State Kirchhoff's rules. Using them, derive the balance condition of a Wheatstone bridge, and explain how a metre bridge finds an unknown resistance.

1. 1 · State both rules.
   • Junction rule: currents into a junction = currents out (conservation of charge).
   • Loop rule: ΣIR = ΣE round any closed loop (conservation of energy).
2. 2 · Draw the bridge. [Diamond ABCD: P on AB, Q on BC, R on AD, S on DC; G with a key between B and D; a cell and key between A and C; currents I₁ and I₃ marked]
3. 3 · Balance. Adjust until G shows no deflection: Ig = 0, so B and D are at the same potential.
4. 4 · Apply the rules. Junction rule: I₁ flows through P and Q, and I₃ through R and S. Loop ABDA: I₁P − I₃R = 0. Loop BCDB: I₁Q − I₃S = 0.
5. 5 · Result. Divide: P/Q = R/S, so S = QR/P.
6. 6 · Metre bridge. A uniform wire 1 m long gives two arms in the ratio l : (100 − l). With R in the left gap, balance at l gives S = R(100 − l)/l.
7. 7 · Real life. Electronic weighing scales and temperature sensors use bridge circuits.
8. 8 · Close in one line. The bridge finds S by a null method, whatever the EMF of the cell.

Scoring tip: Write that the galvanometer current is zero at balance before any equation. That one statement unlocks both loop equations.