Question

A capacitor of capacitance $100\;{\rm{\mu F}}$ is charged by connecting it to a battery of voltage $12\;{\rm{V}}$ with internal resistance $2\;\Omega $. The time after which \[99\% \] of the maximum charge is stored on the capacitor is
A.$0.92\;{\rm{ms}}$
B.$0.72\;{\rm{ms}}$
C.$0.34\;{\rm{ms}}$
D.$0.54\;{\rm{ms}}$

Answer 1

Hint: To solve this problem, we will use the formula of charge on the capacitor. As resistance and capacitance is already given in the question, we can find the time constant. Also, in the given condition, the charge is \[99\% \] of the maximum charge. We will substitute this time constant and charge which is equal to \[99\% \] of the charge in the formula of charge on the capacitor to find the time.

Complete step by step answer:
Given:
The capacitor is of the capacitance $C = 100\;{\rm{\mu F}}$.
The resistance in the circuit is $R = 2\;\Omega $.
The voltage in the circuit is \[V = 12\;{\rm{V}}\].

We will write the expression for the charge in the capacitor at a time $t$, that is given as
$Q = {Q_o}\left( {1 - {e^{\dfrac{{ - t}}{{RC}}}}} \right)$……(i)
Where, $RC$ is the time constant and ${Q_o}$ is the maximum charge stored in the capacitor.

As we have to find out the time when 99% of the maximum charge is stored. Therefore,
$Q = \dfrac{{99}}{{100}}{Q_o} = 0.99{Q_o}$..…(ii)

Now, we will write the expression for the time constant and it can be expressed as,
$ \tau = RC\\$
$\implies \tau = 2\;\Omega \times 100\;{\rm{\mu F}} \times \dfrac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 6}}}}\;{\rm{F}}}}{{{\rm{1}}\;{\rm{\mu F}}}}\\$
$\implies \tau = 2 \times {10^{ - 4}} $……(iii)

On substitute the values of $Q$and $\tau $ that is $RC$ from the equation (ii) and (iii) respectively in the equation (i).
$ 0.99{Q_o} = {Q_o}\left( {1 - {e^{\dfrac{{ - t}}{{2 \times {{10}^{ - 4}}}}}}} \right)\\ $
$\implies 0.99 = 1 - {e^{\dfrac{{ - t}}{{2 \times {{10}^{ - 4}}}}}}\\$
$\implies {e^{\dfrac{{ - t}}{{2 \times {{10}^{ - 4}}}}}} = 1 - 0.99\\$
$\implies {e^{\dfrac{{ - t}}{{2 \times {{10}^{ - 4}}}}}} = 0.01$

Now we will log on both sides of the above expression.
$ \dfrac{{ - t}}{{2 \times {{10}^{ - 4}}}} = \ln \left( {0.01} \right)\\$
$\implies t = - \ln \left( {0.01} \right) \times 2 \times {10^{ - 4}}\\$
$\implies t = 9.2 \times {10^{ - 4}}\;{\rm{s}}\\$
$\therefore t = 0.92\;{\rm{ms}} $

So, the correct answer is “Option A”.

Note:
Make sure to substitute all the values in the standard unit only. The options are in millisecond but we will get the answer in seconds on solving the expression. Therefore, we need to convert the seconds into milliseconds.