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  • A charged capacitor is allowed to discharge through a resistor by closing the key at the instant = 0 (see Fig.). At the instant =(In 4) μs, the reading of the ammeter falls half the initial valu

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

Option: 1

1 \mathrm{~M\Omega}


Option: 2

1 \Omega


Option: 3

2 \Omega


Option: 4

\mathrm{2 M\Omega}


Answers (1)

best_answer

\begin{aligned} &\mathrm{ I=I_0 e^{-t / R C} \text { or } \frac{I_0}{2}=I_0 e^{-t / R C} \Rightarrow t=R C \operatorname{In} 2 }\\ &\mathrm{ \text { or } 10^{-6} \times \operatorname{In} 4=(2+r) \times 0.5 \times 10^{-6} \operatorname{In} 2 }\\ &\mathrm{ \text { Solve to get } r=2 \Omega}\\ \end{aligned}

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Rakesh

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