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(A) The time constant $	au$ for an $RL$ circuit is given by $	au = \frac{L}{R_{eq}}$.When the switch is in position $a$,the inductor $L = 4\,mH$ is

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(A) The time constant $	au$ for an $RL$ circuit is given by $	au = \frac{L}{R_{eq}}$.When the switch is in position $a$,the inductor $L = 4\,mH$ is in series with the parallel combination of resistors $R$ and $500\,\Omega$.The equivalent resistance $R_{eq}$ is the resistance seen by the inductor,which is $R_{eq} = \frac{R 	imes 500}{R + 500}$.Given $	au = 10\,\mu s = 10 	imes 10^{-6}\,s$ and $L = 4\,mH = 4 	imes 10^{-3}\,H$.Substituting these values into the formula:$10 	imes 10^{-6} = \frac{4 	imes 10^{-3}}{\left(\frac{R 	imes 500}{R + 500}ight)}$$10 	imes 10^{-6} = \frac{4 	imes 10^{-3} 	imes (R + 500)}{500R}$$10^{-5} = \frac{4 	imes 10^{-3} 	imes (R + 500)}{500R}$$500R 	imes 10^{-5} = 4 	imes 10^{-3} 	imes (R + 500)$$5 	imes 10^{-3} R = 4 	imes 10^{-3} R + 2000 	imes 10^{-3}$$10^{-3} R = 2$$R = 2000\,\Omega = 2\,K\Omega$.

In the given circuit,the switch is in position $a$. For what value of $R$ will the circuit have a time constant of $10\,\mu s$ in $K\Omega$?

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