(A) For $0 \le t < t_0$,the circuit consists of a battery,resistor $R$,and inductor $L$ in series. The current grows exponentially as $I(t) = \frac{\e

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(A) For $0 \le t At $t = t_0$,the current reaches a value $I_0 = \frac{\epsilon}{R}(1 - e^{-Rt_0/L})$.For $t \ge t_0$,the switch $S_1$ is opened and $S_2$ is closed. The battery is removed,and the inductor $L$ discharges through the resistor $R$. The current decays exponentially as $I(t) = I_0 e^{-R(t-t_0)/L}$.The correct graph shows exponential growth followed by exponential decay to zero,which matches the provided solution image.

Class 12 Physics Electromagnetic Induction R-L D.C. Circuit

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In the circuit shown,the switch $S_1$ is closed at time $t = 0$ and the switch $S_2$ is kept open. At some later time $t_0$,the switch $S_1$ is opened and $S_2$ is closed. The behavior of the current $I$ as a function of time $t$ is given by:

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In the circuit shown,the switch $S_1$ is closed at time $t = 0$ and the switch $S_2$ is kept open. At some later time $t_0$,the switch $S_1$ is opened and $S_2$ is closed. The behavior of the current $I$ as a function of time $t$ is given by:

Similar Questions

The circuit shown consists of a switch $S$,a battery $B$ of emf $E$,a resistance $R$,and an inductor $L$. What is the current in the circuit at the instant the switch $S$ is closed?

In the circuit shown,the key $(K)$ is closed at $t = 0$. Find the current through the key at the instant $t = 10^{-3} \ln 2 \, s$.

In an $L-R$ decay circuit,the initial current at $t = 0$ is $I$. The total charge that has flown through the resistor until the energy in the inductor has reduced to one-fourth of its initial value is:

$Henry/ohm$ can be expressed in

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