In the circuit shown in figure, capacitor was initially uncharged and switch is closed at $t = 0.$ Select the correct alternatives.

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The initial battery current (immediately after switch S is closed) is $40 μ A .$

The time after which current through capacitor becomes half of the initial value is $\ln 4 s .$

The battery current, long time after switch S is closed is $\frac{80}{3} μ A .$

The current through the $600 k Ω$ resistor as a function of time is $\frac{80}{3} (1 - e^{- t}) μ A .$

answer is A, B, D.

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Detailed Solution

At $t = 0 R_{e q} = 1.2 M Ω I = \frac{48}{1.2 \times 10^{6}} = 40 μ A$

$t = \infty R_{e q} = 1.8 M Ω I = \frac{48}{1.8 \times 10^{6}} = \frac{80}{3} μ A$

time constant $τ = R C = \frac{R_{1} R_{2}}{R_{1} + R_{2}} C = 1 \Rightarrow I (t) = \frac{80}{3} (1 - e^{- t}) μ A .$

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