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Q.

The charge developed on $4 μ F$ condenser is

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a

$18 μ c$

b

$4 μ c$

c

$8 μ c$

d

$z e r o$

answer is C.

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Detailed Solution

$\begin{aligned} i = \frac{ε}{R_{2} + r} V_{6 Ω} = 1 \times 6 = 6 v \\ = \frac{10}{10} = 1 A v_{1} : v_{2} = \frac{1}{c_{1}} : \frac{1}{c_{2}} \\ v_{1} = \frac{v}{3} = \frac{1}{3} (6) = 2 v \\ Q_{4 μ F} = C V_{1} = 4 \times 10^{- 6} \times 2 \\ = 8 \times 10^{- 6} \end{aligned}$

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