Question

When a capacitor discharges through a resistance $R$, the time constant is $\tau$ and the maximum current in the circuit is ${i_0}$. (This question has multiple correct options)
A. the initial charge on the capacitor was ${i_0}\tau$
B. the initial charge on the capacitor was $\dfrac{1}{2}{i_0}\tau$
C. the initial energy stored in the capacitor was $i_0^2R\tau$
D. the initial energy stored in the capacitor was $\dfrac{1}{2}i_0^2R\tau$

Answer 1

Use the expression for the instantaneous current in the circuit. Also use the formulae of charge stored on the plates of the capacitor and energy stored in the plates of the capacitor. First determine the initial current in the circuit and then initial charge in the capacitor and then determine the initial energy using all these formulae.

Formulae used:
The instantaneous current $i\left( t \right)$ in the capacitor is given by
$i\left( t \right) = \dfrac{{{V_0}}}{R}{e^{ - t/\tau }}$ …… (1)
Here, ${V_0}$ is the initial potential difference between the plates of the capacitor, $R$ is the resistance, $t$ is the time and $\tau$ is the time constant.
The charge $Q$ stored on the plates of the capacitor is
$Q = CV$ …… (2)
Here, $C$ is the capacitance and $V$ is the potential difference between the plates of the capacitor.
The energy $U$ stored in the plates of the capacitor is
$U = \dfrac{1}{2}C{V^2}$ …… (3)
Here, $C$ is capacitance of the capacitor and $V$ is the potential difference between the plates of the capacitor.

Complete step by step answer:
We have given that the capacitor discharges through a resistance $R$, the time constant is $\tau$ and the maximum current in the circuit is ${i_0}$. Let us first determine the initial charge on the plates of the capacitor. We know that for maximum current in the circuit, we can write using Ohm’s law
${i_0} = \dfrac{{{V_0}}}{R}$
Here, is the initial potential difference between the plates of the capacitor.
Substitute ${i_0}$ for $\dfrac{{{V_0}}}{R}$ in equation (1).
$i\left( t \right) = {i_0}{e^{ - t/\tau }}$

Let us now determine the initial current in the circuit.Substitute $0\,{\text{s}}$ for $t$ in the above equation.
$i\left( {0\,{\text{s}}} \right) = {i_0}{e^{ - \left( {0\,{\text{s}}} \right)/\tau }}$
$\Rightarrow i\left( {0\,{\text{s}}} \right) = {i_0}{e^0}$
$\Rightarrow i\left( {0\,{\text{s}}} \right) = {i_0}\left( 1 \right)$
$\Rightarrow i\left( {0\,{\text{s}}} \right) = {i_0}$
We know that the time constant is given by
$\tau = RC$
$\Rightarrow C = \dfrac{\tau }{R}$

Rewrite equation (2) for the initial charge on the plates of the capacitor.
$Q = C{V_0}$
Substitute $\dfrac{\tau }{R}$ for $C$ and ${i_0}R$for ${V_0}$ in the above equation.
$Q = \left( {\dfrac{\tau }{R}} \right)\left( {{i_0}R} \right)$
$\Rightarrow Q = {i_0}\tau$
Therefore, the initial charge on the plates of the capacitor is ${i_0}\tau$.Hence, option A is correct and option B is incorrect.

Let us now calculate the initial energy stored in the plates of the capacitor. Rewrite equation (3) for the initial energy stored in the plates of the capacitor.
$U = \dfrac{1}{2}CV_0^2$
Substitute $\dfrac{\tau }{R}$ for $C$ and ${i_0}R$for ${V_0}$ in the above equation.
$U = \dfrac{1}{2}\left( {\dfrac{\tau }{R}} \right){\left( {{i_0}R} \right)^2}$
$\therefore U = \dfrac{1}{2}i_0^2R\tau$
Therefore, the initial energy stored in the plates of the capacitor is $\dfrac{1}{2}i_0^2R\tau$. Hence, the option C is incorrect and option D is correct.

Hence, the correct options are A and D.

Note: The students may think that we have not considered the value of the resistance for the initial condition. But the value of the resistance is given in the question that the capacitor discharges through the resistance R. Hence, the resistance is the same for initial and all conditions. Thus, the resistance of the circuit remains the same.