Kirchhoff's rules and the Wheatstone bridge

Series and parallel rules cannot solve every circuit. Kirchhoff's two rules can.

Junction rule: at any junction, current in = current out. Charge cannot pile up. Example: 5 A flows into a junction and 2 A leaves along one wire, so 3 A leaves along the other.

Loop rule: round any closed loop, the sum of the emfs equals the sum of the IR drops. Energy is conserved: a charge going round a loop and back to its start ends up with the same energy.

Example: a 12 V cell with 2 Ω and 4 Ω in series. I = 12 / 6 = 2 A. The drops are 4 V and 8 V, which add up to 12 V.

Method:

• Mark a direction for every unknown current, and use the junction rule so you need as few unknowns as possible.
• Go round each loop, adding emfs and IR drops with a fixed sign rule.
• Solve the equations.
• A negative answer is not a mistake: the current simply flows the other way.

Wheatstone bridge: four resistors P, Q, R, S in a diamond, with a galvanometer across the middle. The bridge is balanced (no current in the galvanometer) when:

P / Q = R / S

Example: P = 2 Ω, Q = 4 Ω, R = 5 Ω. Balance needs S = 10 Ω. This is how an unknown resistance is measured: adjust until the galvanometer reads zero.

Use symmetry where you can: points at the same potential carry no current between them, which removes unknowns.