JEE Main Physics · Chapter 14

Current Electricity

Drift velocity, Ohm’s law, cells, Kirchhoff’s laws, Wheatstone bridge, metre bridge, potentiometer and power.

NCERT Class 12JEE Main: 1–2 Qs per shift 10 min read

1. Chapter at a Glance

Current electricity deals with charges in motion through conductors. It covers electric current and drift velocity, Ohm's law, resistivity and its temperature dependence, combinations of resistors, cells (emf and internal resistance), Kirchhoff's laws, the Wheatstone bridge, metre bridge and potentiometer, and electrical power. It is a high-scoring JEE Main chapter with one or two questions every year.

What JEE asks most
  • Equivalent resistance of networks (including balanced bridges).
  • Kirchhoff's laws for circuits with more than one cell.
  • Cells in series/parallel; terminal voltage; maximum power transfer.
  • Metre bridge and potentiometer readings; power of bulbs.

2. Current, Drift Velocity and Ohm's Law

  • Current I = dQ/dt (ampere). Current density J = I/A.
  • Free electrons drift opposite to the field with a small average speed: I = neAvd, vd = eEτ/m. Typical vd ~ 10⁻⁴ m s⁻¹, yet signals travel almost at the speed of light.
  • Mobility μ = vd/E.
  • Ohm's law: V = IR (at constant temperature). R = ρl/A; conductivity σ = 1/ρ = ne²τ/m.
  • Stretching a wire to n times its length (volume fixed) makes R become n² times.
  • Temperature: R = R₀(1 + αΔT). Metals: α > 0; semiconductors: resistance falls with temperature. Alloys like nichrome and manganin have very small α.

3. Resistors and Cells

FeatureSeriesParallel
Equivalent resistanceR = R₁ + R₂ + …1/R = 1/R₁ + 1/R₂ + …
Same for allCurrentVoltage
DividesVoltage (∝ R)Current (∝ 1/R)
  • A cell of emf E and internal resistance r: I = E/(R + r); terminal voltage V = E − Ir (discharging) or E + Ir (charging).
  • n identical cells in series: emf nE, resistance nr. m cells in parallel: emf E, resistance r/m.
  • Two different cells in parallel: Eeq = (E₁/r₁ + E₂/r₂)/(1/r₁ + 1/r₂), req = r₁r₂/(r₁ + r₂).
  • Maximum power transfer: the external resistor receives maximum power when R = r; Pmax = E²/4r.

4. Kirchhoff's Laws and the Wheatstone Bridge

  • Junction rule (charge conservation): ΣI entering = ΣI leaving.
  • Loop rule (energy conservation): the algebraic sum of potential changes around any closed loop is zero.
  • Wheatstone bridge: balanced (no galvanometer current) when P/Q = R/S. The galvanometer arm can then be removed.
G PQRS balanced if P/Q = R/S
Wheatstone bridge: when P/Q = R/S the two middle junctions are at the same potential, so no current flows through the galvanometer.
  • Metre bridge (a Wheatstone bridge with a 1 m wire): X/R = l/(100 − l), with X in the left gap and balance length l cm from the left end.
  • Potentiometer: potential gradient k = Vwire/L. An emf balances at length l: E = kl. Comparing cells: E₁/E₂ = l₁/l₂. Internal resistance: r = R(l₁/l₂ − 1), where l₁ is the balance length with the cell open and l₂ with a resistor R across it. It draws no current at balance, so it measures true emf.

5. Electrical Energy and Power

P = VI = I²R = V²/R    H = I²Rt (Joule heating)
  • For bulbs rated P at V: R = V²/P. On a lower voltage V′: P′ = P(V′/V)².
  • Bulbs in series: the one with the smaller rated power (larger R) glows brighter. In parallel: the one with the larger rated power glows brighter.
  • 1 kWh (one "unit") = 3.6 × 10⁶ J.

6. Formula Master Table

QuantityFormula
Drift velocityI = neAvd
ResistanceR = ρl/A; R = R₀(1 + αΔT)
CellI = E/(R + r); V = E − Ir
Max powerR = r; Pmax = E²/4r
Wheatstone balanceP/Q = R/S
Metre bridgeX/R = l/(100 − l)
PotentiometerE₁/E₂ = l₁/l₂; r = R(l₁/l₂ − 1)
PowerP = VI = I²R = V²/R

7. Mind Map

Current Electricity
Basics
  • Drift velocity
  • Ohm's law, ρ
  • Temperature effect
Circuits
  • Series, parallel
  • Cells, internal r
  • Kirchhoff's laws
Instruments
  • Wheatstone bridge
  • Metre bridge
  • Potentiometer
Power
  • I²R heating
  • Bulbs
  • Max power transfer

