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Question

(C) During discharging,the current in the circuit is given by $i(t) = i_0 e^{-t / RC}$.Given that at $t = (\ln 4) \, \mu s$,the current $i = i_0 / 2$.

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$2 \, \Omega$

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(C) During discharging,the current in the circuit is given by $i(t) = i_0 e^{-t / RC}$.Given that at $t = (\ln 4) \, \mu s$,the current $i = i_0 / 2$.Substituting these values into the equation: $i_0 / 2 = i_0 e^{-(\ln 4 \, \mu s) / RC}$.This simplifies to $1/2 = e^{-(\ln 4 \, \mu s) / RC}$,or $2 = e^{(\ln 4 \, \mu s) / RC}$.Taking the natural logarithm on both sides: $\ln 2 = (\ln 4 \, \mu s) / RC$.Since $\ln 4 = 2 \ln 2$,we have $\ln 2 = (2 \ln 2 \, \mu s) / RC$.Thus,$RC = 2 \, \mu s$.Given $C = 0.5 \, \mu F$,we find $R = (2 \, \mu s) / (0.5 \, \mu F) = 4 \, \Omega$.The total resistance $R$ is the sum of the resistor and the ammeter resistance $r$: $R = 2 \, \Omega + r$.Therefore,$2 \, \Omega + r = 4 \, \Omega$,which gives $r = 2 \, \Omega$.

$A$ charged capacitor is allowed to discharge through a resistor by closing the key at the instant $t = 0$. At the instant $t = (\ln 4) \, \mu s$,the reading of the ammeter falls to half the initial value. The resistance of the ammeter is equal to

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