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As shown in the figure, in a Wheatstone's bridge, three resistances $\mathrm{P}, \mathrm{Q}$ and $\mathrm{R}$ 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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Verified Answer
The correct answer is:
$\frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{R}\left(\mathrm{S}_1+\mathrm{S}_2\right)}{\mathrm{S}_1 \mathrm{~S}_2}$
Two resistance $\mathrm{S}_1$ and $\mathrm{S}_2$ connected in parallel.
$$
\mathrm{S}=\frac{\mathrm{S}_1 \mathrm{~S}_2}{\mathrm{~S}_1+\mathrm{S}_2}
$$
from the wheat stone bridge
$$
\frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{R}}{\mathrm{S}}=\frac{\mathrm{R}\left(\mathrm{S}_1+\mathrm{S}_2\right)}{\mathrm{S}_1 \mathrm{~S}_2}
$$
$$
\mathrm{S}=\frac{\mathrm{S}_1 \mathrm{~S}_2}{\mathrm{~S}_1+\mathrm{S}_2}
$$
from the wheat stone bridge
$$
\frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{R}}{\mathrm{S}}=\frac{\mathrm{R}\left(\mathrm{S}_1+\mathrm{S}_2\right)}{\mathrm{S}_1 \mathrm{~S}_2}
$$
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