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(C) To find the time constant $	au$ of the charging capacitor circuit,we need to determine the Thevenin equivalent resistance $R_{eq}$ as seen by the

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$\frac{2}{3}RC$

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(C) To find the time constant $	au$ of the charging capacitor circuit,we need to determine the Thevenin equivalent resistance $R_{eq}$ as seen by the capacitor $C$.$1$. First,we remove the capacitor $C$ from the circuit.$2$. We then find the equivalent resistance across the terminals where the capacitor was connected,while replacing the voltage source $E$ with a short circuit (since it is an ideal voltage source).$3$. With the voltage source $E$ shorted,the resistor $R$ and the resistor $2R$ are connected in parallel.$4$. The equivalent resistance $R_{eq}$ is given by: $R_{eq} = \frac{R 	imes 2R}{R + 2R} = \frac{2R^2}{3R} = \frac{2}{3}R$.$5$. The time constant $	au$ is defined as $	au = R_{eq}C$.$6$. Therefore,$	au = \left( \frac{2}{3}R ight)C = \frac{2}{3}RC$.

In the circuit shown in the figure,if the switch $S$ is closed at $t = 0$,then the capacitor charges with a time constant:

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