Question 505: Current Electricity

Question

In a Wheatstone's bridge, the resistances in four arms are as shown in the figure. The balancing condition of the bridge is

Accepted Answer

$$\begin{aligned}
&	ext { For balancing the bridge, }\
&\begin{aligned}
\frac{\mathrm{P}}{\mathrm{Q}} & =\frac{\mathrm{R}}{\mathrm{~S}} \
\mathrm{~S} & =\frac{\mathrm{S}_1 \mathrm{~S}_2}{\mathrm{~S}_1+\mathrm{S}_2} \quad \ldots .\left(\because \mathrm{S}_1, \mathrm{~S}_2 	ext { are in parallel }ight) \
	herefore \quad \frac{\mathrm{P}}{\mathrm{Q}} & =\frac{\mathrm{R}\left(\mathrm{~S}_1+\mathrm{S}_2ight)}{\mathrm{S}_1 \mathrm{~S}_2}
\end{aligned}
\end{aligned}$$

Answer

$\frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{R}}{\mathrm{S}_1+\mathrm{S}_2}$

Answer

$\frac{P}{Q}=\frac{R\left(S_1 S_2\right)}{S_1+S_2}$

Answer

$\frac{P}{Q}=\frac{R\left(S_1+S_2\right)}{2 S_1 S_2}$

Answer

$\frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{R}\left(\mathrm{S}_1+\mathrm{S}_2\right)}{\mathrm{S}_1 \mathrm{~S}_2}$

Current Electricity

In a Wheatstone's bridge, the resistances in four arms are as shown in the figure. The balancing condition of the bridge is

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