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(B) When the switch is closed (charging),the inductor $L$ is in series with the resistor $2R$. The resistor $R$ is in parallel with the series combina

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$3 : 2$

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(B) When the switch is closed (charging),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 time constant for the charging circuit is determined by the equivalent resistance $R_{eq}$ seen by the inductor. When the battery is shorted (or considered ideal),the equivalent resistance is $R_{eq} = 2R + R = 3R$. Thus,the charging time constant is $	au_{1} = \frac{L}{3R}$.When the switch is opened (discharging),the inductor $L$ forms a closed loop with the resistors $2R$ and $R$ in series. The total resistance in this loop is $R_{eq} = 2R + R = 3R$. Thus,the discharging time constant is $	au_{2} = \frac{L}{3R}$.Wait,re-evaluating the circuit: When the switch is closed,the inductor is in series with $2R$. The resistor $R$ is in parallel with the branch containing $2R$ and $L$. The effective resistance seen by the inductor is $R_{eq} = 2R + 0 = 2R$ (since the battery is ideal,the parallel branch $R$ does not affect the time constant). So $	au_{1} = \frac{L}{2R}$.When the switch is opened,the inductor discharges through the series combination of $2R$ and $R$. So $R_{eq} = 2R + R = 3R$. Thus,$	au_{2} = \frac{L}{3R}$.The ratio $\frac{	au_{1}}{	au_{2}} = \frac{L/2R}{L/3R} = \frac{3}{2}$.

The ratio of the time constant in charging and discharging in the circuit shown in the figure is

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