Question: A capacitor C has been charged up to potential ${\mathit{V}}_{\mathbf{0}}$at time $\mathbf{}\mathit{t}\mathbf{=}\mathbf{0}$, it is connected to a resistor R, and begins to discharge (Fig. 7.5a).

(a) Determine the charge on the capacitor as a function of time,$\mathit{Q}\mathbf{\left(}\mathit{t}\mathbf{\right)}$What is the current through the resistor,$\mathit{l}\mathbf{\left(}\mathit{t}\mathbf{\right)}$?

(b) What was the original energy stored in the capacitor (Eq. 2.55)? By integrating Eq. 7.7, confirm that the heat delivered to the resistor is equal to the energy lost by the capacitor.

Now imagine charging up the capacitor, by connecting it (and the resistor) to a battery of voltage localid="1657603967769" ${\mathit{V}}_{\mathbf{0}}$, at time t = 0 (Fig. 7.5b).

(c) Again, determine localid="1657603955495" $\mathit{Q}\mathbf{\left(}\mathit{t}\mathbf{\right)}$and $\mathit{l}\mathbf{\left(}\mathit{t}\mathbf{\right)}$.

(d) Find the total energy output of the battery $\left(\int Vldt\right)$. Determine the heat delivered to the resistor. What is the final energy stored in the capacitor? What fraction of the work done by the battery shows up as energy in the capacitor? [Notice that the answer is independent of R!]

The capacitor is $C$charged up to potential $V_0$at time t = 0

The resistance is$R$ .

The current in the circuit$l$ .

The equation to calculate the potential difference across the capacitor is given as follows.

$\mathbf{V}\mathbf{=}\frac{\mathbf{Q}}{\mathbf{C}}$ …… (1)

Here, is$\mathit{Q}$ the charge of the capacitor.

The relationship between the current, voltage and resistance is given as follows.

$\mathbf{V}\mathbf{=}\mathbf{IR}$ …… (2)

The equation to calculate the current through the resistor is given as follows.

$\mathbf{I}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\frac{\mathbf{9}\mathbf{dQ}\mathbf{\left(}\mathbf{t}\mathbf{\right)}}{\mathbf{dt}}$

The equation to calculation the initial charge in the capacitor is given as follows.

${\mathbf{Q}}_{\mathbf{0}}\mathbf{=}{\mathbf{CV}}_{\mathbf{0}}$

The equation to calculate the work done is given as follows.

$\mathbf{W}\mathbf{=}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{\infty }}{\mathbf{V}}_{\mathbf{0}}\mathbf{l}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{dt}$

(a)

Equate the equations (1) and (2).

$$\begin{gathered} IR=\frac{Q}{C} \\ I=\frac{Q}{RC} \end{gathered}$$

Here l is positive, therefore the charge of capacitor is decreasing.

The current is defined as the rate of charge of moving charge.

$\frac{dQ}{dt}=-l$

Substitute $\frac{Q}{RC}$for l into above equation,

$\frac{dQ}{dt}=-\frac{Q}{RC}$

By integrating the above equation from $Q_0$to $Q\left(t\right)$.

$$\begin{gathered} {\int }_{Q_{\circ }}^{Q\left(t\right)}\frac{dQ}{Q}={\int }_0^t-\frac{dt}{RC} \\ \mathrm{In}{\left(Q\right)}_{Q_{\circ }}^{Q\left(t\right)}=-\frac{1}{RC}{\left(t\right)}_0^t+\mathrm{InA} \\ \mathrm{In}(Q\left(t\right)-Q_0)=-\frac{1}{RC}(t-0)+\mathrm{InA} \\ \mathrm{In}\left(\frac{Q\left(t\right)}{Q_0}\right)=-\frac{1}{RC}\left(t\right)+\mathrm{InA} \end{gathered}$$ ….. (3)

Here is A the integration constant.

