In the circuit, switch $S$ is closed at time $t = 0$ , Let $I_{1}$ and $I_{2}$ be the currents at any finite time $t$ , then the ratio of $\frac{I_{1}}{I_{2}}$

is constant

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increases with time

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decreases with time

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first increases, then decreases

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Solution

The correct option is C decreases with time
For the given circuit at any instant time $t$ ,

$I_{1} = \frac{V}{R} e^{(- \frac{t}{R C})} . . . (1)$

$I_{2} = \frac{V}{R} e^{(- \frac{t}{2 R C})} . . . (2)$

Fro equations (1) and (2),

$\frac{I_{1}}{I_{2}} = e^{(- \frac{t}{R C} + \frac{t}{2 R C})}$

$\Rightarrow \frac{I_{1}}{I_{2}} = e^{(- \frac{t}{2 R C})}$

$\Rightarrow \frac{I_{1}}{I_{2}} = \frac{1}{e^{(\frac{t}{2 R C})}}$

From this expression, it is clear that as $t$ increases, ratio of $\frac{I_{1}}{I_{2}}$ decreases.

Hence, option (c) is the correct answer.

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Similar questions

In the circuit shown if switch S is closed at $t = 0$ , then $i_{1}$ and $i_{2}$ be the current at any finite time $t$ . The ratio $\frac{(i_{1})}{(i_{2})}$ is

Q. In the circuit shown, switch S is closed at

t = 0

. Let

i_{1}

and

i_{2}

be the current at any finite time t, then the ratio

i_{1} / i_{2}

In the circuit shown if switch S is closed at $t = 0$ , then $i_{1}$ and $i_{2}$ be the current at any finite time $t$ . The ratio $\frac{(i_{1})}{(i_{2})}$ is