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Current Electricity: concepts, formulas & solved PYQ-style examples

Current electricity is a high-scoring chapter: the questions are predictable once you can redraw a circuit cleanly. Most mistakes come from misreading which resistors are in series and which are in parallel.

Redraw the circuit, label junction potentials, then apply Kirchhoff’s laws or the shortcut that fits (balanced bridge, symmetry).

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Key concepts

Ohm’s law and resistivity

R = ρL/A; stretching a wire changes both L and A.

EMF and internal resistance

Terminal voltage V = E − Ir while discharging; maximum power transfer when R = r.

Series and parallel

Same current in series, same voltage in parallel; the smallest parallel resistor dominates.

Kirchhoff’s laws

Junction rule (charge) and loop rule (energy) — solve any circuit.

Wheatstone bridge and meter bridge

Balanced when P/Q = R/S; the galvanometer arm then carries no current.

Formula sheet

Ohm’s law, resistivity

V=IR,R=ρLAV = IR,\quad R = \dfrac{\rho L}{A}

Cell with internal resistance

I=ER+r,V=E−IrI = \dfrac{E}{R + r},\quad V = E - Ir

Combinations

Rs=R1+R2,1Rp=1R1+1R2R_s = R_1 + R_2,\quad \dfrac{1}{R_p} = \dfrac{1}{R_1} + \dfrac{1}{R_2}

Power

P=VI=I2R=V2RP = VI = I^2R = \dfrac{V^2}{R}

Balanced Wheatstone bridge

PQ=RS\dfrac{P}{Q} = \dfrac{R}{S}

What PYQs keep asking

Patterns that repeat across previous-year JEE and NEET papers for this chapter.

  • Terminal voltage and power with internal resistance of a cell.
  • Equivalent resistance of networks that hide a balanced Wheatstone bridge.
  • Current through one branch of a series-parallel circuit.
  • Meter bridge and potentiometer balance-length problems.

Solved examples

Example 1

A cell of EMF 12 V and internal resistance 1 Ω is connected to a 5 Ω resistor. Find the current, terminal voltage and power in the resistor.

  1. 1Current: I=ER+r=125+1=2I = \dfrac{E}{R + r} = \dfrac{12}{5 + 1} = 2 A.
  2. 2Terminal voltage: V=E−Ir=12−2(1)=10V = E - Ir = 12 - 2(1) = 10 V.
  3. 3Power in the resistor: P=I2R=22×5=20P = I^2R = 2^2 \times 5 = 20 W.
Answer: I = 2 A, V = 10 V, P = 20 W
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Example 2

A 6 Ω and a 3 Ω resistor are in parallel, and the pair is in series with 4 Ω across a 12 V supply. Find the current in the 6 Ω resistor.

  1. 1Parallel pair: Rp=6×36+3=2 ΩR_p = \dfrac{6 \times 3}{6 + 3} = 2\ \Omega; total R=2+4=6 ΩR = 2 + 4 = 6\ \Omega.
  2. 2Total current: I=12/6=2I = 12/6 = 2 A, so the voltage across the pair is 2×2=42 \times 2 = 4 V.
  3. 3Current in 6 Ω: 4/64/6 A.
Answer: 23\tfrac{2}{3} A ≈ 0.67 A
Watch a whiteboard video solution of this example →

Common traps

  • Using the EMF instead of the terminal voltage across the external resistor.
  • Missing a wire that shorts out a resistor completely.
  • Taking the larger resistor in a parallel pair as carrying more current — it carries less.

Practise with

ElectrostaticsModern PhysicsRay OpticsCircuit builder simulation380 Current Electricity practice questionsVideo solutions

FAQ

What is the difference between EMF and terminal voltage?

EMF is the energy given per unit charge by the cell; terminal voltage is what the external circuit gets, E − Ir while the cell supplies current.

How do I spot a balanced Wheatstone bridge?

If the ratio of resistances on one side equals the ratio on the other (P/Q = R/S), the middle resistor carries no current and can be removed.

Is current electricity easy scoring for NEET?

Yes — most NEET questions are direct applications of Ohm’s law, combinations, Kirchhoff’s laws and the meter bridge.

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