At t = 0,Q(t) = 0

$$\begin{gathered} \mathrm{In}\left(\frac{Q\left(0\right)}{Q_0}\right)=-\frac{1}{RC}\left(0\right)+\mathrm{InA} \\ \mathrm{In}\left(\frac{0}{Q_0}\right)=-\frac{1}{RC}\left(0\right)+\mathrm{InA} \\ \mathrm{InA}=0 \end{gathered}$$

Substitute 0 for A into equation (3)

$$\begin{gathered} \mathrm{In}\left(\frac{Q\left(t\right)}{Q_0}\right)=-\frac{1}{RC}\left(t\right)+0 \\ \left(\frac{Q\left(t\right)}{Q_0}\right)=e^{-\frac{1}{RC}\left(t\right)} \\ Q\left(t\right)=Q_0{e}^{-\frac{1}{RC}} \end{gathered}$$

Substitute ${CV}_0$for $Q_0$in above equation.

$Q\left(t\right)=C{V}_0{e}^{-\frac{1}{RC}}$

Therefore, the charge on the capacitor is $Q\left(t\right)=C{V}_0{e}^{-\frac{1}{RC}}$.

Calculate the current through the resistor.

Substitute$C{V}_0{e}^{-\frac{1}{RC}}$for Q(t) into equation (3).

$$\begin{gathered} l\left(t\right)=-\frac{d\left(C{V}_0{e}^{-\frac{1}{RC}}\right)}{dt} \\ l\left(t\right)=-C{V}_0\frac{d}{dt}\left(e^{-\frac{1}{RC}}\right) \\ l\left(t\right)=-C{V}_0\left(-\frac{1}{RC}\right)e^{-\frac{1}{RC}} \\ l\left(t\right)=\frac{V_0}{R}{e}^{-\frac{1}{RC}} \end{gathered}$$

Therefore, the current through the resistor is $l\left(t\right)=\frac{V_0}{R}{e}^{-\frac{1}{RC}}$.

Hence the charge on the capacitor Q(t) is$C{V}_0{e}^{-\frac{1}{RC}}$ and current through resistor is$\frac{V_0}{R}{e}^{-\frac{1}{RC}}$ .

(b)

By the equation 7.7,

$P=l^{2}R$

Integrate the above equation.0.

${\int }_0^{\infty }Pdt={\int }_0^{\infty }{l}^{2}Rdt$

Substitute$\frac{V_0}{R}{e}^{-\frac{t}{RC}}$ for l into above equation.

$$\begin{gathered} {\int }_0^{\infty }Pdt={\int }_0^{\infty }{\left(\frac{V_0}{R}{e}^{\frac{t}{RC}}\right)}^{2}Rdt \\ {\int }_0^{\infty }Pdt={\left(\frac{V_0}{R}\right)}^{2}R{\int }_0^{\infty }\left(e^{-\frac{2t}{RC}}\right)dt \\ {\int }_0^{\infty }Pdt=\frac{V_0^2}{R}{\left(-\frac{RC}{2}{e}^{-\frac{2t}{RC}}\right)}_0^{\infty } \end{gathered}$$

Apply the limits,

$$\begin{gathered} {\int }_0^{\infty }Pdt=-\frac{C{V}_0^2}{2}{\left(e^{\infty }-e^0\right)}_0^{\infty } \\ {\int }_0^{\infty }Pdt=-\frac{C{V}_0^2}{2}\left(0-1\right) \\ {\int }_0^{\infty }Pdt=\frac{1}{2}C{V}_0^2 \end{gathered}$$

Hence the original energy stored in the capacitor is$\frac{1}{2}C{V}_0^2$ and by integrating the above equation 7.7, it is confirmed that the heat delivered to the resistor is equal to the energy lost by the capacitor.

(c)

The total charge on capacitor is given by,

$$\begin{gathered} V_0=\frac{Q\left(t\right)}{C}+l\left(t\right)r \\ {V}_0-\frac{Q\left(t\right)}{C}=l\left(t\right)r \\ \frac{C{V}_0-Q\left(t\right)}{C}=l\left(t\right)r \\ l\left(t\right)=\frac{1}{RC}\left[C{V}_0-Q\left(t\right)\right] \end{gathered}$$

The current is given by,

$l=\frac{dQ}{dt}$

Substitute $\frac{1}{RC}\left[C{V}_0-Q\left(t\right)\right]$for l(t) into above equation.

$$\begin{gathered} \frac{dQ}{dt}=\left\{\frac{1}{RC}\left[C{V}_0-Q\left(t\right)\right]\right\} \\ \frac{1}{C{V}_0-Q\left(t\right)}dQ=\frac{1}{RC}dt \end{gathered}$$

