In this figure the resistance of the coil of galvanometer G is 2Ω. The emf of the cell is 4 V. The ratio of potential difference across C1 and C2 is
- \(\frac{4}{5}\)
- 1
- \(\frac{5}{4}\)
- \(\frac{3}{4}\)
The Correct Option is A
Solution and Explanation
Step 1: Apply Kirchhoff’s laws.- Using the given circuit, calculate the currents and potential drops across C1 and C2.- The ratio is determined as:
$\frac{V_{C1}}{V_{C2}} = \frac{4}{5}$.
Final Answer: The ratio of potential differences is $\frac{4}{5}$
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