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(A) The current in a discharging $RC$ circuit is given by $I(t) = I_0 e^{-t/	au}$,where $	au = R_{eq}C$ is the time constant.Here,$R_{eq} = R + r$,w

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$0 \, \Omega$

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(A) The current in a discharging $RC$ circuit is given by $I(t) = I_0 e^{-t/	au}$,where $	au = R_{eq}C$ is the time constant.Here,$R_{eq} = R + r$,where $R = 2 \, \Omega$ is the external resistance and $r$ is the resistance of the ammeter.The capacitance is $C = 0.5 \, \mu F = 0.5 	imes 10^{-6} \, F$.Given that at $t = \ln 2 \, \mu s = \ln 2 	imes 10^{-6} \, s$,the current $I(t) = I_0 / 2$.Substituting these values into the equation: $I_0 / 2 = I_0 e^{-t/	au} \Rightarrow 1/2 = e^{-t/	au}$.Taking the natural logarithm on both sides: $\ln(1/2) = -t/	au \Rightarrow -\ln 2 = -t/	au \Rightarrow 	au = t / \ln 2$.Substituting $t = \ln 2 \, \mu s$: $	au = (\ln 2 \, \mu s) / \ln 2 = 1 \, \mu s = 10^{-6} \, s$.Since $	au = (R + r)C$,we have $10^{-6} = (2 + r) 	imes 0.5 	imes 10^{-6}$.Dividing both sides by $10^{-6}$: $1 = (2 + r) 	imes 0.5$.$2 = 2 + r \Rightarrow r = 0 \, \Omega$.

$A$ charged capacitor is allowed to discharge through a resistance of $2 \, \Omega$ by closing the switch $S$ at the instant $t = 0$. At time $t = \ln 2 \, \mu s$,the reading of the ammeter falls to half of its initial value. The resistance of the ammeter is equal to:

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