Chapter 25, Problem 118P

(a)

To determine

The comparison of total energy stored in the two capacitors.

Answer to Problem 118P

The comparison of total energy stored in the two capacitors is $\frac{U_{\text{i}}}{U_{\text{f}}}=1+\frac{C_2}{C_1}$ .

Explanation of Solution

Formula Used:

The expression for the charge on the capacitor is given by,

  $Q=C{V}_0$

The expression for the potential of the capacitor of the final stage.

  $\frac{Q_1}{C_1}=\frac{Q_2}{C_2}$

The expression for the total charge on the capacitor is given by,

  $Q=Q_1+Q_2$

The expression for the initial energies stored in the capacitor is given by,

  $U_{\text{i}}=\frac{1}{2}{C}_1{V}_0^2$ …… (I)

The expression for the final energies stored in the capacitor is given by,

  $U_{\text{f}}=\frac{1}{2}\frac{Q_1^2}{C_1}+\frac{1}{2}\frac{Q_2^2}{C_2}$ …… (II)

The expression for the charge $Q_1$ is given by,

  $Q_1=\frac{C_1^2}{C_1+C_2}{V}_0$

The expression for charge $Q_2$ is given by,

  $Q_2=\frac{C_1^2}{C_1+C_2}{V}_0$

Equation (II) can be re written as,

  $\begin{array}{c}{U}_{\text{f}}=\frac{1}{2}\frac{Q_1^2}{C_1}+\frac{1}{2}\frac{Q_2^2}{C_2}\\ =\frac{1}{2}\frac{{\left(\frac{C_1^2}{C_1+C_2}{V}_0\right)}^2}{C_1}+\frac{1}{2}\frac{{\left(\frac{C_1{C}_2}{C_1+C_2}{V}_0\right)}^2}{C_2}\\ =\frac{1}{2}\frac{C_1^2}{C_1+C_2}{V}_0^2\end{array}$

The expression for the ratios of initial and final energies is given by,

  $\begin{array}{c}\frac{U_{\text{i}}}{U_{\text{f}}}=\frac{\frac{1}{2}{C}_1{V}_0^2}{\frac{1}{2}\frac{C_1^2}{C_1+C_2}{V}_0^2}\\ =1+\frac{C_2}{C_1}\end{array}$

Conclusion:

Therefore, the comparison of total energy stored in the two capacitors is $\frac{U_{\text{i}}}{U_{\text{f}}}=1+\frac{C_2}{C_1}$ .

(b)

To determine

The current through $R$ as a function of $t$ .

Answer to Problem 118P

The current through $R$ as a function of $t$ is $I=\frac{V_0}{R}{e}^{-\frac{t}{\tau }}$ .

Explanation of Solution

Formula Used:

The expression for the current passing through the circuit is given by,

  $I=\frac{d{q}_2}{dt}$

The expression Kirchhoff’s loop rule to circuit when the switch is at $a$ is given by,

  $\begin{array}{c}\frac{q_1}{C_1}-IR-\frac{q_2}{C_2}=0\\ \frac{q_1}{C_1}-R\frac{d{q}_2}{dt}-\frac{q_2}{C_2}=0\end{array}$ ……. (III)

The expression for the law of the charge on the capacitors is given by,

  $\begin{array}{c}{q}_1=Q-q_2\\ =C_1{V}_0-q_2\end{array}$

Equation (III) can be re written as,

  $\begin{array}{c}\frac{q_1}{C_1}-R\frac{d{q}_2}{dt}-\frac{q_2}{C_2}=0\\ V_0-\frac{q_2}{C_1}-R\frac{d{q}_2}{dt}-\frac{q_2}{C_2}=0\\ V_0=\frac{q_2}{C_1}+R\frac{d{q}_2}{dt}+\frac{q_2}{C_2}\end{array}$ …… (IV)

Let the solution for the differential equation (IV) be,

  $q_2\left(t\right)=a+b{e}^{-\frac{t}{\tau }}$

So,

  $\begin{array}{c}\frac{d{q}_2\left(t\right)}{dt}=\frac{d}{dt}\left(a+b^{-\frac{t}{\tau }}\right)\\ =-\frac{b}{\tau }{e}^{-\frac{t}{\tau }}\end{array}$

Equation (IV) can be rewritten as,

  $R\left(-\frac{b}{\tau }{e}^{-\frac{t}{\tau }}\right)+\left(\frac{C_1+C_2}{C_1{C}_2}\right)\left(a+b{e}^{-\frac{t}{\tau }}\right)=V_0$

Compare with solution of the equation.

