Using balanced Wheatstone bridge principle. l/(100-l) = R/S.

Meter Bridge — How to Find Unknown Resistance

Question

In a meter bridge experiment, a wire of unknown resistance $S$ is placed in the right gap. A known resistance $R = 30\ \Omega$ is in the left gap. The balance point (null point) is found at $l = 60$ cm from the left end. Find $S$ .

Solution — Step by Step

The meter bridge works on the Wheatstone bridge principle. When the galvanometer reads zero, the bridge is balanced. The standard result for a balanced meter bridge is:

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

where $l$ is the balance length measured from the end where $R$ is connected.

From the problem: $R = 30\ \Omega$ , $l = 60$ cm, so $100 - l = 40$ cm.

We need to find $S$ .

Rearranging for $S$ :

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

Plug in the numbers:

S = 30 \times \frac{40}{60} = 30 \times \frac{2}{3} = 20\ \Omega

The unknown resistance $S = 20\ \Omega$ .

This makes sense — $l > 50$ cm means the balance point has shifted toward the right, which means $S < R$ . Our answer of $20\ \Omega < 30\ \Omega$ is consistent.

Why This Works

The meter bridge is a practical form of the Wheatstone bridge. The resistance wire (100 cm long) acts as the two ratio arms — the left segment of length $l$ acts as $P$ , and the right segment $(100 - l)$ acts as $Q$ . At balance, no current flows through the galvanometer, which means $P/Q = R/S$ .

Since the wire is uniform, resistance is directly proportional to length. So $P/Q = l/(100 - l)$ , which gives us the formula directly. This is why the wire must be uniform — any variation in cross-section breaks the linearity.

The physical intuition: if the unknown resistance is smaller than the known one, the balance point shifts closer to the unknown resistance side (right side), giving $l > 50$ cm. Conversely, $l < 50$ cm means $S > R$ .

Alternative Method — Cross Multiplication

Instead of rearranging first, you can directly cross-multiply from the balanced condition:

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

R \times (100 - l) = S \times l

30 \times 40 = S \times 60

S = \frac{1200}{60} = 20\ \Omega

Same answer, slightly different working. In board exams, showing the cross-multiplication step explicitly often fetches the method marks even if the final answer has an arithmetic slip.

If you’re asked to find $R$ when $S$ is known, the formula flips: $R = S \times \frac{l}{100 - l}$ . Write this form down explicitly — CBSE 2024 asked for $R$ in one question and $S$ in another, and students who memorised only one form dropped marks.

Common Mistake

The most common error: using $l$ and $(100 - l)$ in the wrong positions. Students write $R/S = (100-l)/l$ instead of $l/(100-l)$ . Remember — $l$ corresponds to the side where $R$ is connected. If $R$ is in the left gap and the jockey reads 60 cm from the left, then $l = 60$ goes with $R$ . Drawing a quick diagram with labels takes 10 seconds and prevents this swap entirely. This exact swap appeared as the distractor in CBSE 2024 marking schemes.

Question

Solution — Step by Step

Why This Works

Alternative Method — Cross Multiplication

Common Mistake

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