Question

A charged capacitor is allowed to discharge through a resistance $2\Omega$ by closing the switch $S$ at the instant $t=0$. At time$t=\ln 2\mu \text{s}$, the reading of the ammeter falls half of its initial value. the resistance of the ammeter equals to

• A. $0\Omega$
• B. $2\Omega$
• C. $\infty$
• D. $2M\Omega$

Answer 1

The correct option is A $0\Omega$

While discharging, current in the circuit after time $t$ is,
$$i=i_0{e}^{\frac{-t}{R_{eq}C}}$$ Given, $i=\frac{i_0}{2}$

Substituting the values from the figure, we get

$\frac{1}{2}=e^{-\frac{\left(\ln 2\right)\times {10}^{-6}}{R_{eq}\times 0.5\times {10}^{-6}}}$

Rearranging the equation and taking $\ln$ both sides,

$\Rightarrow \ln 2=\frac{\ln 2}{R_{eq}\times 0.5}$

$\Rightarrow R_{eq}=2\Omega$

Here,

$R_{eq}=\left(\text{Ammeter resistance}\right)R_A+R=2$
$[\because R=2\Omega ]$

$\therefore R_A=0\Omega$

Hence, option (a) is the correct answer.