Question

In a circuit, three resistance $\text{P, Q}$ and $\text{R}$ are connected in the three arms and the fourth arm is formed by two resistance $S_1$ and $S_2$ connected in parallel. The condition for which the circuit will become a balanced wheatstone bridge is

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

Arrangement of Wheatstone bridge mention in question will be as shown below.


Effective resistance of $S_1$ and $S_2$ will be

$S_{eq}=\frac{S_1{S}_2}{S_1+S_2}$

Since, Wheatstone bridge is balanced, so

$\therefore \frac{P}{R}=\frac{Q}{S_{eq}}$

$\Rightarrow \frac{P}{Q}=\frac{R}{S_{eq}}=\frac{R}{S_1{S}_2/(S_1+S_2)}$

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

Hence, option $\left(d\right)$ is correct.