How many time constants will elapse before a current in a charging RC circuit drops to half of its initial value? Answer the same question for a charging RC circuit.
A. 6.9, 1.23
B. 1.23, 0.69
C. 0.69, 1.23
D. 0.69, 0.69

Hint:Use the expression for the instantaneous electric current in a RC charging circuit. Substitute the given condition for the electric current at a time t in this expression and solve this expression for the value of time which is equal to a multiple of the time constant. This multiple of the time constant is the final required answer. Check the expression of the instantaneous electric current for the RC discharging circuit and solve it in the same manner.

Formula used:
The expression for an instantaneous electric current \[i\] in a RC charging circuit is given by
\[i = {i_0}{e^{ - \dfrac{t}{{RC}}}}\] …… (1)
Here, \[{i_0}\] is the initial peak current in the RC circuit, \[t\] is the time, \[R\] is resistance and \[C\] is capacitance.

Complete step by step answer:
We have given that the circuit is a charging RC circuit.We know that the instantaneous electric current \[i\] in a RC charging circuit is given by
\[i = {i_0}{e^{ - \dfrac{t}{{RC}}}}\]
The time constant in the above equation is \[RC\].We have asked to calculate the integral number which is multiple of the time constant before the electric current in the circuit becomes half of its initial value.
\[i = \dfrac{{{i_0}}}{2}\]
Substitute \[\dfrac{{{i_0}}}{2}\] for \[i\] in equation (1).
\[\dfrac{{{i_0}}}{2} = {i_0}{e^{ - \dfrac{t}{{RC}}}}\]
\[ \Rightarrow \dfrac{1}{2} = {e^{ - \dfrac{t}{{RC}}}}\]
\[ \Rightarrow {e^{\dfrac{t}{{RC}}}} = 2\]
\[ \Rightarrow \dfrac{t}{{RC}} = {\log _e}2\]
\[ \Rightarrow \dfrac{t}{{RC}} = 0.69\]
\[ \therefore t = 0.69RC\]
Hence, the number of time constants that will elapse before the electric current in the RC charging circuit drops to half of its initial value is 0.69.The value of the instantaneous current for the RC discharging circuit is the same as that of the instantaneous current for the RC charging circuit.Thus, we will obtain the same value for the number of time constant that will elapse for the RC discharging circuit.Therefore, the number of time constant that will elapse before the electric current in the RC charging and RC discharging circuit drops to half of its initial value is 0.69.

Hence, the correct option is D.

Note: One can also solve the same question by another method. One can use the expression for the instantaneous charge for the RC charging and RC discharging circuit and solve these two expressions for the half of the value of the initial charge as the electric current and charge increase or decrease simultaneously. Although the value of instantaneous electric current for the RC charging and discharging circuit is the same, it is different for the charge.