  $\begin{array}{c}\left(\frac{C_1+C_2}{C_1{C}_2}\right)a=V_0\\ a=V_0\left(\frac{C_1{C}_2}{C_1+C_2}\right)\\ =V_0{C}_{\text{eq}}\end{array}$

And,

  $\begin{array}{c}R\left(-\frac{b}{\tau }{e}^{-\frac{t}{\tau }}\right)+\left(\frac{C_1+C_2}{C_1{C}_2}\right)\left(b{e}^{-\frac{t}{\tau }}\right)=0\\ -\frac{R}{\tau }+\frac{C_1+C_2}{C_1{C}_2}=0\\ \tau =R\left(\frac{C_1{C}_2}{C_1+C_2}\right)\\ =R{C}_{\text{eq}}\end{array}$

From the initial condition,

  $\begin{array}{l}a+b=\\ b=-a\\ =-V_0{C}_{\text{eq}}\end{array}$

The solution of the differential equation is,

  $\begin{array}{c}{q}_2\left(t\right)=a+b{e}^{-\frac{t}{\tau }}\\ =C_{\text{eq}}{V}_0+\left(-C_{\text{eq}}{V}_0\right)e^{-\frac{t}{\tau }}\\ =C_{\text{eq}}{V}_0\left(1-e^{-\frac{t}{\tau }}\right)\end{array}$

The current in the circuit is given by,

  $\begin{array}{c}I=\frac{d}{dt}\left(C_{\text{eq}}{V}_0\left(1-e^{-\frac{t}{\tau }}\right)\right)\\ =\frac{C_{\text{eq}}{V}_0}{\tau }{e}^{-\frac{t}{\tau }}\\ =\frac{V_0}{R}{e}^{-\frac{t}{\tau }}\end{array}$

Conclusion:

Therefore, the current through $R$ as a function of $t$ is $I=\frac{V_0}{R}{e}^{-\frac{t}{\tau }}$ .

(c)

To determine

The energy delivered to resistor as a function of $t$ .

Answer to Problem 118P

The energy delivered to resistor as a function of $t$ is $P\left(t\right)=\frac{V_0^2}{R}{e}^{-\frac{t}{\tau }}$ .

Explanation of Solution

Formula Used:

The current through $R$ as a function of $t$ is given by,

  $I=\frac{V_0}{R}{e}^{-\frac{t}{\tau }}$

The expression for the energy dissipated through the resistor is given by,

  $\begin{array}{c}P\left(t\right)=I^{2}R\\ ={\left(\frac{V_0}{R}{e}^{-\frac{t}{\tau }}\right)}^{2}R\\ =\frac{V_0^2}{R}{e}^{-\frac{t}{\tau }}\end{array}$

Conclusion:

Therefore, the energy delivered to resistor as a function of $t$ is $P\left(t\right)=\frac{V_0^2}{R}{e}^{-\frac{t}{\tau }}$ .

(d)

To determine

The total energy dissipated.

Answer to Problem 118P

The total energy dissipated is $E=\frac{1}{2}\left(\frac{C_1{C}_2}{C_1+C_2}\right)V_0^2$ .

Explanation of Solution

Formula Used:

The expression for the energy dissipated through the resistor is given by,

  $P\left(t\right)=\frac{V_0^2}{R}{e}^{-\frac{t}{\tau }}$

The expression for the total energy dissipated through the resistor is given by,

  $\begin{array}{c}E={\displaystyle {\int }_0^{\infty }P\left(t\right)dt}\\ ={\displaystyle {\int }_0^{\infty }\frac{V_0^2}{R}{e}^{-\frac{t}{\tau }}}\\ =\frac{1}{2}{V}_0^2{C}_{\text{eq}}\\ =\frac{1}{2}\left(\frac{C_1{C}_2}{C_1+C_2}\right)V_0^2\end{array}$

Conclusion:

Therefore, the total energy dissipated is $E=\frac{1}{2}\left(\frac{C_1{C}_2}{C_1+C_2}\right)V_0^2$ .