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In a Wheatstone's bridge, there resistances $\mathrm{P}, \mathrm{Q}$ and $\mathrm{R}$ connected in the three arms and the fourth arm is formed by two resistances $\mathrm{S}_1$ and $\mathrm{S}_2$ connected in parallel. The condition for bridge to be balanced will be
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$\frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{R}\left(\mathrm{S}_1+\mathrm{S}_2\right)}{\mathrm{S}_1 \mathrm{~S}_2}$
$\frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{R}\left(\mathrm{S}_1+\mathrm{S}_2\right)}{\mathrm{S}_1 \mathrm{~S}_2}$
$\frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{R}\left(\mathrm{S}_1+\mathrm{S}_2\right)}{\mathrm{S}_1 \mathrm{~S}_2}$
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