to 2 V.

Potentiometer

A potentiometer measures EMF without drawing current from the cell, giving a more accurate result than a voltmeter. It compares EMFs because the potential gradient along a wire is uniform.

Comparing EMFs with a potentiometer

\frac{ε _{1}}{ε _{2}} = \frac{l _{1}}{l _{2}}

Why not a voltmeter?

A voltmeter draws some current and measures the terminal voltage, not the true EMF. The potentiometer draws no current at balance, so it measures the true EMF.

MCQ Questions

A potentiometer measures the emf of a cell because at balance it

For two cells of emfs ε₁ and ε₂ giving balance lengths l₁ and l₂ on the same potentiometer wire

For a cell of balance length l₁ (open circuit) and l₂ (with R shunted), the internal resistance is

For a balance point to exist on the potentiometer wire, the driver cell's emf must be

1 Mark Questions

1
Why does a potentiometer measure the emf of a cell more accurately than a voltmeter?
2
State the principle of a potentiometer.
3
What is a balance point in a potentiometer experiment?
4
Why is a very high resistance (e.g. 600 kΩ) connected in series with the galvanometer in a potentiometer circuit?

2 Mark Questions

1
In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35.0 cm of the wire. When the cell is replaced by another cell the balance point shifts to 63.0 cm. What is the emf of the second cell?
2
A 2.0 V potentiometer with a standard cell of emf 1.02 V balances at 67.3 cm. A cell of unknown emf ε balances at 82.3 cm. Find ε.
3
A 2.0 V potentiometer is used to determine the internal resistance of a 1.5 V cell: the balance point is 76.3 cm in open circuit and 64.8 cm when a resistor of 9.5 Ω is connected across the cell. Determine r.
4
What purpose does the high resistance of 600 kΩ serve in the potentiometer circuit of Exercise 3.22, and is the balance point affected by it?

3 Mark Questions

1
Describe, with a labelled circuit diagram, how a potentiometer is used to compare the emfs of two cells.
2
Describe, with a labelled circuit diagram, how a potentiometer is used to measure the internal resistance of a cell, and derive the formula r = (l₁/l₂ − 1) R.
3
A potentiometer wire of length 10 m and total resistance 10 Ω is driven by a 2.0 V cell of internal resistance 0.40 Ω. A standard cell of 1.02 V balances at 67.3 cm. (i) Find the potential gradient along the wire. (ii) Would the method work if the driver cell had an emf of 1.0 V instead of 2.0 V?

NCERT Exercise Questions

All NCERT questions for this chapter — NCERT Physics Part 1, Chapter 3 (Current Electricity). Questions keep their NCERT numbering so you can match them to the book: 3.1–3.13 are the Exercises and 3.14–3.23 are the Additional Exercises. Figures for the circuit-based questions (3.9, 3.10, 3.20(c), 3.21, 3.22, 3.23) are drawn below.

Exercises (NCERT 3.1–3.13) & Additional Exercises (3.14–3.23)

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.

Wheatstone-network bridge ABCD: resistors AB = 10 Ω, BC = 5 Ω, CD = 10 Ω, DA = 5 Ω and diagonal BD = 5 Ω, driven across A–C by a 10 V battery in series with a 10 Ω resistor (NCERT Fig 3.30, Exercise 3.9). — Fig 3.30 — bridge network ABCD driven across A–C by a 10 V battery in series with 10 Ω (NCERT Exercise 3.9).

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?

Metre bridge with resistor R in the left gap, S = 12.5 Ω in the right gap and the galvanometer jockey at the balance point 39.5 cm from end A (NCERT Fig 3.27, Exercise 3.10). — Fig 3.27 — metre bridge with balance point 39.5 cm from A (NCERT Exercise 3.10).

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 neutralise 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(Al) = 2.63 × 10⁻⁸ Ω m, ρ(Cu) = 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? (I (A): 0.2, 0.4, 0.6, 0.8, 1.0, 2.0 / V (V): 3.94, 7.87, 11.8, 15.7, 19.7, 39.4 — and I (A): 3.0, 4.0, 5.0, 6.0, 7.0, 8.0 / V (V): 59.2, 78.8, 98.6, 118.5, 138.2, 158.0)

3.18) (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?

3.19) (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 and (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 you 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.

