Question

The capacitor gets almost fully charged after time t equal to
(A) RC
(B) 3RC
(C) 4RC
(D) 2RC

Answer 1

Hint: Use the formula for rise in the voltage across the capacitor $v = V\left( {1 - {e^{ - \dfrac{t}{{RC}}}}} \right)$ and keep substituting t as $RC,2RC....$ and so on, until you get $v \approx V$, i.e. an almost fully charged capacitor.

Complete step by step solution
Let the voltage required to charge a capacitor to the maximum value or to get the capacitor almost fully charged be V.
Taking RC to be the time constant, the expression for instantaneous rise in the voltage across the capacitor is given by:
$v = V\left( {1 - {e^{ - \dfrac{t}{{RC}}}}} \right){\text{ }} - - - - (1)$
Consider the first time constant, i.e. for $t = RC$ . On putting this in equation (1), we obtain
$
   \Rightarrow v = V\left( {1 - {e^{ - \dfrac{{RC}}{{RC}}}}} \right) \\
   \Rightarrow v = V\left( {1 - {e^{ - 1}}} \right) \\
 $
Substituting the value of $e = 2.718$ in the above equation yields,
$
   \Rightarrow v = V\left( {1 - 0.368} \right) \\
   \Rightarrow v = V\left( {0.632} \right) \\
   \Rightarrow v \simeq 0.63V \\
 $
This means that after the first time constant, i.e. $t = RC$, we have managed to raise the voltage to 0.63 times of V. So, we have increased the voltage to $0.63 \times 100 = 63\% $ of V.
We require to add another time constant. Taking second time constant, i.e. $t = 2RC$, and substituting this t in equation (1), we obtain
$
   \Rightarrow v = V\left( {1 - {e^{ - \dfrac{{2RC}}{{RC}}}}} \right) \\
   \Rightarrow v = V\left( {1 - {e^{ - 2}}} \right) \\
 $
Again substituting $e = 2.718$,
$
   \Rightarrow v = V\left( {1 - 0.1353} \right) \\
   \Rightarrow v = V\left( {0.8646} \right) \\
   \Rightarrow v \simeq 0.865V \\
 $
This means that after the second time constant, i.e. $t = 2RC$, we have managed to raise the voltage to 0.865 times of V. So, this time we have increased the voltage to $0.865 \times 100 = 86.5\% $ of V. We again need to move to subsequent time constants.
Taking third time constant, i.e. $t = 3RC$, and substituting this t in equation (1), we obtain
$
   \Rightarrow v = V\left( {1 - {e^{ - \dfrac{{3RC}}{{RC}}}}} \right) \\
   \Rightarrow v = V\left( {1 - {e^{ - 3}}} \right) \\
 $
Again substituting $e = 2.718$,
$
   \Rightarrow v = V\left( {1 - 0.0497} \right) \\
   \Rightarrow v = V\left( {0.95} \right) \\
   \Rightarrow v = 0.95V \\
 $
This means, that after the third time constant, i.e. $t = 3RC$, we have managed to rise the voltage to 0.95 times of V. So, this time we have increased the voltage to $0.95 \times 100 = 95\% $ of V. Although this is close to fully charged but we again need to move to subsequent time constant.
Taking fourth time constant, i.e. $t = 4RC$, and substituting this t in equation (1), we obtain
\[
   \Rightarrow v = V\left( {1 - {e^{ - \dfrac{{4RC}}{{RC}}}}} \right) \\
   \Rightarrow v = V\left( {1 - {e^{ - 4}}} \right) \\
 \]
Again substituting $e = 2.718$,
$
   \Rightarrow v = V\left( {1 - 0.0183} \right) \\
   \Rightarrow v = V\left( {0.9816} \right) \\
   \Rightarrow v \simeq 0.98V \\
 $
This means, that after the fourth time constant, i.e. $t = 4RC$, we have managed to rise the voltage to 0.98 times of V. So, this time we have increased the voltage to $0.98 \times 100 = 98\% $ of V. Here, we can conclude that this is the maximum of the voltage V that we can develop across the capacitor.
Therefore, the capacitor gets almost fully charged during the fourth time constant, i.e. $t = 4RC$.

Option (C) is correct.

Note: As we subsequently keep on increasing time constants to further values, i.e. 5RC, 6RC and so on, the value of rise in voltage across the capacitor will keep on increasing. But keeping in mind the options, 4 times constants or 4RC will be the time required for the capacitor to become almost fully charged.