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Self Induction

In the circui...

Question

In the circuit shown in Figure, 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 $I_{1} / I_{2}$

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 D decreases with time
The equation of current in RC series circuit is $I = (V_{0} / R) e^{- t / R C}$
here, $I_{1} = (V / R) e^{- t / R C}$ and $I_{2} = (V / R) e^{- t / R (2 C)}$
Thus, $\frac{I_{1}}{I_{2}} = e^{- t / R C + t / 2 R C} = e^{- t / 2 R C}$
From this expression, it is clear that when $t$ increases, ratio decreases.

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In the circuit shown in Figure, switch S is closed at time t=0. Let I1 and I2 be the currents at any finite time t, then the ratio I1/I2

The correct option is D decreases with timeThe equation of current in RC series circuit is I=(V0/R)e−t/RChere, I1=(V/R)e−t/RC and I2=(V/R)e−t/R(2C)Thus, I1I2=e−t/RC+t/2RC=e−t/2RCFrom this expression, it is clear that when t increases, ratio decreases.

In the circuit shown in Figure, 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 $I_{1} / I_{2}$