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 half the initial value. The resistance of the ammeter is equal to
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During discharging, current, $i(t)=i_{0} e^{-t / R C}$
If $r$ be the resistance of ammeter, net resistance, $R=2+r$
when $t=\ln 4, i=i_{0} / 2 \Rightarrow i_{0} e^{-(l n 4 / R C)}=i_{0} / 2$
or $e^{(l n 4 / R C)}=2$
take In on both sides, $\ln 4 / R C=\ln 2$
or $R C=\frac{l n 4}{l n 2}=\frac{2 l n 2}{l n 2}=2$
or $R=\frac{2}{C}=2 / 0.5=4$
or $2+r=4 \Rightarrow r=2 \Omega$
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