In the circuit shown below,the switch is clos | vedclass

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(D) The correct answer is $IV$.At $t=0$,when the switch is closed,the current in the circuit is zero because the inductor opposes any instantaneous ch

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$IV$

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(D) The correct answer is $IV$.At $t=0$,when the switch is closed,the current in the circuit is zero because the inductor opposes any instantaneous change in current. Therefore,the potential drop across the resistor $(V_R = iR)$ is zero.By Kirchhoff's voltage law,the entire source voltage of $10 \, V$ appears across the inductor at $t=0$. Thus,the initial voltage across the inductor is $10 \, V$.As time passes,the current $i$ in the circuit grows exponentially according to the equation $i(t) = \frac{V}{R} (1 - e^{-Rt/L})$.The voltage across the inductor is given by $V_L = L \frac{di}{dt} = V e^{-Rt/L}$.As $t$ increases,the current increases,and the potential drop across the resistor increases,while the potential drop across the inductor decreases exponentially from $10 \, V$ to $0 \, V$.When the current reaches its maximum steady value,the rate of change of current becomes zero,and the potential drop across the inductor tends to zero. This exponential decay is best represented by graph $IV$.

In the circuit shown below,the switch is closed at time $t=0$. Which of the graphs shown below best represents the voltage across the inductor,as seen on an oscilloscope?

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