Question

In a Wheatstone's bridge, three resistances P,Q and R are connected in 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. $\frac{P}{Q}=\frac{2R}{S_1+S_2}$
• B. $\frac{P}{Q}=\frac{R(S_1+S_2)}{S_1{S}_2}$
• C. $\frac{P}{Q}=\frac{R(S_1+S_2)}{2S_1{S}_2}$
• D. $\frac{P}{Q}=\frac{R}{S_1+S_2}$

Answer 1

Answer: (B)

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

The condition for a balanced Wheatstone bridge is:

$\frac{P}{Q}=\frac{R}{S}$

$\Rightarrow S=\frac{S_1{S}_2}{S_1+S_2}[\because S_1,S_2$ in parallel]

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

Hence, option B is correct.