In the transient circuit shown,the time const | vedclass

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

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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 determine the equivalent resistance $R_{eq}$ as seen by the capacitor $C$ when the voltage sources are short-circuited.$1$. Short-circuit the voltage sources $V_1$ and $V_2$. The resistor $2R$ and $R$ (in the left branch) are in series,giving $3R$.$2$. This $3R$ is in parallel with the middle resistor $R$. The equivalent resistance of this part is $R_{th} = \frac{(3R 	imes R)}{(3R + R)} = \frac{3R^2}{4R} = \frac{3}{4}R$.$3$. The capacitor $C$ is in series with the resistor $R$ on the right side. Thus,the total equivalent resistance $R_{eq}$ seen by the capacitor is $R_{eq} = R_{th} + R = \frac{3}{4}R + R = \frac{7}{4}R$.$4$. Therefore,the time constant is $	au = R_{eq}C = \frac{7}{4}RC$.

In the transient circuit shown,the time constant of the circuit is:

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