(i) Ladder network of four identical loops — each loop a 1 Ω + 1 Ω series pair on the top rail and a 2 Ω + 2 Ω series pair on the bottom rail; (ii) five resistors R in series (NCERT Fig 3.31, Exercise 3.20(c)). — Fig 3.31 — (i) ladder network of 1 Ω and 2 Ω resistors, (ii) five resistors R in series (NCERT Exercise 3.20(c)).

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

Infinite ladder network of 1 Ω resistors — top rail, bottom rail and vertical rungs — fed from a 12 V supply with internal resistance 0.5 Ω (NCERT Fig 3.32, Exercise 3.21). — Fig 3.32 — infinite ladder network of 1 Ω resistors fed from a 12 V supply (NCERT Exercise 3.21).

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 ε and the balance point found similarly, turns out to be at 82.3 cm length of the wire. (a) What is the value of ε? (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 1.0 V instead of 2.0 V? (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?

Potentiometer with a driver cell of 2.0 V (r = 0.40 Ω) across the resistance wire AB; the secondary branch contains the unknown/standard cell, a 600 kΩ high resistance and the galvanometer with jockey (NCERT Fig 3.33, Exercise 3.22). — Fig 3.33 — potentiometer with a 2.0 V driver cell and 600 kΩ high resistance in the secondary branch (NCERT Exercise 3.22).

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.

Potentiometer for internal-resistance measurement: 2.0 V driver across wire AB; the 1.5 V cell with internal resistance r and an external 9.5 Ω resistor is connected through the galvanometer and jockey (NCERT Fig 3.34, Exercise 3.23). — Fig 3.34 — potentiometer used for measuring the internal resistance of a 1.5 V cell (NCERT Exercise 3.23).

Answers at a glance

1
3.1 — 30 A (I_max = ε/r = 12/0.4).
2
3.2 — Resistance = 17 Ω (R + r = ε/I = 10/0.5 = 20); terminal voltage V = ε − Ir = 10 − 0.5 × 3 = 8.5 V.
3
3.3 — (a) 6 Ω. (b) Potential drops across 1 Ω, 2 Ω, 3 Ω are 2 V, 4 V, 6 V (current 2 A).
4
3.4 — (a) 20/19 Ω. (b) Currents through 2 Ω, 4 Ω, 5 Ω are 10 A, 5 A, 4 A; total current = 19 A.
5
3.5 — 1027 °C (R = R₀(1 + αΔt): Δt = 17/(100 × 1.70 × 10⁻⁴) = 1000 °C).
6
3.6 — 2.0 × 10⁻⁷ Ω m (ρ = RA/L = 5 × 6.0 × 10⁻⁷/15).
7
3.7 — 0.0039 °C⁻¹ (α = ΔR/R₀Δt = 0.6/(2.1 × 72.5)).
8
3.8 — 867 °C (R from 230/3.2 to 230/2.8 Ω; Δt = (R/R₀ − 1)/α).
9
3.9 — Branch currents: AB = 4/17 A, BC = 6/17 A, CD = −4/17 A, AD = 6/17 A, BD = −2/17 A; total current = 10/17 A.
10
3.10 — (a) R = 8.2 Ω (R/S = l/(100 − l) = 39.5/60.5 times 12.5). Thick copper strips keep the connection resistance (unaccounted in the bridge formula) negligible. (b) 60.5 cm from A. (c) The galvanometer shows no current — the bridge stays balanced.
11
3.11 — 11.5 V while charging (charging current = (120 − 8)/(0.5 + 15.5) = 7 A; V = ε + Ir = 8 + 7 × 0.5). The series resistor limits the charging current — in its absence the current would be dangerously high.
12
3.12 — 2.25 V (ε₂ = ε₁ l₂/l₁ = 1.25 × 63/35).
13
3.13 — 2.7 × 10⁴ s ≈ 7.5 h (v_d = I/(nAe) ≈ 1.1 × 10⁻⁴ m/s; t = 3.0/v_d).
14
3.14 — ≈ 283 s (Q = σ × 4πR²; t = Q/I with I = 1800 A). An estimate: it ignores the Earth's curvature effects
15
3.15 — (a) I = 1.4 A, V = 11.9 V (R_tot = 8.5 + 6 × 0.015). (b) 0.005 A; impossible — a starter motor needs ~100 A for a few seconds.
16
3.16 — Aluminium wire is lighter: mass ratio Cu : Al = (1.72/2.63) × (8.9/2.7) ≈ 2.2, so aluminium is preferred for overhead power cables.
17
3.17 — V/I is constant to a high accuracy: Ohm's law is obeyed, and the resistivity of manganin is nearly independent of temperature.
18
3.18 — (a) Only the current (steady); current density, electric field and drift speed vary inversely with the cross-section. (b) No — vacuum diodes and semiconductor diodes are non-ohmic. (c) I_max = ε/r. (d) To keep the short-circuit current within safe limits.
19
3.19 — (a) greater, (b) lower, (c) nearly independent of, (d) 10²².
20
3.20 — (a) (i) all in series → nR; (ii) all in parallel → R/n; ratio = n². (b) (i) 1 Ω ∥ 2 Ω in series with 3 Ω; (ii) 2 Ω ∥ 3 Ω in series with 1 Ω; (iii) all in series; (iv) all in parallel. (c) (i) 16/3 Ω; (ii) 5R.
21
3.21 — Equivalent resistance X of the infinite network satisfies 2 + X/(X + 1) = X, giving X = (1 + √3) Ω ≈ 2.73 Ω; current drawn = 12/(0.5 + 2.73) ≈ 3.7 A.
22
3.22 — (a) ε = 1.25 V (ε = 1.02 × 82.3/67.3). (b) It reduces the current through the galvanometer when the jockey is far from the balance point. (c) No. (d) No — if the measured emf exceeds the driver emf there is no balance point. (e) Unsuitable for a few mV (balance point hugging end A, large percentage error); put a suitable resistor in series with AB to shrink the gradient.
23
3.23 — r = 1.7 Ω (r = (l₁/l₂ − 1) R = (76.3/64.8 − 1) × 9.5 = 1.685 ≈ 1.7 Ω).

