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Byju's Answer

Standard XII

Physics

Inductors in Circuits

In the circui...

Question

In the circuit shown,

the switch $S_{1}$ is closed at time $t = 0$ and the switch $S_{2}$ is kept open. At some time $(t_{0})$ later, the switch $S_{1}$ is opened and $S_{2}$ is closed. The behaviour of the current $I$ as a function of time $t$ is given by:

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Solution

The correct option is B
There will be growth of current from $t = 0$ to $t = t_{0}$ and it is represented by the expression,

$I = I_{0} (1 - e^{- \frac{t R}{L}})$

After $t = t_{0}$ , there will be decay of current represented by the expression,

$I = I_{0} e^{- \frac{t R}{L}}$

The above condition is represented as

Hence, option $(B)$ is correct.

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Similar questions

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In the circuit shown, the switch S1 is closed at time t=0 and the switch S2 is kept open. At some time (t0) later, the switch S1 is opened and S2 is closed. The behaviour of the current I as a function of time t is given by:

The correct option is B There will be growth of current from t=0 to t=t0 and it is represented by the expression, I=I0(1−e−tRL) After t=t0, there will be decay of current represented by the expression, I=I0e−tRL The above condition is represented as Hence, option (B) is correct.

In the circuit shown,

the switch $S_{1}$ is closed at time $t = 0$ and the switch $S_{2}$ is kept open. At some time $(t_{0})$ later, the switch $S_{1}$ is opened and $S_{2}$ is closed. The behaviour of the current $I$ as a function of time $t$ is given by: