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In a Wheatstone bridge, $R_{1} = 15 Ω$ , $R_{2} = 10 Ω$ , $R_{3} = 30 Ω$ and the unknown resistor $R_{x}$ is connected opposite $R_{3}$ . (i) Find the value of $R_{x}$ that balances the bridge. (ii) If $R_{2}$ is increased by $15%$ while $R_{1}$ and $R_{3}$ remain unchanged, what must the new value of $R_{x}$ be to keep the bridge balanced?

hard 3 marks Class 12 · Physics · Current Electricity

Answer

(i)

R_{x} = 20 Ω

; (ii)

R_{x} = 23 Ω

(rounded to nearest ohm)

The correct answer is (i) $R_{x} = 20 Ω$ ; (ii) $R_{x} = 23 Ω$ (rounded to nearest ohm) — Round to the nearest ohm: $R_{x}^{'} \approx 23 Ω$ .

Step-by-step solution

For a balanced Wheatstone bridge, the condition is $\frac{R _{1}}{R _{2}} = \frac{R _{3}}{R _{x}}$ .
(i) Substitute the given values: $\frac{15}{10} = \frac{30}{R _{x}}$ .
Simplify the left side: $\frac{15}{10} = 1.5$ .
Cross‑multiply: $1.5 R_{x} = 30$ .
Solve for $R_{x}$ : $R_{x} = \frac{30}{1.5} = 20 Ω$ .
(ii) Increase $R_{2}$ by $15%$ : $R_{2}^{'} = 10 \times (1 + 0.15) = 11.5 Ω$ .
Apply the balance condition again with the new $R_{2}^{'}$ : $\frac{R _{1}}{R _{2}^{'}} = \frac{R _{3}}{R _{x}^{'}}$ .
Plug the numbers: $\frac{15}{11.5} = \frac{30}{R _{x}^{'}}$ .
Compute the left fraction: $\frac{15}{11.5} \approx 1.30435$ .
Cross‑multiply: $1.30435 R_{x}^{'} = 30$ .
Solve for $R_{x}^{'}$ : $R_{x}^{'} = \frac{30}{1.30435} \approx 23.0 Ω$ .
Round to the nearest ohm: $R_{x}^{'} \approx 23 Ω$ .

Common mistake: Using

R_{1} R_{3} = R_{2} R_{x}

instead of the correct ratio form, leading to inverted results.

Concept tested: Balancing condition of Wheatstone bridge

Published 18 September 2026 · How this solution was prepared and checked

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