Current Electricity: Step-by-Step Worked Examples (12)

A balanced Wheatstone bridge satisfies:

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

Plugging in:

$\frac{10}{15} = \frac{20}{S} \implies S = 30 \,\Omega$

Since the bridge is balanced, no current flows through the galvanometer — it’s effectively absent. The bridge becomes two parallel branches:

• Branch 1: $P + Q = 10 + 15 = 25\,\Omega$
• Branch 2: $R + S = 20 + 30 = 50\,\Omega$

Parallel combination:

$R_{\text{eq}} = \frac{25 \cdot 50}{25 + 50} = \frac{1250}{75} \approx 16.67\,\Omega$

Battery current:

$I = \frac{V}{R_{\text{eq}}} = \frac{10}{16.67} = 0.6 \text{ A}$

Current through branch 1: $I_1 = 10/25 = 0.4$ A. Current through branch 2: $I_2 = 10/50 = 0.2$ A. Check: $I_1 + I_2 = 0.6$ A. Good.

Potential at junction of $P$ and $Q$ (taking battery negative terminal as $0$ V): drop across $P$ is $0.4 \cdot 10 = 4$ V, so junction is at $10 - 4 = 6$ V. Same junction at the other branch: drop across $R$ is $0.2 \cdot 20 = 4$ V, junction is at $10 - 4 = 6$ V. Both junctions at 6 V — confirms the bridge is balanced. (a) $S = 30\,\Omega$, (b) $I = 0.6$ A, (c) both galvanometer junctions at 6 V.