Kirchhoff's laws — solve a Wheatstone bridge circuit for balanced condition

Question

In a Wheatstone bridge, the four resistances are $P = 100\,\Omega$ , $Q = 200\,\Omega$ , $R = 150\,\Omega$ , and $S$ is unknown. Find the value of $S$ for which the bridge is balanced. What is the current through the galvanometer when the bridge is balanced?

(CBSE 2023, similar pattern)

Solution — Step by Step

A Wheatstone bridge is balanced when no current flows through the galvanometer. This happens when the potential difference across the galvanometer is zero.

The balance condition is:

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

\frac{100}{200} = \frac{150}{S}

\frac{1}{2} = \frac{150}{S}

S = 150 \times 2 = \mathbf{300\,\Omega}

When the bridge is balanced, the potential at points B and D (the two junctions connected to the galvanometer) are equal. Since there is no potential difference across the galvanometer:

I_g = \mathbf{0}

This is the whole point of a balanced Wheatstone bridge — the galvanometer reads zero.

Why This Works

The Wheatstone bridge works by comparing two voltage dividers in parallel. The top branch divides the battery voltage in the ratio $P:Q$ , and the bottom branch divides it in the ratio $R:S$ . When $P/Q = R/S$ , both dividers produce the same intermediate voltage — so the galvanometer sees no potential difference.

This is why the Wheatstone bridge is used for precision resistance measurement. At balance, the measurement is independent of the battery voltage and galvanometer sensitivity — it depends only on the ratio of known resistances.

Alternative Method

You can also derive the balance condition using Kirchhoff’s laws directly. Apply KVL (Kirchhoff’s Voltage Law) around the loops and KCL (Kirchhoff’s Current Law) at the junctions. Set $I_g = 0$ and solve the system of equations.

Let current $I_1$ flow through $P$ and $Q$ , and current $I_2$ flow through $R$ and $S$ (since $I_g = 0$ , current in each branch is continuous).

KVL around loop ABDA: $I_1 P - I_2 R = 0 \Rightarrow I_1 P = I_2 R$

KVL around loop BCDB: $I_1 Q - I_2 S = 0 \Rightarrow I_1 Q = I_2 S$

Dividing: $\dfrac{P}{Q} = \dfrac{R}{S}$ — the same result.

For CBSE board exams, always state Kirchhoff’s two laws explicitly before applying them. Examiners give marks for stating KCL (sum of currents at a junction = 0) and KVL (sum of EMFs = sum of potential drops in a loop) separately.

Common Mistake

Students often confuse which resistances are “opposite” in the Wheatstone bridge. The balance condition is $P/Q = R/S$ where $P$ and $R$ are on the same side (both connected to the same battery terminal), and $Q$ and $S$ are on the other side. If you mix up the arrangement — say writing $P/R = Q/S$ — you will get the wrong answer. Always label the circuit diagram first before writing the ratio.

Another frequent error: assuming the Wheatstone bridge formula works when the bridge is NOT balanced. The simple ratio $P/Q = R/S$ applies ONLY at balance. For an unbalanced bridge, you must use the full Kirchhoff’s law analysis.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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