8. Tips and Tricks

Shortcuts
  • n equal resistors: series nR, parallel R/n.
  • A wire of resistance R bent into a circle: between diametrically opposite points R/4.
  • Before solving a bridge, check whether it is balanced — if so, drop the middle resistor.
  • In Kirchhoff problems, assume current directions; a negative answer just means the real direction is opposite.
Mistakes students make
  • Using the rated power of a bulb when it runs on a different voltage.
  • Forgetting internal resistance when finding terminal voltage.
  • Assuming the brighter bulb in series is the higher-wattage one.

9. Important One-liners

  • Resistivity depends on material and temperature, not on the shape of the conductor.
  • Kirchhoff's junction rule expresses conservation of charge; the loop rule, conservation of energy.
  • A potentiometer measures emf accurately because it draws no current at balance.
  • The resistance of a metal increases with temperature.
  • Maximum power is delivered when the load equals the internal resistance.
  • Drift speed of electrons is very small, of the order of 10⁻⁴ m s⁻¹.

10. Solved Numericals: Average to Advanced

Five average-level and five advanced-level problems. Click to reveal each answer and its full working.

Average level

N1Average

A copper wire of cross-section 1 mm² carries 1.5 A. If n = 8.5 × 10²⁸ m⁻³, find the drift velocity.

Show answer & solution

Concept: vd = I/(neA).

  1. neA = 8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 10⁻⁶ = 1.36 × 10⁴.
  2. vd = 1.5/(1.36 × 10⁴) ≈ 1.1 × 10⁻⁴ m s⁻¹.

Answer: about 1.1 × 10⁻⁴ m s⁻¹

N2Average

A 10 Ω wire is stretched uniformly to twice its length. Find its new resistance.

Show answer & solution

Concept: volume is constant, so A halves when l doubles; R ∝ l/A ∝ l².

  1. R′ = R × (2)² = 4R.
  2. = 40 Ω.

Answer: 40 Ω

N3Average

A 12 V cell with internal resistance 1 Ω is connected to a 5 Ω resistor. Find the current, terminal voltage and power in the resistor. What load would receive maximum power, and how much?

Show answer & solution

Concept: I = E/(R + r); Pmax at R = r.

  1. I = 12/6 = 2 A.
  2. V = E − Ir = 12 − 2 = 10 V; P = I²R = 4 × 5 = 20 W.
  3. Maximum power when R = 1 Ω: Pmax = E²/4r = 144/4 = 36 W.

Answer: 2 A; 10 V; 20 W; 1 Ω gives 36 W

N4Average

A coil has resistance 50 Ω at 20 °C and temperature coefficient 0.004 °C⁻¹. Find its resistance at 70 °C.

Show answer & solution

Concept: R = R₀(1 + αΔT).

  1. ΔT = 50 °C.
  2. R = 50(1 + 0.004 × 50) = 50 × 1.2 = 60 Ω.

Answer: 60 Ω

N5Average

A bulb is rated 100 W, 220 V. Find its resistance and the power it consumes on a 110 V supply.

Show answer & solution

Concept: R = V²/P (assumed constant); P′ = V′²/R.

  1. R = 220²/100 = 484 Ω.
  2. P′ = 110²/484 = 12 100/484 = 25 W (one-quarter).

Answer: 484 Ω; 25 W

Advanced level

N6Advanced

Two cells — 6 V with 1 Ω and 3 V with 1 Ω — are connected in parallel (positive to positive) across a 2 Ω resistor. Find the current in the resistor and in each cell.

Show answer & solution

Concept: equivalent cell, then Kirchhoff to find branch currents.

  1. Eeq = (6/1 + 3/1)/(1 + 1) = 4.5 V; req = 0.5 Ω.
  2. IR = 4.5/(2 + 0.5) = 1.8 A; voltage across R = 3.6 V.
  3. 6 V cell: I₁ = (6 − 3.6)/1 = 2.4 A (discharging).
  4. 3 V cell: I₂ = (3 − 3.6)/1 = −0.6 A, i.e. 0.6 A flows into it (it is being charged).
  5. Check: 2.4 − 0.6 = 1.8 A ✓.

Answer: 1.8 A; 2.4 A out of the 6 V cell; 0.6 A into the 3 V cell

N7Advanced

In a metre bridge, an unknown resistance X is in the left gap and 30 Ω in the right gap. The balance point is at 40 cm from the left end. Find X. Where will the balance point be if X and 30 Ω are interchanged?

