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(C) To find the time constant $	au$ of an $LR$ circuit,we need to determine the equivalent resistance $R_{eq}$ as seen from the terminals of the indu

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$\frac{2}{3}\frac{L}{R}$

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(C) To find the time constant $	au$ of an $LR$ circuit,we need to determine the equivalent resistance $R_{eq}$ as seen from the terminals of the inductor $L$ by short-circuiting the voltage source $E$.$1$. Short-circuit the voltage source $E$. The circuit now consists of two resistors $R$ in parallel with each other,and this combination is in series with the third resistor $R$.$2$. The two resistors $R$ connected in parallel have an equivalent resistance of $R_p = \frac{R 	imes R}{R + R} = \frac{R}{2}$.$3$. This parallel combination is in series with the remaining resistor $R$. Therefore,the total equivalent resistance $R_{eq}$ seen by the inductor is $R_{eq} = R + \frac{R}{2} = \frac{3}{2}R$.$4$. The time constant $	au$ for an $LR$ circuit is given by $	au = \frac{L}{R_{eq}}$.$5$. Substituting the value of $R_{eq}$,we get $	au = \frac{L}{\frac{3}{2}R} = \frac{2L}{3R}$.

Find the time constant of the following circuit.

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