A charged capacitor is allowed to discharge through a resistance $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 half of its initial value. The resistance of the ammeter equal to
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$q=q_{0} \cdot e^{-t / \tau} \quad$ where $\tau=(2+r) 0.5 \mu F$
$\Rightarrow \frac{\mathrm{dq}}{\mathrm{dt}}=\frac{\mathrm{q}_{0}}{\mathrm{z}} \cdot \mathrm{e}^{-\mathrm{t} / \tau}$
$\mathrm{e}^{-\mathrm{t} / \tau}=1 / 2 \Rightarrow \ln 2=\mathrm{t} / \tau \Rightarrow \frac{\ln 2 \times \mu \mathrm{s}}{\mathrm{z}}=\ln 2$
$\Rightarrow \tau=1 u s=(2+r) \times 1 / 2 \Rightarrow r=0$
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