The ratio of time constants during growth and | vedclass

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(B) For the growth of current,the switch is closed. The inductor $L$ is in series with the resistor $2R$. The resistor $R$ is in parallel with the ser

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$3 : 2$

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(B) For the growth of current,the switch is closed. The inductor $L$ is in series with the resistor $2R$. The resistor $R$ is in parallel with the series combination of $2R$ and $L$. The equivalent resistance $R_{eq}$ seen by the inductor during growth is $2R$ (since the battery acts as a short circuit for the ideal case or is considered part of the loop). Thus,the time constant for growth is $	au_{growth} = \frac{L}{2R}$.For the decay of current,the switch is opened. The inductor $L$ now forms a closed loop with the resistors $R$ and $2R$ in series. The equivalent resistance $R_{eq}$ seen by the inductor during decay is $R + 2R = 3R$. Thus,the time constant for decay is $	au_{decay} = \frac{L}{3R}$.The ratio of time constants during growth and decay is $\frac{	au_{growth}}{	au_{decay}} = \frac{L/2R}{L/3R} = \frac{3R}{2R} = \frac{3}{2}$ or $3 : 2$.

The ratio of time constants during growth and decay of current in the circuit is

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