In a Wheatstone bridge, three resistances are $R_1=12,Omega$, $R_2=8,Omega$ and $R_3=18,Omega$. The bridge is balanced when a galvanometer shows zero current. (i) Find the value of the unknown resistance $R_x$ connected across the arm opposite $R_3$. (ii) If $R_2$ is increased by 20% while $R_1$ and $R_3$ remain unchanged, what must $R_x$ be changed to so that the bridge remains balanced?

Step-by-step solution

1. Use the balance condition for a Wheatstone bridge: $R_1/R_2=R_3/R_x$.
2. Solve for $R_x$: $R_x=R_3\times R_2/R_1=18\times 8/12=12,\Omega$ for part (i).
3. For part (ii), increase $R_2$ by 20%: $R_2^{\prime }=8\times 1.20=9.6,\Omega$.
4. Apply the balance condition again: $R_x^{\prime }=R_3\times R_2^{\prime }/R_1=18\times 9.6/12=14.4,\Omega$.

Concept tested: Wheatstone bridge balance condition

Published 10 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked