Question

A resistor with resistance of $10\text{M}\Omega$ is connected in series with a capacitor of capacitance $1\mu \text{F}$ and a battery with e.m.f. $12\text{V}$. Before the switch is closed at time $t=0,$ the capacitor is uncharged. What fraction of the initial current remains at time, $t=46\text{s}$? $(e^{-4.6}=0.01)$

• A. $0.09$
• B. $0.01$
• C. $0.02$
• D. $0.05$

Answer 1

The correct option is B $0.01$

Given,
Resistance, $R=10\text{M}\Omega =10\times {10}^6\Omega$
Capacitance, $C=1\mu \text{F}=1\times {10}^{-6}\text{F}$

The time constant is,

$\tau =RC=10\times {10}^6\times 1\times {10}^{-6}=10\text{s}$

As we know that,
$I=I_0{e}^{-\frac{t}{\tau }}$

$\Rightarrow \frac{I}{I_0}=e^{-4.6}=0.01$

Hence, option (b) is the correct answer.