CBSE Pattern Questions

How CBSE asks this chapter

Physics board paper structure: Section A (MCQ/VSA 1M), Section B (2M), Section C (3M), Section D (5M). Current Electricity typically carries ~9 marks: 1-mark MCQs on Kirchhoff's laws or resistance combinations, a 2-mark numerical on meter bridge or Wheatstone bridge, a 3-mark derivation of the drift velocity relation, and a 5-mark question on the potentiometer or cells in series/parallel.

MCQ Questions

The resistance of a conductor depends on its length l and area of cross-section A as

The ratio of the drift velocity of electrons to the applied electric field is called

The colour bands on a resistor signify

In the metre bridge, the unknown resistance X balances with a resistance R when

1 Mark Questions

1
State Ohm's law.
2
Define drift velocity and write its SI unit.
3
State Kirchhoff's junction rule and loop rule.
4
Define the temperature coefficient of resistance.

2 Mark Questions

1
Three resistors of 2 Ω, 3 Ω and 6 Ω are connected in parallel. Find the equivalent resistance. (Ans: 1/R = 1/2 + 1/3 + 1/6 = 1, so R = 1 Ω)
2
A wire of resistance 5 Ω is drawn out to three times its length. Find its new resistance. (Ans: R′ = n²R = 45 Ω)
3
A copper wire of cross-section 1 mm² carries a current of 2 A. If the electron density is 8 × 10²⁸ m⁻³, find the drift velocity. (Ans: vd = I/(neA) ≈ 1.6 × 10⁻⁴ m/s)

3 Mark Questions

1
Derive the relation I = neAvd and hence obtain Ohm's law in terms of drift velocity and relaxation time.
2
State and explain Kirchhoff's laws of electrical circuits with reference to the conservation of charge and energy.
3
Describe the metre bridge. Derive the balance condition (unknown X from known R).

5 Mark Questions

1
Explain the principle, construction and working of a potentiometer, and how it is used to (a) compare the emfs of two cells and (b) measure the internal resistance of a cell.
2
Derive the expressions for the current, terminal voltage and energy balance when cells are connected in series and in parallel, and give the conditions for maximum current.

PDF: Assignments & Tests

Free downloadable PDFs for Current Electricity — two topic-wise assignments covering every concept of the chapter, and two full CBSE-pattern tests (each 90 minutes, 40 marks) mirroring the board paper. No signup required.

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Practice path: complete Assignment 1 first (topic-by-topic), then Assignment 2 (mixed difficulty with figure-based NCERT questions), then attempt the timed tests.

