A$\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\mu F}$capacitor with an initial stored energy of$\mathbf{0}\mathbf{.}\mathbf{50}\mathbf{\text{J}}$is discharged through a$\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{M\Omega }$resistor. (a) What is the initial charge on the capacitor? (b) What is the current through the resistor when the discharge starts? Find an expression that gives, as a function of time$\mathit{t}$, (c) the potential difference${\mathbf{V}}_{\mathbf{c}}$across the capacitor, (d) the potential difference${\mathit{V}}_{\mathbf{R}}$across the resistor, and (e) the rate at which thermal energy is produced in the resistor.

a) Capacitance of the capacitor,$C=1.0\times {10}^{-6}\text{F}$

b) Initial stored energy, $U=0.50\text{J}$

c) Resistance of the resistor,$R=1.0\times {10}^6\Omega$

Using the concept of energy being stored within the plates of the capacitor, the charge within the plates can be calculated. Now, the charging and discharging of the capacitor plates and the amount of charge flowing through the resistor given in the capacitor branch, the voltage equation of the capacitor and the resistor can be given individually. Now, the rate at which the energy is generated thermally can be calculated using the current equation across the resistor.

Formulae:

The energy stored in the capacitor, $U=\frac{q^2}{2C}$ (i)

The charge equation of a RC circuit, $q=q_0{e}^{-t/\tau }$ (ii)

The time constant of the RC circuit, $\tau =RC$ (iii)

The current flowing within the given time, $i=\frac{dq}{dt}$ (iv)

The charge stored within the plates of a capacitor, $q=CV$ (v)

The voltage equation using Ohm’s law, $V=IR$ (vi)

The power generated in the resistor, $P=i^{2}R$ (vii)

(a)

Using equation (i), the initial charge present within the plates of the capacitor can be given as follows:

$$\begin{gathered} q_0=\sqrt{2UC} \\ =\sqrt{2\left(0.50\text{J}\right)\left(1.0\times {10}^{-6}\text{F}\right)} \\ =1.0\times {10}^{-3}\text{C} \end{gathered}$$

Hence, the value of the charge is$1.0\times {10}^{-3}\text{C}$.

(b)

Using equation (ii) in equation (iv), the current expression through the resistor can be given as follows:

$$\begin{gathered} i=\frac{d\left(q_0{e}^{-t/\tau }\right)}{dt} \\ =\frac{q_0}{\tau }{e}^{-t/\tau }.....................\left(a\right) \end{gathered}$$

Now, the time constant of the capacitor can be given using the given data in equation (iii) as follows:

$$\begin{gathered} \tau =\left(1.0\times {10}^6\Omega \right)\left(1.0\times {10}^{-6}\text{F}\right) \\ =1.0\text{s} \end{gathered}$$

Now, using the above data and the given data in equation (iv), the initial current value through the resistor can be calculated as follows:

$$\begin{gathered} i_0=\frac{q_0}{\tau } \\ =\frac{\left(1.0\times {10}^{-3}\text{C}\right)}{1.0\text{s}} \\ =1.0\times {10}^{-3}\text{A} \end{gathered}$$

Hence, the expression of the current through the resistor is$i=-\frac{q_0}{\tau }{e}^{-t/\tau }$ and the value of the initial current is$1.0\times {10}^{-3}\text{A}$.

(c)

Substituting equation (ii) in equation (v), the potential difference across the capacitor can be given as follows:

$$\begin{gathered} V_C=\frac{q_0}{C}{e}^{-t/\tau } \\ =\frac{\left(1.0\times {10}^{-3}\text{C}\right)}{\left(1.0\times {10}^{-6}\text{F}\right)}{e}^{-t/1.0\text{s}} \\ =\left(1.0\times {10}^3\text{V}\right)e^{-t} \end{gathered}$$

Hence, the value of the potential difference is$\left(1.0\times {10}^3\text{V}\right)e^{-t}$.

(d)

Now, the voltage expression across the resistor can be given using equation (a) in equation (vi) as follows:

$$\begin{gathered} V_R=\left(\frac{q_0}{\tau }{e}^{-t/\tau }\right)R \\ =\frac{\left(1.0\times {10}^{-3}C\right){\displaystyle \left(1.0\times {10}^5\Omega \right)}}{\left(1.0_S\right)}{e}^{-t/1.0\text{s}} \\ =\left(1.0\times {10}^3\text{V}\right)e^{-t} \end{gathered}$$

Hence, the potential difference across the resistor is$\left(1.0\times {10}^3\text{V}\right)e^{-t}$.

(e)

Now, the rate of the thermal energy produced in the resistor can be given using equation (a) in equation (vii) as follows:

$$\begin{gathered} P={\left(\frac{q_0}{\tau }{e}^{-t/\tau }\right)}^{2}R \\ =\frac{{\displaystyle {\left(1.0\times {10}^{-3}C\right)}^2}{\displaystyle \left(1.0\times {10}^6\Omega \right)}}{\left(1.0s\right)}{e}^{-2t/1.0\text{s}} \\ =\left(1.0\text{W}\right)e^{-2t} \end{gathered}$$

Hence, the value of the power generated is$\left(1.0\text{W}\right)e^{-2t}$.