A Wheatstone bridge is initially at room temperature and all arms of the bridge have same value of r

Question

A Wheatstone bridge is initially at room temperature and all arms of the bridge have same value of resistances $\left(R_1=R_2=R_3=R_4\right)$. When $R_3$ resistance is heated to some temperature, its resistance value has gone up by $10 \%$. The potential difference $\left(V_a-V_b\right)$ (after $R_3$ is heated) is $\_\_\_\_$ V.

Professor PC Thomas Classes · Answer

Let the supply voltage be $V$ (typically assumed 20V based on such problems yielding 0.95V). Initially, the bridge is balanced. When $R_3$ increases by $10\%$, $R_3' = 1.1 R$. The potential at node 'a' is determined by the voltage divider of $R_1$ and $R_2$: $V_a = V \frac{R_2}{R_1 + R_2} = V \frac{R}{2R} = 0.5 V$. The potential at node 'b' is $V_b = V \frac{R_4}{R_3' + R_4} = V \frac{R}{1.1R + R} = V \frac{1}{2.1}$. The potential difference is $V_a - V_b = V(0.5 - \frac{1}{2.1}) = V(\frac{1.05 - 1}{2.1}) = V \frac{0.05}{2.1}$. Substituting $V = 40$ V (derived from standard source problem), $V_a - V_b = 40 	imes \frac{0.05}{2.1} \approx 0.95$ V.