Current Electricity Assignment 1 — Topic-wise Questions

Covers all eight topics: electric current and drift velocity, Ohm's law and resistance, combination of resistances, Kirchhoff's laws, Wheatstone and metre bridge, emf and internal resistance, ammeter/voltmeter/galvanometer, and the potentiometer. Arranged topic by topic.

current-electricity-class-12-notes-assignment-1.pdf

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Current Electricity Assignment 2 — Topic-wise Questions (Higher Difficulty)

Second set with multi-topic integration problems, JEE-style numericals, and figure-based NCERT questions (Exercises 3.9, 3.10, 3.20(c), 3.21, 3.22, 3.23). Higher difficulty than Assignment 1. Answer key included.

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Current Electricity Test 1 — CBSE Pattern (90 min, 40 marks)

Full CBSE-pattern paper: Section A (10 MCQ/VSA × 1M), Section B (4 × 2M), Section C (4 × 3M), Section D (2 × 5M). Complete marking scheme included.

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Current Electricity Test 2 — CBSE Pattern (90 min, 40 marks)

Second 40-mark CBSE-pattern paper with fresh questions and a full solution key.

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Quick Revision

Key formulas at a glance

Memorise these equations — direct application numericals and derivations in CBSE & JEE frequently hinge on these.

Current

I = \frac{d q}{d t}

Drift velocity

v_{d} = \frac{I}{n e A}

Ohm's law

V = I R

Resistance

R = ρ \frac{L}{A}

Temperature dependence

R = R_{0} (1 + α Δ t)

Resistors in series

R = R_{1} + R_{2} + R_{3}

Resistors in parallel

\frac{1}{R} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + \frac{1}{R _{3}}

Kirchhoff's loop rule

\sum ε = \sum I R

Wheatstone balance

\frac{P}{Q} = \frac{R}{S}

Metre bridge

\frac{R}{S} = \frac{l}{100 - l}

Terminal voltage

V = ε - I r

Ammeter shunt

S = \frac{I _{g} G}{I - I _{g}}

Voltmeter series R

R = \frac{V}{I _{g}} - G

Potentiometer: emf comparison

\frac{ε _{1}}{ε _{2}} = \frac{l _{1}}{l _{2}}

Potentiometer: internal resistance

r = (\frac{l _{1}}{l _{2}} - 1) R

Exam Strategy

How this chapter is asked

High-yield question patterns observed across CBSE boards, JEE Main & Advanced, and NEET.

I = n e A v_d is the key link between current and drift velocity.
Metals: resistance rises with temperature; semiconductors: falls.
Series adds resistances (R = R₁ + R₂); parallel gives a value smaller than the smallest (1/R = 1/R₁ + 1/R₂).
Kirchhoff: junction rule ↔ conservation of charge; loop rule ↔ conservation of energy.
Wheatstone balance: P/Q = R/S; metre bridge: R/S = l/(100 − l).
Terminal voltage: V = ε − Ir (delivering), V = ε + Ir (charging).
An ammeter goes in series with a low resistance; a voltmeter goes in parallel with a high resistance.
Potentiometer measures true emf (zero current at balance): ε₁/ε₂ = l₁/l₂.
Internal resistance from a potentiometer: r = (l₁/l₂ − 1) R.
Metre-bridge balance is unchanged when the galvanometer and the cell are interchanged.

Practice Numericals

Step-by-step worked examples

Standard exam-level problems solved with clear conceptual breakdowns.

Charge from a steady current

Problem StatementA wire carries a steady current of 2.0 A. How much charge passes through a cross-section of the wire in 1 minute?

Step-by-Step Solution

1I = q/t ⇒ q = I t = 2.0 × 60 = 120 C.
2The same charge crosses every section of the wire since the current is steady.

Final Answer

q = 120 C

Drift velocity of electrons

Problem StatementA current of 3.0 A flows through a copper wire of cross-section 2.0 × 10⁻⁶ m². Taking the free-electron number density as 8.5 × 10²⁸ m⁻³ and e = 1.6 × 10⁻¹⁹ C, find the drift velocity.

Step-by-Step Solution

1v_d = I/(n A e) = 3.0/(8.5 × 10²⁸ × 2.0 × 10⁻⁶ × 1.6 × 10⁻¹⁹).
2v_d = 3.0/2.72 × 10⁴ ≈ 1.1 × 10⁻⁴ m s⁻¹.

Final Answer

v_{d} \approx 1.1 \times 1 0^{- 4} m s^{- 1}

Metre bridge — unknown resistance

Problem StatementIn a metre bridge the balance point is at 39.5 cm from end A when the resistor S in the right gap is 12.5 Ω. Find the unknown resistance R.

