In a Wheatstone bridge, the known resistances are $R_1=16,Omega$, $R_2=12,Omega$, $R_3=24,Omega$ and an unknown resistor $R_4$ is connected opposite $R_3$. (i) Find the value of $R_4$ that balances the bridge. (ii) If $R_2$ is increased by 25 % while $R_1$ and $R_3$ remain unchanged, what must $R_4$ be changed to so that the bridge remains balanced?

Step-by-step solution

1. For a balanced Wheatstone bridge, the ratio of resistances in one branch equals that in the other: $\frac{R_1}{R_2}=\frac{R_4}{R_3}$.
2. Solve for $R_4$: $R_4=R_3\frac{R_1}{R_2}=24\times \frac{16}{12}=32,\Omega$.
3. Increase $R_2$ by 25 %: new $R_2^{\prime }=12\times 1.25=15,\Omega$.
4. Re‑apply the balance condition with the new $R_2^{\prime }$: $R_4^{\prime }=R_3\frac{R_1}{R_2^{\prime }}=24\times \frac{16}{15}=25.6,\Omega$.

Concept tested: Wheatstone bridge balance

Published 15 September 2026 · How this solution was prepared and checked