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

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$2 MΩ$

$2 Ω$

$1 MΩ$

$1 Ω$

answer is C.

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Detailed Solution

$I = I_{0} e^{- t / RC} \Rightarrow \frac{I_{0}}{2} = I_{0} e^{- t / RC} \Rightarrow t = RCln 2$
$\Rightarrow$ $10^{- 6} \times \ln 4 = (2 + r) \times 0 .5 \times 10^{- 6} \ln 2$
on solving we get $r = 2 Ω$

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