Meter bridge experiment — find unknown resistance using balanced condition

Question

In a meter bridge experiment, the null point is obtained at 40 cm from the left end. If a known resistance of 12 $\Omega$ is in the left gap, find the unknown resistance in the right gap. What happens to the null point if the two resistances are interchanged?

(CBSE 2023, similar pattern)

Solution — Step by Step

A meter bridge is essentially a Wheatstone bridge. At the null point (galvanometer reads zero), the bridge is balanced:

\frac{R}{S} = \frac{l}{100 - l}

where $R$ is the resistance in the left gap, $S$ is in the right gap, and $l$ is the null point distance from the left end in cm.

Given: $R = 12\,\Omega$ , $l = 40$ cm.

\frac{12}{S} = \frac{40}{100 - 40} = \frac{40}{60} = \frac{2}{3}

S = 12 \times \frac{3}{2}

\boxed{S = 18\,\Omega}

After interchanging, $R' = 18\,\Omega$ is in the left gap and $S' = 12\,\Omega$ is in the right gap:

\frac{18}{12} = \frac{l'}{100 - l'}

\frac{3}{2} = \frac{l'}{100 - l'}

300 - 3l' = 2l' \Rightarrow l' = 60 \text{ cm}

The null point shifts from 40 cm to 60 cm — exactly $(100 - 40)$ cm. This symmetry always holds when you interchange the resistances.

Why This Works

The meter bridge wire has uniform resistance per unit length. So the resistance of a length $l$ of wire is proportional to $l$ . The Wheatstone bridge balance condition $R_1/R_2 = R_3/R_4$ translates directly to $R/S = l/(100-l)$ .

At the null point, no current flows through the galvanometer — the potential at both junctions is equal. This makes the measurement independent of the galvanometer sensitivity and the battery EMF, which is why the meter bridge gives accurate results.

Alternative Method

You can also use the end-correction method for better accuracy. If the bridge wire has end corrections $\alpha$ and $\beta$ at the two ends:

\frac{R}{S} = \frac{l + \alpha}{(100 - l) + \beta}

In CBSE problems, end corrections are usually ignored unless explicitly mentioned.

For NEET, remember the quick check: if $l = 50$ cm, then $R = S$ . If $l < 50$ cm, then $R < S$ . If $l > 50$ cm, then $R > S$ . This helps eliminate wrong options without full calculation.

Common Mistake

The most frequent error: writing the formula as $R/S = (100-l)/l$ instead of $R/S = l/(100-l)$ . Remember that $R$ is in the left gap and $l$ is measured from the left end. The resistance in the left gap corresponds to the wire length on the left side. Drawing the circuit diagram first (with R on the left) prevents this mix-up.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

Try These Next