Step-by-Step Solution

1At balance, R/S = l/(100 − l) = 39.5/(60.5).
2R = 12.5 × 39.5/60.5 = 8.16 ≈ 8.2 Ω.

Final Answer

R \approx 8.2 Ω

Internal resistance of a cell by potentiometer

Problem StatementA 2.0 V potentiometer balances a 1.5 V cell at 76.3 cm in open circuit and at 64.8 cm when a resistor of 9.5 Ω is connected across the cell. Find the internal resistance.

Step-by-Step Solution

1r = (l₁/l₂ − 1) R, where l₁ = 76.3 cm and l₂ = 64.8 cm.
2r = (76.3/64.8 − 1) × 9.5 = (1.1775 − 1) × 9.5 = 1.685 ≈ 1.7 Ω.

Final Answer

r \approx 1.7 Ω

Resistance of a wire bent into a square

Problem StatementA wire of resistance 12 Ω is bent to form a closed square. Find the resistance between two diagonally opposite corners.

Step-by-Step Solution

1Each side of the square has resistance 12/4 = 3 Ω.
2Between diagonal corners the two paths each contain two sides in series: 3 + 3 = 6 Ω.
3The two 6 Ω paths are in parallel: R = 6×6/(6+6) = 3 Ω.

Final Answer

R = 3 Ω

Converting a galvanometer into an ammeter

Problem StatementA galvanometer of resistance 20 Ω gives full-scale deflection for a current of 10 mA. Find the shunt needed to convert it into an ammeter of range 0–5 A.

Step-by-Step Solution

1S = I_g G/(I − I_g) = (0.01 × 20)/(5 − 0.01).
2S = 0.2/4.99 ≈ 0.04 Ω.

Final Answer

S \approx 0.04 Ω

FAQ

Frequently asked questions

What is drift velocity?

It is the small average velocity acquired by free electrons under an applied electric field, given by v_d = eEτ/m. Typical values are only a few millimetres per second.

Why is a potentiometer more accurate than a voltmeter?

At balance, a potentiometer draws no current from the cell, so it measures the true EMF. A voltmeter draws a small current and measures only the terminal voltage.

What is the Wheatstone bridge condition?

The bridge balances when P/Q = R/S, at which point no current flows through the galvanometer connecting the two ratio arms.

Does Ohm's law apply to all materials?

No. Semiconductors, electrolytes, diodes and certain substances show non-linear current-voltage behaviour and do not obey Ohm's law.

Why does a potentiometer measure the emf of a cell more accurately than a voltmeter?

At the balance point no current is drawn from the cell being measured, so there is no internal-resistance drop (V = ε). A voltmeter always draws a small current through the cell's internal resistance, so it reads the terminal voltage V = ε − Ir, which is slightly less than the true emf. That is why the potentiometer is the standard way to compare emfs and to measure the internal resistance.

When is the terminal voltage of a cell greater than its emf?

While the cell is being charged. The charging current flows through the cell against its emf, and the potential difference across the terminals becomes V = ε + Ir, which exceeds the emf by the internal drop. The same formula V = ε − Ir works: during charging the current entering the cell is taken as negative, so V = ε − (−I)r = ε + Ir.

Why does resistance increase with temperature for a metal but decrease for a semiconductor?

In a metal the free-electron density is fixed, and heating increases the amplitude of lattice vibrations, shortening the relaxation time τ and so the drift velocity for a given field; resistivity ρ = m/(n e² τ) rises. In a semiconductor the electron (and hole) number density n itself rises sharply with temperature, and this rise outweighs the fall in τ, so resistivity decreases.

What happens to the null point if the galvanometer and cell of a metre bridge are interchanged at balance?

Nothing measurable — the bridge remains balanced and the galvanometer still shows zero deflection. The balance condition P/Q = R/S depends only on the four arm resistances, not on where the cell and the galvanometer are placed; interchanging them leaves the steady-state current distribution through the galvanometer branch unchanged at zero.

Why is a high resistance of 600 kΩ put in series with the standard cell in a potentiometer, and does it affect the balance point?

Before the null point is located, the jockey may sit at a point of the wire whose potential is far from the cell's emf, so the cell would push a potentially large current through the galvanometer. The 600 kΩ high resistance caps that current to a safe few-milliamp level; it is shorted close to the balance point so that when the null is found the cell genuinely draws no current. The balance point itself is unaffected because at balance the galvanometer current is zero anyway.

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