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Q.

In this figure the resistance of the coil of galvanometer G is 2 $Ω$ . The emf of the cell is 4 V. The ratio of potential difference across C₁ and C₂ is

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a

1

b

$\frac{3}{4}$

c

$\frac{5}{4}$

d

$\frac{4}{5}$

answer is D.

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$\begin{array}{l} i = \frac{4}{6 + 2 + 8} = \frac{1}{4} A \\ V_{A} = 4 V, V_{D} = 0 V \\ V_{B} = V_{A} - i (6) = 4 - \frac{1}{4} \times 6 = 2.5 V \\ V_{C} = V_{B} - i (2) = 2.5 - \frac{1}{4} \times 2 = 2 V \\ Across C_{1}, V_{1} = V_{A} - V_{C} = 4 - 2 = 2 V \\ Across C_{2}, V_{2} = V_{B} - V_{D} = 2.5 - 0 = 2.5 V \\ ∴ \frac{V_{1}}{V_{2}} = \frac{2}{2.5} = \frac{4}{5} \end{array}$

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