In the transient current circuit shown,the ti | vedclass

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(C) To find the time constant $	au = R_{eq} C$,we need to find the equivalent resistance $R_{eq}$ seen by the capacitor $C$ when the voltage sources

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$\frac{7}{4} RC$

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(C) To find the time constant $	au = R_{eq} C$,we need to find the equivalent resistance $R_{eq}$ seen by the capacitor $C$ when the voltage sources are short-circuited.$1$. Short-circuit the voltage sources $V_1$ and $V_2$.$2$. The resistance $2R$ and $R$ in the first branch are in series,giving $3R$.$3$. This $3R$ is in parallel with the resistance $R$ of the second branch.$4$. The equivalent resistance of these two parallel branches is $r_{eq} = \frac{3R 	imes R}{3R + R} = \frac{3R^2}{4R} = \frac{3}{4} R$.$5$. This $r_{eq}$ is in series with the resistance $R$ connected in series with the capacitor.$6$. Therefore,the total equivalent resistance $R_{eq} = R + r_{eq} = R + \frac{3}{4} R = \frac{7}{4} R$.$7$. The time constant is $	au = R_{eq} C = \frac{7}{4} RC$.

In the transient current circuit shown,the time constant is

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