Wheatstone Bridge — Condition for Balance

Question

Four resistors P, Q, R, and S are arranged in a Wheatstone bridge as shown. Under what condition is the bridge balanced, and what does “balanced” mean physically?

Also: In a Wheatstone bridge, P = 100 Ω, Q = 200 Ω, R = 150 Ω. Find S for balance.

Solution — Step by Step

Why This Works

The key insight is that a balanced Wheatstone bridge is really just two potential dividers in parallel, and we’re asking: do both dividers give the same output voltage?

The left arm (P and Q in series) divides the supply voltage. The fraction that appears across Q is $\frac{Q}{P+Q}$. The right arm (R and S in series) divides the same supply. The fraction across S is $\frac{S}{R+S}$. For the midpoints to be at equal potential, these fractions must be equal — which is exactly what $P/Q = R/S$ says when you cross-multiply.

This is why the balance condition is independent of the supply voltage. Whether you apply 2 V or 12 V, if the ratios match, the galvanometer reads zero. That’s what makes the Wheatstone bridge so useful for precision resistance measurement.

Alternative Method — Using Kirchhoff’s Laws

Instead of the ratio approach, we can use KVL directly. At balance, no current flows through the galvanometer, so we treat it as an open circuit.

Assign current $I_1$ through the P-Q arm and $I_2$ through the R-S arm (they’re independent since the galvanometer branch is open).

Applying KVL to the loop A→B→C: $\varepsilon = I_1(P + Q)$

Applying KVL to the loop A→D→C: $\varepsilon = I_2(R + S)$

For $V_B = V_D$: the voltage at B (measured from C) is $I_1 Q$, and at D is $I_2 S$.

Setting $I_1 Q = I_2 S$ and substituting $I_1 = \varepsilon/(P+Q)$ and $I_2 = \varepsilon/(R+S)$:

Cross-multiplying and simplifying gives us $PS = QR$, which is the same as $P/Q = R/S$.

Common Mistake