Question

In a Wheatstone’s 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{2R}{S_{1} +S_{2}} $
• B. $ \frac{P}{Q} =\frac{R\left(S_{1} +S_{2}\right)}{S_{1} S_{2}} $
• C. $ \frac{P}{Q} =\frac{R\left(S_{1} +S_{2}\right)}{2S_{1} S_{2}} $
• D. $ \frac{P}{Q} = \frac{R}{S_{1} +S_{2}} $

Answer 1

Answer: (B)

For balanced Wheatstone's bridge
$\frac{P}{Q}=\frac{R}{S}$ here $S=S_{1} \| S_{2}=\frac{S_{1} S_{2}}{S_{1}+S_{2}}$
$\Rightarrow \frac{P}{Q}=\frac{R\left(S_{1}+S_{2}\right)}{S_{1} S_{2}}$