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In a Wheatstone bridge, three resistance $P, Q$ 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 is
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The correct answer is:
$\frac{P}{Q}=\frac{R\left(S_1+S_2\right)}{S_1 S_2}$
For balanced Wheatstone's bridge we have
$$
\frac{P}{Q}=\frac{R}{S}
$$
Where $S$ is the equivalent resistance of $S_1$ and $S_2$ in parallel
$$
S=\frac{S_1 S_2}{S_1+S_2}
$$
$$
\therefore \frac{P}{Q}=\frac{R\left(S_1+S_2\right)}{S_1 S_2}
$$
$$
\frac{P}{Q}=\frac{R}{S}
$$
Where $S$ is the equivalent resistance of $S_1$ and $S_2$ in parallel
$$
S=\frac{S_1 S_2}{S_1+S_2}
$$
$$
\therefore \frac{P}{Q}=\frac{R\left(S_1+S_2\right)}{S_1 S_2}
$$
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