JEE Physics — Current Electricity Deep Dive

Chapter Overview & Weightage

Current Electricity is JEE’s bread-and-butter chapter from Class 12 Physics. Problems range from simple Ohm’s law to nasty Wheatstone bridge variants and Kirchhoff’s law networks.

Year	JEE Main Qs	JEE Advanced Qs
2024	2	1
2023	3	2
2022	2	1
2021	2	1

Roughly 8-10% weightage in JEE Main. JEE Advanced tends to combine this chapter with capacitors or EMI for multi-step problems.

Key Concepts You Must Know

Ohm’s law: $V = IR$ , valid for ohmic conductors at constant temperature.
Resistivity: $\rho = RA/L$ , intrinsic property of material.
Drift velocity: $v_d = I/(neA) = eE\tau/m$ .
Current density: $\vec{J} = nev_d = \sigma\vec{E}$ (conductivity form of Ohm’s law).
Series and parallel: $R_{series} = R_1 + R_2 + \ldots$ ; $1/R_{parallel} = 1/R_1 + 1/R_2 + \ldots$ .
Kirchhoff’s laws: KCL (junction rule, charge conservation); KVL (loop rule, energy conservation).
Wheatstone bridge: balance condition $P/Q = R/S$ .
Internal resistance: $V_{terminal} = \varepsilon - Ir$ .
Power dissipation: $P = VI = I^2R = V^2/R$ .
Cells in series/parallel: emf and internal resistance combine differently.

Important Formulas

$\rho = \rho_0[1 + \alpha(T - T_0)]$

When to use: resistance of metallic conductor changes with temperature; $\alpha$ is the temperature coefficient.

$v_d = \frac{eE\tau}{m} = \frac{I}{neA}$

When to use: connecting microscopic (electrons) and macroscopic (current) descriptions.

$\frac{P}{Q} = \frac{R}{S} \implies I_g = 0$

When to use: meter bridge, slide wire problems, finding unknown resistance.

Maximum power delivered to load: $R_{load} = r$ (internal resistance).

$P_{max} = \frac{\varepsilon^2}{4r}$

When to use: optimising battery output power.

Solved Previous Year Questions

PYQ 1 (JEE Main 2024)

A wire of length $L$ and resistance $R$ is stretched to twice its original length. Find its new resistance.

Solution: Volume conserved: $A \cdot L = A' \cdot 2L \implies A' = A/2$ .

$R' = \rho \cdot \frac{2L}{A/2} = 4 \cdot \frac{\rho L}{A} = 4R$

Stretching by factor $n$ increases resistance by factor $n^2$ . Memorise.

PYQ 2 (JEE Main 2023)

A 12 V battery with internal resistance $0.5\,\Omega$ is connected to two resistors $4\,\Omega$ and $6\,\Omega$ in parallel. Find current through each resistor.

Solution: Parallel combination: $1/R = 1/4 + 1/6 = 5/12 \implies R = 2.4\,\Omega$ .

Total resistance with internal: $R_{total} = 2.4 + 0.5 = 2.9\,\Omega$ .

Total current: $I = 12/2.9 \approx 4.14$ A.

Voltage across parallel combination: $V = 4.14 \times 2.4 = 9.93$ V.

Current through $4\,\Omega$ : $9.93/4 \approx 2.48$ A.

Current through $6\,\Omega$ : $9.93/6 \approx 1.66$ A.

PYQ 3 (JEE Advanced 2022)

In a meter bridge, the null point is found at 40 cm from the left when a $5\,\Omega$ resistor is in the left gap. Find the resistance in the right gap.

Solution: Bridge balance: $P/Q = l_1/l_2 = 40/60 = 2/3$ .

$\frac{5}{R} = \frac{2}{3} \implies R = \frac{15}{2} = 7.5\,\Omega$

Difficulty Distribution

Easy (35%): Series/parallel combinations, Ohm’s law applications, basic power calculations.
Medium (45%): Kirchhoff’s laws on 2-loop networks, Wheatstone bridge, internal resistance.
Hard (20%): Symmetry-based circuit analysis, infinite ladder networks, combined R-C transient questions.

Expert Strategy

For complex networks, look for symmetry first. Equipotential nodes can be merged or split to simplify the topology dramatically. Cube of resistors and infinite ladder problems are solved entirely by symmetry.

For Kirchhoff’s law problems with 3+ loops, set up matrix form: $RI = V$ where $R$ is the resistance matrix, $I$ is the unknown current vector, $V$ is the EMF vector. Cramer’s rule or Gaussian elimination gives all currents in one shot.

The “stretched wire” problem is a JEE classic. If a wire is stretched to $n$ times its length, resistance becomes $n^2 R$ . If compressed to $1/n$ length, resistance becomes $R/n^2$ . Volume conservation is the key.

Common Traps

Trap 1: Treating EMF as terminal voltage. EMF is the “force” the battery generates internally; terminal voltage = EMF − $Ir$ . When current flows, terminal voltage is always less than EMF.

Trap 2: Power in series vs parallel resistors. In series, more resistance → more power. In parallel, more resistance → less power. JEE asks “which bulb glows brighter” — answer depends on circuit configuration.

Trap 3: Sign convention in KVL. When traversing a battery from $-$ to $+$ , EMF is $+$ . Going across a resistor in the direction of current, voltage drops by $IR$ . Mixing these up gives wrong currents.

JEE Advanced often combines current electricity with capacitors. Remember: in steady state, no current flows through a capacitor branch, so treat that branch as broken when finding currents elsewhere. The capacitor voltage equals the voltage across whatever branch it’s in parallel with.