Chapter 25, Problem 115P

(a)

To determine

The time constant for the charging of capacitor.

Answer to Problem 115P

The time constant for the charging of capacitor is $10.0\text{ms}$ .

Explanation of Solution

Given:

The given values in the circuit are,

  $\begin{array}{c}\epsilon =800\text{V}\\ V_{\text{c}}=4.20\text{V}\\ R_2=1.00\text{m}\Omega \\ R_1=0.500\text{M}\Omega \end{array}$

And,

  $C=20.0\text{nF}$

Formula Used:

The expression for the time constant for charging the capacitor is given by,

  $\tau =R_{1}C$

Calculation:

The time constant for charging the capacitor is calculated as,

  $\begin{array}{c}\tau =R_{1}C\\ =\left(0.500\text{M}\Omega \times \frac{{10}^6\Omega }{1\text{M}\Omega }\right)\left(20.0\text{nF}\times \frac{{10}^{-9}\text{F}}{1\text{nF}}\right)\\ =10.0\times {10}^{-3}\text{s}\times \frac{{10}^3\text{ms}}{1\text{s}}\\ =10.0\text{ms}\end{array}$

Conclusion:

Therefore, the time constant for the charging of capacitor is $10.0\text{ms}$ .

(b)

To determine

The proof that the potential across switch increases from $0.200\text{V}$ to $4.20\text{V}$ .

Answer to Problem 115P

The proof that as the potential across switch increases from $0.200\text{V}$ to $4.20\text{V}$ , the potential across the capacitor increases linearly with time is stated below.

Explanation of Solution

Given:

The given values in the circuit are,

  $\begin{array}{c}\epsilon =800\text{V}\\ V_{\text{c}}=4.20\text{V}\\ R_2=1.00\text{m}\Omega \\ R_1=0.500\text{M}\Omega \end{array}$

And,

  $C=20.0\text{nF}$

Formula Used:

The expression for the voltage across the charging capacitor is given by,

  $\begin{array}{c}V\left(t\right)=\epsilon \left(1-e^{-t/\tau }\right)\\ e^{-t/\tau }=1-\frac{V\left(t\right)}{\epsilon }\end{array}$ ....... (I)

Now, $V\left(t\right)<<\epsilon$ .

Let ,

  $\eta =\frac{V\left(t\right)}{\epsilon }$

Rewrite equation (I) as,

  $\begin{array}{c}{e}^{-t/\tau }=1-\eta \\ e^{t/\tau }={\left(1-\eta \right)}^{-1}\\ \approx 1+\eta \end{array}$ ....... (II)

Expand the term $e^{-t/\tau }$ as,

  $\begin{array}{c}{e}^{t/\tau }=1+\frac{1}{\tau }+\frac{1}{2!}{\left(\frac{t}{\tau }\right)}^2+\mathrm{...}\\ \approx 1+\frac{1}{\tau }t\end{array}$

Equation (II) can be written as,

  $\begin{array}{c}1+\frac{1}{\tau }t=1+\eta \\ =1+\frac{V\left(t\right)}{\epsilon }\\ V\left(t\right)=\frac{\epsilon }{\tau }t\end{array}$

This is a linear function.

Conclusion:

Therefore, the proof that as the potential across switch increases from $0.200\text{V}$ to $4.20\text{V}$ , the potential across the capacitor increases linearly with time is stated above.

(c)

To determine

The changed value of $R_1$ so that the capacitor charges from $0.200\text{V}$ to $4.20\text{V}$ in $0.100\text{s}$ .

Answer to Problem 115P

The changed value of $R_1$ so that the capacitor charges from $0.200\text{V}$ to $4.20\text{V}$ in $0.100\text{s}$ is $1.00\text{G}\Omega$ .

Explanation of Solution

Given:

The given values in the circuit are,

  $\begin{array}{c}\epsilon =800\text{V}\\ V_{\text{c}}=4.20\text{V}\\ R_2=1.00\text{m}\Omega \\ R_1=0.500\text{M}\Omega \end{array}$

And,

  $C=20.0\text{nF}$

Formula Used:

The expression for the potential across the capacitor is given by

  $\begin{array}{c}\Delta V\left(t\right)=\frac{\epsilon }{\tau }\Delta t\\ =\frac{\epsilon }{R_{1}C}\Delta t\\ R_1=\frac{\epsilon }{C\Delta V\left(t\right)}\Delta t\end{array}$

Calculation:

The changed value of $R_1$ is calculated as,

  $\begin{array}{c}{R}_1=\frac{\epsilon }{C\Delta V\left(t\right)}\Delta t\\ =\frac{\left(800\text{V}\right)\left(0.10\text{s}\right)}{\left(20.0\text{nF}\right)\left(4.2\text{V}-0.2\text{V}\right)}\\ =1\times {10}^9\Omega \times \frac{{10}^{-9}\text{GΩ}}{1\Omega }\\ =1.00\text{G}\Omega \end{array}$

Conclusion:

Therefore, the changed value of $R_1$ so that the capacitor charges from $0.200\text{V}$ to $4.20\text{V}$ in $0.100\text{s}$ is $1.00\text{G}\Omega$ .

(d)

To determine

The time elapse during the discharge of capacitor when the switch $S$ closes.

