Class 12 — Current Electricity

Chapter Overview & Weightage

Current Electricity is one of the highest-weightage chapters in CBSE Class 12 Physics — typically 8-10 marks out of 70. The chapter blends conceptual depth (drift velocity, internal resistance) with numerical practice (Kirchhoff’s laws, Wheatstone bridge, potentiometer).

For JEE/NEET aspirants, this chapter is doubly important — about 5-7 marks in JEE Main and 4 marks in NEET every year.

The chapter typically includes one 5-mark long answer (Wheatstone bridge derivation, potentiometer principle, or Kirchhoff’s laws application) plus 2-3 mark short answers.

Key Concepts You Must Know

Prioritized by exam frequency:

Ohm’s law and resistivity — $V = IR$ , $\rho = RA/L$ , dependence on temperature.
Drift velocity and current density — $I = nAev_d$ , link to mobility.
Internal resistance and EMF — $V = E - Ir$ , terminal voltage in circuits.
Kirchhoff’s laws — junction rule (conservation of charge), loop rule (conservation of energy).
Wheatstone bridge — derivation of balance condition, sensitivity.
Potentiometer — comparing EMFs, measuring internal resistance.
Combination of resistors and cells — series, parallel, mixed.

Important Formulas

$V = IR$

Use when: any simple circuit problem.

$R = \rho \frac{L}{A}$

Use when: comparing resistors of different geometry or material.

$I = nAev_d$

Use when: relating macroscopic current to microscopic charge motion. $n$ is electron density, $A$ is area, $e$ is electron charge, $v_d$ is drift velocity.

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

Use when: bridge circuit with known three resistors and one unknown.

$V = E - Ir$

Use when: terminal voltage less than EMF due to internal resistance.

Solved Previous Year Questions

PYQ 1 (CBSE 2023)

A wire of resistance $R$ is stretched to double its length. Find its new resistance.

Solution: Volume conservation: $A_1 L_1 = A_2 L_2$ , so $A_2 = A_1/2$ when $L_2 = 2L_1$ . New resistance $R' = \rho L_2/A_2 = \rho(2L)/(A/2) = 4\rho L/A = 4R$ .

PYQ 2 (CBSE 2022)

Two cells of EMF $1.5 \, \text{V}$ each and internal resistance $0.5 \, \Omega$ each are connected in parallel and the combination is connected to an external resistance of $2 \, \Omega$ . Find the current through the external resistor.

Solution: Equivalent EMF $= 1.5 \, \text{V}$ (same EMF, parallel). Equivalent internal resistance $= 0.5/2 = 0.25 \, \Omega$ . Total resistance $= 2 + 0.25 = 2.25 \, \Omega$ . Current $I = 1.5/2.25 = 2/3 \, \text{A} \approx 0.67 \, \text{A}$ .

PYQ 3 (CBSE 2024)

State the principle of a potentiometer. Why is it preferred over a voltmeter for measuring EMF?

Solution: Principle: potential drop across a uniform wire is directly proportional to its length, when current flowing through it is constant. Preferred because at balance, no current flows through the cell being measured — so it gives true EMF, not terminal voltage. A voltmeter draws current and gives $V = E - Ir$ (less than EMF).

Difficulty Distribution

Easy (1-2 marks): Definitions, simple Ohm’s law, formula recall — ~30%
Medium (3 marks): Combinations of resistors, Kirchhoff applications — ~50%
Hard (5 marks): Wheatstone derivation, potentiometer applications, multi-cell circuits — ~20%

Expert Strategy

Strategy 1: Master the Wheatstone derivation cold. This is a recurring 3-5 mark question. Practice writing it in under 5 minutes with a clean diagram.

Strategy 2: Always draw the circuit before applying Kirchhoff. Mark currents with arrows, choose loop directions, and stick to them throughout the problem.

Strategy 3: For drift velocity numericals, the trick is unit conversion. $n$ is given in $/m^3$ , area in $m^2$ , but current and electron charge are in everyday units. Plug in carefully.

The “stretched wire” trick (PYQ 1 above) appears every alternate year. Memorise: stretching by factor $k$ multiplies resistance by $k^2$ . This single fact handles a 2-mark question in seconds.

Common Traps

Trap 1: Forgetting that internal resistance is in SERIES with external circuit. The total current is $I = E/(R + r)$ , not $E/R$ .

Trap 2: Computing parallel resistance as the sum. Always use $1/R_p = 1/R_1 + 1/R_2$ . For two resistors specifically: $R_p = R_1 R_2/(R_1 + R_2)$ .

Trap 3: Sign errors in Kirchhoff’s loop rule. Pick a direction (say clockwise) and stick to it; voltage drops are negative going through resistors in direction of assumed current.

Trap 4: Mixing up EMF and terminal voltage. EMF is the cell’s “natural” voltage with no current; terminal voltage drops when current flows due to internal resistance.

This chapter rewards consistent practice. Aim for 90%+ accuracy on numericals — they are pure plug-and-chug once the circuit is sketched cleanly.