Integrate the above equation,

$$\begin{gathered} {\int }_0^Q\frac{1}{C{V}_0-Q\left(t\right)}dQ=\frac{1}{RC}{\int }_0^{t}dt \\ \mathrm{In}\left(Q\right(t)-C{V}_0)=-\frac{1}{RC}t \\ \mathrm{In}\left(Q\right(t)-C{V}_0)=-\frac{1}{RC}t+\mathrm{In}k \\ \mathrm{In}\left(Q\right(t)-C{V}_0)-\mathrm{In}k=-\frac{1}{RC} \end{gathered}$$

Here In k is the integration constant.

$$\begin{gathered} \mathrm{In}\left(\frac{\mathrm{Q}\left(\mathrm{t}\right)-{\mathrm{CV}}_0}{\mathrm{k}}\right)=-\frac{1}{RC}t \\ \frac{\mathrm{Q}\left(\mathrm{t}\right)-{\mathrm{CV}}_0}{\mathrm{k}}=e^{-\frac{t}{RC}} \\ \mathrm{Q}\left(\mathrm{t}\right)-{\mathrm{CV}}_0=k{e}^{-\frac{t}{RC}} \\ \mathrm{Q}\left(\mathrm{t}\right)=k{e}^{-\frac{t}{RC}}+C{V}_0 \end{gathered}$$ ….. (4)

For t = 0 , Q(0) = 0

$$\begin{gathered} Q\left(0\right)=k{e}^{-\frac{0}{RC}}+C{V}_0 \\ 0=k+C{V}_0 \\ k=-C{V}_0 \end{gathered}$$

Substitute $-C{V}_0$for k into equation (4).

$$\begin{gathered} Q\left(t\right)=-C{V}_0{e}^{-\frac{0}{RC}}+C{V}_0 \\ Q\left(t\right)=C{V}_0\left(1-e^{-\frac{0}{RC}}\right) \end{gathered}$$

Therefore, the charge on the capacitor is $Q\left(t\right)=C{V}_0\left(1-e^{-\frac{0}{RC}}\right)$
.

The current can be calculated as,

$l\left(t\right)=\frac{dq\left(t\right)}{dt}$

Substitute $C{V}_0\left(1-e^{-\frac{t}{RC}}\right)$for Q(t) into above equation.

$$\begin{gathered} l\left(t\right)=\frac{d}{dt}\left[C{V}_0\left(1-e^{-\frac{t}{RC}}\right)\right] \\ l\left(t\right)=-C{V}_0\left(-\frac{1}{RC}{e}^{-\frac{t}{RC}}\right) \\ l\left(t\right)=\frac{V_0}{C}{e}^{-\frac{t}{RC}} \end{gathered}$$

Therefore, the expression for the current is$l\left(t\right)=\frac{V_0}{C}{e}^{-\frac{t}{RC}}$ .

Hence the charge on the capacitor Q(t) is$C{V}_0\left(1-e^{-\frac{t}{RC}}\right)$ and expression for current is$l\left(t\right)=\frac{V_0}{C}{e}^{-\frac{t}{RC}}$ .

(d)

The work done is given by,

$W={\int }_0^{\infty }{V}_{0}l\left(t\right)dt$

Substitute$\frac{V_0}{R}{e}^{-\frac{t}{Rc}}$for l(t) into above equation.

$$\begin{gathered} W={\int }_0^{\infty }\frac{V_0}{R}{e}^{-\frac{t}{Rc}}dt \\ W=\frac{V_0^2}{R}{\int }_0^{\infty }{e}^{-\frac{t}{Rc}}dt \\ W=\frac{V_0^2}{R}{\left(-RC{e}^{-\frac{t}{Rc}}\right)}_0^{\infty } \end{gathered}$$

Apply the limits,

$$\begin{gathered} W=-C{V}_0^2\left(e^{-\frac{\infty }{RC}}-e{e}^{-\frac{0}{Rc}}\right) \\ W=-C{V}_0^2(0-1) \\ W=C{V}_0^2 \end{gathered}$$

The total output energy is given by,

$E=\frac{1}{2}C{V}_0^2$

Substitute W for$C{V}_0^2$ into above equation.

$E=\frac{1}{2}W$

Hence, the total output energy of the battery is $\frac{1}{2}C{V}_0^2$and half of the work done is equal to the total energy of the battery.