Answer to Problem 115P

The time elapse during the discharge of capacitor when the switch $S$ closes is $60.9\text{ps}$ .

Explanation of Solution

Given:

The given values in the circuit are,

  $\begin{array}{c}\epsilon =800\text{V}\\ V_{\text{c}}=4.20\text{V}\\ R_2=1.00\text{m}\Omega \\ R_1=0.500\text{M}\Omega \end{array}$

And,

  $C=20.0\text{nF}$

Formula Used:

The expression for the potential difference across the capacitor is given by,

  $V_{\text{C}}\left(t\right)=V_{\text{c0}}{e}^{-t/{\tau }^{\prime }}$ ....... (III)

The expression for ${\tau }^{\prime }$ is given by,

  ${\tau }^{\prime }=R_{2}C$

Rewrite equation (III) as,

  $\begin{array}{c}{V}_{\text{C}}\left(t\right)=V_{\text{c0}}{e}^{-t/{\tau }^{\prime }}\\ V_{\text{C}}\left(t\right)=V_{\text{c0}}{e}^{-t/R_{2}C}\\ t=-R_{2}C\ln \left(\frac{V_{\text{C}}\left(t\right)}{V_{\text{C0}}}\right)\end{array}$

Calculation:

The time elapsed during the discharge of capacitor is calculated as,

  $\begin{array}{c}t=-R_{2}C\ln \left(\frac{V_{\text{C}}\left(t\right)}{V_{\text{C0}}}\right)\\ =-\left(0.001\Omega \right)\left(20.0\text{nF}\right)\ln \left(\frac{0.2\text{V}}{4.2\text{V}}\right)\\ =0.06089\times {10}^{-9}\text{s}\times \frac{{10}^{12}\text{ps}}{1\text{s}}\\ =60.9\text{ps}\end{array}$

Conclusion:

Therefore, the time elapse during the discharge of capacitor when the switch $S$ closes is $60.9\text{ps}$ .

(e)

To determine

The average rate at which the energy is delivered to resistor $R_1$ and to discharge the resistance $R_2$ .

Answer to Problem 115P

The average rate at which the energy is delivered to resistor $R_1$ is $6.17\times {10}^{-9}\text{W}$ and to switch the resistance $R_2$ to discharge is $2.89\text{kW}$ .

Explanation of Solution

Given:

The given values in the circuit are,

  $\begin{array}{c}\epsilon =800\text{V}\\ V_{\text{c}}=4.20\text{V}\\ R_2=1.00\text{m}\Omega \\ R_1=0.500\text{M}\Omega \end{array}$

And,

  $C=20.0\text{nF}$

Formula Used:

The expression for the rate at which energy is dissipated in $R_1$ is given by.

  $P_1=\frac{\Delta E_1}{\Delta t}=I^2{R}_1$

The current varies with time. So, integrate over time.

  $\begin{array}{c}\Delta E_1={\displaystyle \int I^2{R}_{1}dt}\\ ={\displaystyle \int {\left(\frac{V\left(t\right)}{R_1}\right)}^{2}dt}\\ ={\displaystyle \int {\left(\frac{\epsilon t}{\tau R_1}\right)}^2{R}_{1}dt}\\ ={\left(\frac{\epsilon }{\tau }\right)}^2\left(\frac{1}{R_1}\right){\displaystyle {\int }_{0.005}^{0.105}{t}^{2}dt}\end{array}$

The expression for the rate at which energy is dissipated in the switch resistance is given by,

  $P_2=\frac{1}{2}C\left(\frac{V_{\text{i}}{}^2-V_{\text{f}}^2}{\Delta t}\right)$

Calculation:

The value of $\Delta E_1$ is calculated as,

  $\begin{array}{c}\Delta E_1=\left(\frac{\epsilon }{\tau }\right)\left(\frac{1}{R_1}\right){\displaystyle {\int }_{0.005}^{0.105}{t}^{2}dt}\\ ={\left(\frac{800\text{V}}{20\text{s}}\right)}^2\left(\frac{1}{1\text{G}\Omega \times \frac{{10}^9\Omega }{1\text{G}\Omega }}\right){\left(\frac{t^3}{3}\right)}_{0.005}^{0.105}\\ =6.17\times {10}^{-10}\text{J}\end{array}$

The value of $P_1$ is calculated as,

  $\begin{array}{c}{P}_1=\frac{\Delta E_1}{\Delta t}\\ =\frac{6.17\times {10}^{-10}\text{J}}{0.1\text{s}}\\ =6.17\times {10}^{-9}\text{W}\end{array}$

The rate at which energy is dissipated in the switch resistance is calculated as,

  $\begin{array}{c}{P}_2=\frac{1}{2}\left(20.0\text{nF}\times \frac{{10}^{-9}\text{F}}{1\text{nF}}\right)\left(\frac{{\left(4.2\text{V}\right)}^2-{\left(0.2\text{V}\right)}^2}{60.9\text{ps}\times \frac{{10}^{-12}\text{s}}{1\text{s}}}\right)\\ =2.89\times {10}^3\text{W}\end{array}$

Conclusion:

Therefore, the average rate at which the energy is delivered to resistor $R_1$ is $6.17\times {10}^{-9}\text{W}$ and to switch the resistance $R_2$ to discharge is $2.89\text{kW}$ .