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. $E=12\text{V}$. Before the switch is closed at time $t=0,$ the capacitor is uncharged. What fraction of the final charge is on the plates at time, $t=46\text{s}$? $(e^{-4.6}=0.01)$

• A. $99\%$
  
• B. $10\%$
• C. $20\%$
  
• D. $50\%$
  

Answer 1

The correct option is A $99\%$


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,

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

The fraction $f$ of the final charge on capacitor is $\frac{q}{Q}$ .

$\frac{q}{Q}=1-{\exp }^{-t/RC}$

$\Rightarrow \frac{q}{Q}=1-{\exp }^{-46/10}$

As, $e^{-4.6}=0.01$

$\Rightarrow \frac{q}{Q}=0.99$

Hence, option (a) is the correct answer.