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

In a Wheatstone's bridge, the resistances in four arms are as shown in the figure. The balancing condition of the bridge is
  1. $\quad \frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{R}}{\mathrm{S}_1+\mathrm{S}_2}$
  2. $\frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{R}\left(\mathrm{S}_{\mathrm{i}} \mathrm{S}_2\right)}{\mathrm{S}_1+\mathrm{S}_2}$
  3. $\frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{R}\left(\mathrm{S}_1+\mathrm{S}_2\right)}{2 \mathrm{~S}_1 \mathrm{~S}_2}$
  4. $\frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{R}\left(\mathrm{S}_1+\mathrm{S}_2\right)}{\mathrm{S}_1 \mathrm{~S}_2}$

Solution

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

Asked in: MHT CET 2024 (04 May Shift 2)

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