Current electricity for JEE Main

Current electricity is one of the most dependable scoring chapters in JEE Main Physics. The ideas are few and the questions repeat in shape: reduce a network, apply Kirchhoff's laws, account for a cell's internal resistance, or balance a bridge. What separates a quick solve from a slow one is spotting symmetry before you start writing equations.

The official 2026 syllabus covers current, drift velocity and mobility, Ohm's law, resistance and resistivity, I–V characteristics of ohmic and non-ohmic conductors, energy and power, series and parallel resistors, temperature dependence of resistance, emf and internal resistance, cells in series and parallel, Kirchhoff's laws, the Wheatstone bridge and the metre bridge. The metre bridge also appears in the experimental skills list.

Drift velocity, mobility and Ohm's law

In a wire, free electrons move randomly at high speed, but an applied field gives them a small average drift. If n is the number of free electrons per m³:

• I = neAv_d
• v_d = eEτ/m, where τ is the mean time between collisions
• mobility μ = v_d/E = eτ/m
• current density J = I/A = σE, with conductivity σ = ne²τ/m and resistivity ρ = 1/σ

Drift speeds are tiny, around a tenth of a millimetre per second. A bulb lights at once because the field is set up along the wire almost instantly, not because electrons race from the switch.

Ohm's law, V = IR, holds for ohmic conductors, whose I–V graph is a straight line through the origin at fixed temperature. A diode, a bulb filament (whose resistance rises as it heats) and an electrolyte are non-ohmic.

Resistance, resistivity and temperature

R = ρL/A. Resistivity depends on the material and temperature, not on shape.

Stretching a wire. The volume stays the same, so if the length becomes n times, the area becomes 1/n times and R becomes n² times.

Over moderate ranges, R_T = R₀(1 + αΔT).

• Metals: α is positive (more collisions as the lattice vibrates harder).
• Semiconductors: α is negative (more carriers are freed as temperature rises).
• Alloys such as manganin and constantan: α is very small, so they are used for standard resistors.

Power and combinations

Power is P = VI = I²R = V²/R. In series, the same current flows, so the larger resistance dissipates more power. In parallel, the same voltage acts, so the smaller resistance dissipates more.

Series resistances add; for parallel resistances, the reciprocals add. For two in parallel, R = R₁R₂/(R₁ + R₂).

Cells, emf and internal resistance

A cell of emf E and internal resistance r driving an external resistance R:

• current I = E/(R + r)
• terminal voltage V = E − Ir while discharging, and E + Ir while being charged
• power to R is greatest when R = r, and then P_max = E²/(4r)

A cell connected the wrong way round in a series string subtracts its emf but still adds its internal resistance.

Kirchhoff's laws

• Junction rule: the currents into a junction equal the currents out. This is conservation of charge.
• Loop rule: around any closed loop, the sum of emfs equals the sum of IR drops. This is conservation of energy.

Assign a direction to each unknown current and keep it. If a current comes out negative, it simply flows the other way. For circuits with one or two nodes, the node-potential method is often faster: set one node at 0 V, call another V, and write the junction rule in terms of V.

Wheatstone bridge and metre bridge

Four resistances P, Q, R and S form a bridge with a galvanometer across the middle. When P/Q = R/S, the bridge is balanced: no current flows through the galvanometer, and that branch can be removed.

A metre bridge is a Wheatstone bridge whose ratio arm is a uniform 100 cm wire. With an unknown R in the left gap, a known S in the right gap and the balance point at l cm:

R/S = l/(100 − l)

The result is most accurate when the balance point is near the middle. Swapping R and S and averaging cancels end errors.

Worked problems

Problem 1: Kirchhoff with a cell being charged. Cell A (10 V) and cell B (4 V), each with negligible internal resistance, are each in series with a 2 Ω resistor. Both branches are connected in parallel across a 4 Ω load, positive terminals joined. Find the current in each branch.

Take the lower rail as 0 V and the upper as V. Junction rule: (10 − V)/2 + (4 − V)/2 = V/4.
So 7 − V = V/4, giving V = 5.6 V.
Cell A: (10 − 5.6)/2 = 2.2 A, out of its positive terminal.
Cell B: (4 − 5.6)/2 = −0.8 A, so 0.8 A flows into B; it is being charged.
Load: 5.6/4 = 1.4 A. Check: 2.2 − 0.8 = 1.4 A.

Problem 2: bulbs in series. A 100 W and a 60 W bulb, both rated for 220 V, are joined in series across 220 V. Which glows brighter?

R = V²/P: the 100 W bulb has 484 Ω and the 60 W bulb about 807 Ω. The current is the same, so P = I²R is larger for the larger resistance. The 60 W bulb glows brighter.

Problem 3: metre bridge. With S = 6 Ω in the right gap, the balance point is at 40 cm. Find the unknown R and the new balance point if R and S are swapped.

R = 6 × 40/60 = 4 Ω. After swapping, 6/4 = l/(100 − l), so 600 − 6l = 4l and l = 60 cm.

Problem 4 (numerical answer): the cube. Twelve resistors of 12 Ω each form the edges of a cube. Find the resistance, in Ω, between two opposite corners of a body diagonal.

A current I entering one corner splits equally into three edges (I/3 each), by symmetry. Each of those three corners sends it on through two edges, so six edges carry I/6. The current then recombines through three edges of I/3 into the far corner.
Along any path: V = (I/3)R + (I/6)R + (I/3)R = (5/6)IR. So R_eq = 5R/6 = 10 Ω.

Practice set

1. A wire is stretched to twice its length. What happens to its resistance?
2. Find the equivalent of three 6 Ω resistors in parallel and in series.
3. A cell has E = 10 V and r = 2 Ω. What is the maximum power it can deliver to an external resistor?
4. A cell of E = 12 V and r = 1 Ω is connected to 5 Ω. Find the current and the terminal voltage.
5. A copper wire of cross-section 1 mm² carries 1.6 A. Taking n = 10²⁹ m⁻³, find the drift speed.
6. A wire has a resistance of 10 Ω at 0 °C and α = 4 × 10⁻³ per °C. Find its resistance at 100 °C.
7. Two cells, 2 V and 4 V, each with r = 1 Ω, are joined in parallel with like terminals together. Find the equivalent emf and internal resistance.
8. Two resistors give 25 Ω in series and 4 Ω in parallel. Find them.

Answers

1. It becomes 4 times (R ∝ L² at constant volume).
2. 2 Ω and 18 Ω.
3. At R = r: I = 2.5 A and P = 2.5² × 2 = 12.5 W.
4. I = 12/6 = 2 A; V = 12 − 2 × 1 = 10 V.
5. v_d = 1.6/(10²⁹ × 1.6 × 10⁻¹⁹ × 10⁻⁶) = 10⁻⁴ m/s (0.1 mm/s).
6. 10(1 + 0.4) = 14 Ω.
7. E = (2 + 4)/(1 + 1) = 3 V; r = 0.5 Ω.
8. R₁R₂ = 25 × 4 = 100 and R₁ + R₂ = 25, so 20 Ω and 5 Ω.

What to do next

• Solve 10 two-loop circuits by Kirchhoff's laws and again by the node-potential method; compare the time taken.
• Learn the cube results by deriving them: 5R/6 across a body diagonal, 3R/4 across a face diagonal, 7R/12 across an edge.
• Practise metre bridge readings, including the swap-and-average method.
• Revise capacitors, then move on to magnetic effects of current.