Wheatstone Bridge Condition

At balance (no current through galvanometer): P/Q = R/S. Used to find unknown resistance S when P, Q, R are known. Balance is independent of EMF and galvanometer resistance.

Circuit

Four resistors $P$, $Q$, $R$, $S$ in a diamond arrangement. Battery between vertices A and C. Galvanometer between B and D.

At balance: no current through the galvanometer ($I_g = 0$), so $V_B = V_D$.

Applying KCL and KVL at balance

Since $I_g = 0$: current through $P$ = current through $Q$ = $I_1$, and current through $R$ = current through $S$ = $I_2$.

Potential drop from A to B: $V_A - V_B = I_1 P$

Potential drop from A to D: $V_A - V_D = I_2 R$

Since $V_B = V_D$:

$$I_1 P = I_2 R \quad \cdots (1)$$

Similarly, B to C: $V_B - V_C = I_1 Q$, and D to C: $V_D - V_C = I_2 S$:

$$I_1 Q = I_2 S \quad \cdots (2)$$

Dividing (1) by (2):

$$\frac{P}{Q} = \frac{R}{S}$$ $$\boxed{\frac{P}{Q} = \frac{R}{S}}$$

Finding unknown resistance

Unknown $S = R \cdot Q/P$. $P$, $Q$, $R$ are known standard resistances. $R$ is varied until the galvanometer reads zero.