Question

In a Wheatstone bridge, three resistances P, Q and R are connected in the three arms and the fourth arm is formed by two resistances $S_1$ and $S_2$ connected in parallel. The condition for the bridge to be balanced will be

• A. $\frac{P}{Q}=\frac{R(S_1+S_2)}{2S_1{S}_2}$
• B. $\frac{P}{Q}=\frac{R}{S_1+S_2}$
• C. $\frac{P}{Q}=\frac{2R}{S_1+S_2}$
• D. $\frac{P}{Q}=\frac{R(S_1+S_2)}{S_1{S}_2}$

Answer 1

The correct option is D $\frac{P}{Q}=\frac{R(S_1+S_2)}{S_1{S}_2}$

The balanced condition for Wheatstone bridge is $\frac{P}{Q}=\frac{R}{S}$

Here $S=\frac{S_1{S}_2}{S_1+S_2}$ (as $S_1$ and $S_2$ are connected in parallel in S arm)

so, $\frac{P}{Q}=\frac{R(S_1+S_2)}{S_1{S}_2}$