A capacitor charges from a cell through a resistance whose time constant is T. In how much time will the capacitor collect 10% of its charge?
(A) T log$_e$0.1
(B) T log$_e$0.9
(C) T log$_e\left( {\dfrac{{10}}{9}} \right)$
(D) T log$_e\left( {\dfrac{{11}}{{10}}} \right)$

Hint: The charge on capacitor, Q is given by Q = ${Q_0}\left( {1 - {e^{ - \dfrac{t}{T}}}} \right)$, where Q$_0$ is the final charge on the capacitor, T is the time constant and t is the time elapsed. We know that Q = 0.1Q$_0$.

Complete step-by-step answer:
When a capacitor is connected to a cell, it does not gain all the charge all of a sudden. Sometime is elapsed between the capacitor being connected to the cell and the capacitor being fully charged. The charge accumulated on the capacitor is given by the equation,
$Q = {Q_0}\left( {1 - {e^{ - \dfrac{t}{T}}}} \right)$ ………………………….(1)
Where Q$_0$is the final charge on the capacitor, t is the time elapsed from connecting the capacitor to the cell and T is the time constant which is equal to the time taken for the capacitor to reach 63% of its maximum possible fully charged voltage.
From the given question, we know that the charge accumulated on the capacitor is 10% of the maximum possible charge on the capacitor.
$ \Rightarrow Q = 0.1{Q_0}$ ……………………. (2)
Now, we substitute the value of equation (2) in equation (1), we obtain,
$0.1{Q_0} = {Q_0}\left( {1 - {e^{ - \dfrac{t}{T}}}} \right)$
Upon simplifying, we find out the value of t which is the time taken by the capacitor to accumulate 10% of the maximum charge.
$0.9 = {e^{ - \dfrac{t}{T}}}$
Taking log of the equation we obtain,
$ - \dfrac{t}{T} = {\log _e}\left( {0.9} \right) = {\log _e}\left( {\dfrac{9}{{10}}} \right)$
On removing the minus sign, the term inside the log function is reciprocated. Hence, we find that,
$\dfrac{t}{T} = {\log _e}\left( {\dfrac{{10}}{9}} \right)$
Hence, the time taken by the capacitor to accumulate 10% of its maximum charge is t = T${\log _e}\left( {\dfrac{{10}}{9}} \right)$

Therefore, option C is the correct answer.

Note: We do not need the emf of the cell or the maximum charge that can be accumulated in the capacitor to solve this question. If this data is given in the question, you can ignore it if there is not any other part of the question which requires the use of that data.