Question

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

• A. $1\text{M}\Omega$
• B. $1\Omega$
• C. $2\Omega$
• D. $2\text{M}\Omega$

Answer 1

The correct option is C $2\Omega$

While discharging, the current at time $t$ is,$$I=I_0{e}^{\frac{-t}{RC}}$$Given, at time $t=\ln 4\mu \text{s}$, $I$ becomes $\frac{I_0}{2}$.

$\Rightarrow \frac{I_0}{2}=I_0{e}^{\frac{-t}{RC}}$

$\Rightarrow 2=e^{\frac{t}{RC}}$

$\Rightarrow t=RC\ln 2$

Substituting the values considering $r$ as the resistance of the ammeter.

$\Rightarrow \ln 4\times {10}^{-6}=(2+r)\times 0.5\times {10}^{-6}\times \ln 2$

$\Rightarrow 2\ln 2=(2+r)\times 0.5\times \ln 2$

$\therefore r=2\Omega$

Hence, option (c) is the correct answer.