MHT CET · Physics · 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
- A \(\frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{R}}{\mathrm{S}_1+\mathrm{S}_2}\)
- B \(\frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{R}\left(\mathrm{S}_{\mathrm{i}} \mathrm{S}_2\right)}{\mathrm{S}_1+\mathrm{S}_2}\)
- C \(\frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{R}\left(\mathrm{S}_1+\mathrm{S}_2\right)}{2 \mathrm{~S}_1 \mathrm{~S}_2}\)
- D \(\frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{R}\left(\mathrm{S}_1+\mathrm{S}_2\right)}{\mathrm{S}_1 \mathrm{~S}_2}\)
Answer & Solution
Correct Answer
(D) \(\frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{R}\left(\mathrm{S}_1+\mathrm{S}_2\right)}{\mathrm{S}_1 \mathrm{~S}_2}\)
Step-by-step Solution
Detailed explanation
For balancing the bridge,
\(\begin{aligned}
& \frac{P}{Q}=\frac{R}{S} \\
& \mathrm{~S}=\frac{\mathrm{S}_1 \mathrm{~S}_2}{\mathrm{~S}_1+\mathrm{S}_2} \quad \ldots .\left(\because \mathrm{S}_1, \mathrm{~S}_2 \text { are in parallel }\right) \\
& \therefore \quad \frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{R}\left(\mathrm{~S}_1+\mathrm{S}_2\right)}{\mathrm{S}_1 \mathrm{~S}_2}
\end{aligned}\)
\(\begin{aligned}
& \frac{P}{Q}=\frac{R}{S} \\
& \mathrm{~S}=\frac{\mathrm{S}_1 \mathrm{~S}_2}{\mathrm{~S}_1+\mathrm{S}_2} \quad \ldots .\left(\because \mathrm{S}_1, \mathrm{~S}_2 \text { are in parallel }\right) \\
& \therefore \quad \frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{R}\left(\mathrm{~S}_1+\mathrm{S}_2\right)}{\mathrm{S}_1 \mathrm{~S}_2}
\end{aligned}\)
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