The time required for a \[50\,{\text{Hz}}\] alternating current to increases from zero to \[70.7\% \] of its peak value is:
A. \[2.5\,{\text{ms}}\]
B. \[10\,{\text{ms}}\]
C. \[20\,{\text{ms}}\]
D. \[14.14\,{\text{ms}}\]

Hint: Use the formula for the instantaneous current in a circuit. This formula gives the relation between the instantaneous current in the circuit, peak current, angular frequency and time. Substitute the value of the 70.7% of the peak value current in this formula and then determine the time required for the current to increase to this value of current.

Formula used:
The instantaneous current \[I\] in a circuit is given by
\[I = {I_0}\sin \left( {\omega t} \right)\] …… (1)
Here, \[{I_0}\] is the peak value of the current, \[\omega \] is angular frequency and \[t\] is time.

Complete step by step answer:
We have given that the frequency of the alternating current is \[50\,{\text{Hz}}\].
\[f = 50\,{\text{Hz}}\]
The current increases from zero to \[70.7\% \] of the peak value of the current.We have asked to calculate the time required for the current to reach \[70.7\% \] of the peak value of current from zero.The \[70.7\% \] of the peak value of the current is
\[I = \dfrac{{70.7}}{{100}}{I_0}\]

Let us calculate the time required for the current to reach the required value using equation (1).Substitute \[\dfrac{{70.7}}{{100}}{I_0}\] for \[I\] in equation (1).
\[\dfrac{{70.7}}{{100}}{I_0} = {I_0}\sin \left( {\omega t} \right)\]
\[ \Rightarrow \dfrac{{70.7}}{{100}} = \sin \left( {\omega t} \right)\]
\[ \Rightarrow \sin \left( {\omega t} \right) = 0.707\]
\[ \Rightarrow \omega t = {\sin ^{ - 1}}\left( {0.707} \right)\]
\[ \Rightarrow \omega t = {\sin ^{ - 1}}\left( {\dfrac{1}{{\sqrt 2 }}} \right)\]
\[ \Rightarrow \omega t = \dfrac{\pi }{4}\]
\[ \Rightarrow t = \dfrac{\pi }{{4\omega }}\]

Substitute \[2\pi f\] for \[\omega \] in the above equation.
\[ \Rightarrow t = \dfrac{\pi }{{4\left( {2\pi f} \right)}}\]
\[ \Rightarrow t = \dfrac{1}{{8f}}\]
Substitute \[50\,{\text{Hz}}\] for \[f\] in the above equation.
\[ \Rightarrow t = \dfrac{1}{{8\left( {50\,{\text{Hz}}} \right)}}\]
\[ \Rightarrow t = 0.0025\,{\text{s}}\]
\[ \Rightarrow t = 2.5 \times {10^{ - 3}}\,{\text{s}}\]
\[ \therefore t = 2.5\,{\text{ms}}\]
Therefore, the time required for the current to increase to the required value is \[2.5\,{\text{ms}}\].

Hence, the correct option is A.

Note:The students should keep in mind that we should use the correct instantaneous value of the current in the circuit. If the formula for the instantaneous current is not used correctly then we will end with the incorrect value of the time in which the current increases from zero to the required value of the current.