Show answer & solution

Concept: X/R = l/(100 − l).

  1. X/30 = 40/60 ⇒ X = 20 Ω.
  2. After interchange: 30/20 = l/(100 − l) ⇒ 3(100 − l) = 2l ⇒ l = 60 cm.

Answer: 20 Ω; 60 cm

N8Advanced

A 10 m potentiometer wire has a steady 2 V across it. (a) A cell balances at 4 m — find its emf. (b) Another cell balances at 400 cm on open circuit and at 320 cm when a 10 Ω resistor is connected across it. Find its internal resistance.

Show answer & solution

Concept: E = kl with k = V/L; r = R(l₁/l₂ − 1).

  1. (a) k = 2/10 = 0.2 V m⁻¹ ⇒ E = 0.2 × 4 = 0.8 V.
  2. (b) r = 10 × (400/320 − 1) = 10 × 0.25.
  3. r = 2.5 Ω.

Answer: 0.8 V; 2.5 Ω

N9Advanced

In a Wheatstone network, P = 10 Ω, Q = 20 Ω, R = 15 Ω and S = 30 Ω, with a 5 Ω resistor in the galvanometer arm. Find the equivalent resistance between the input corners.

Show answer & solution

Concept: check balance first; if P/Q = R/S, the middle arm carries no current and can be removed.

  1. P/Q = 10/20 = 0.5 and R/S = 15/30 = 0.5 ⇒ balanced.
  2. Remove the 5 Ω arm. Upper path: 10 + 20 = 30 Ω; lower path: 15 + 30 = 45 Ω.
  3. Req = 30 × 45/(30 + 45) = 1350/75 = 18 Ω.

Answer: 18 Ω

N10Advanced

A 1000 W electric kettle heats 1 kg of water from 20 °C to 100 °C. Assuming no heat loss (c = 4200 J kg⁻¹ K⁻¹), find the time taken. What is the resistance of its element on 220 V?

Show answer & solution

Concept: electrical energy Pt = mcΔT; R = V²/P.

  1. Heat needed = 1 × 4200 × 80 = 336 000 J.
  2. t = 336 000/1000 = 336 s = 5.6 min.
  3. R = 220²/1000 = 48.4 Ω.

Answer: 336 s; 48.4 Ω

11. PYQ-pattern Questions with Solutions

These questions are original and modelled on concepts frequently tested in JEE Main.

Q1Easy

The resistivity of a metal wire depends on:

  1. Its length
  2. Its area of cross-section
  3. Its material and temperature
  4. The current through it
Show solution

Resistivity is a property of the material at a given temperature; resistance depends on the dimensions.

Answer: (C)

Q2Medium

A 100 W and a 60 W bulb (both rated for 220 V) are connected in series to 220 V. Which glows brighter?

  1. 100 W
  2. 60 W
  3. Both equally
  4. Neither glows
Show solution

The 60 W bulb has the larger resistance (V²/P). In series, the same current flows, so I²R is larger for it.

Answer: (B)

Q3Numerical

Three 6 Ω resistors are connected in parallel. Find the equivalent resistance in ohms.

Show solution

R = 6/3 = 2 Ω.

Answer: 2

Q4Easy

Kirchhoff's junction rule is based on conservation of:

  1. Energy
  2. Momentum
  3. Charge
  4. Mass
Show solution

Charge cannot pile up at a junction, so current in = current out.

Answer: (C)

Q5Hard

n identical cells, each of emf E, are connected in series, but one cell is connected the wrong way round. The net emf is:

  1. nE
  2. (n − 1)E
  3. (n − 2)E
  4. (n − 1)E/2
Show solution

The reversed cell subtracts E instead of adding it: (n − 1)E − E = (n − 2)E.

Answer: (C)

Q6Numerical

A 10 Ω wire is bent into a circle. Find the resistance (in Ω) between two diametrically opposite points.

Show solution

Two 5 Ω halves in parallel: 5 × 5/10 = 2.5 Ω.

Answer: 2.5

12. Quick Revision Summary

Remember before the exam
  • I = neAvd; R = ρl/A; stretched wire R ∝ l².
  • Series R adds; parallel 1/R adds.
  • Cell: I = E/(R + r); V = E − Ir; Pmax = E²/4r at R = r.
  • Kirchhoff: junction (charge), loop (energy); bridge balance P/Q = R/S.
  • Potentiometer: E ∝ l; r = R(l₁/l₂ − 1).
  • P = VI = I²R = V²/R.
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