Chapter 18, Problem 92QAP

To determine

(a)

The new capacitance of a capacitor when it is changed to make an ultracapacitor.

Answer to Problem 92QAP

New capacitance of the capacitor will increase $1\times {10}^8$ times to its initial value.

Explanation of Solution

Given:

Initial value of capacitance of a normal capacitor = $C=10\text{μF}=1\times {10}^{-5}\text{F}$

Relation between area of plate of ultra-capacitor and normal capacitor,

  $A_{ultra}=100000A$

Relation between distances for ultra-capacitor and normal capacitor,

  $d_{ultra}=\frac{d}{1000}$

Formula used:

Capacitance of the parallel plate capacitor is defined as,

  $\begin{array}{l}C=\frac{A{\epsilon }_o}{d}\\ \text{Here, }A\text{ = area of plate}\\ d\text{ = seperation of plates}\\ {\epsilon }_o=\text{ permittivity in free space}\end{array}$

Calculation:

Area of ultra-capacitor = $A_{ultra}=100000A$

Separation of plates for ultra-capacitor = $d_{ultra}=\frac{d}{1000}$

Capacitance of ultra-capacitor = $C_{ultra}=\frac{A_{ultra}{\epsilon }_o}{d_{ultra}}=\frac{100000A{\epsilon }_o}{\frac{d}{1000}}=\frac{{10}^{8}A{\epsilon }_o}{d}$

  $C_{ultra}={10}^8\times 1\times {10}^{-5}\text{F}=1000\text{F}$

Then, $\frac{C_{ultra}}{C}=\frac{\frac{{10}^{8}A{\epsilon }_o}{d}}{\frac{A{\epsilon }_o}{d}}={10}^8=1\times {10}^8$

Conclusion:

Thus, capacitance of the ultra-capacitor is $1000\text{F}$ and it is $1\times {10}^8$ times larger than normal capacitor.

To determine

(b)

The energy stored in ultra-capacitor when it is charged to $1.5\text{V}$ and its comparison to the energy stored in the normal capacitor.

Answer to Problem 92QAP

The energy stored in ultra-capacitor is $1.1\times {10}^3\text{J}$ and energy stored in the ultra-capacitor is $1\times {10}^8$ times larger than energy stored in normal capacitor.

Explanation of Solution

Given:

Potential difference = $\Delta V=1.5\text{V}$

Capacitance of ultra-capacitor = $C_{ultra}=1000\text{F}$

Capacitance of normal capacitor = $C=1\times {10}^{-5}\text{F}$

Formula used:

Energy stored in the capacitor is defined as,

  $U=\frac{1}{2}C\Delta V^2$

Calculation:

The energy stored in the ultra-capacitor,

  $\begin{array}{l}{U}_{ultra}=\frac{1}{2}{C}_{ultra}\Delta V^2\\ U_{ultra}=\frac{1}{2}(1000\text{F}){(}^{1.5}=1.125\times {10}^3\text{J}\end{array}$

Energy stored in normal capacitor,

  $\begin{array}{l}U=\frac{1}{2}C\Delta V^2\\ U=\frac{1}{2}(1\times {10}^{-5}\text{F}){(}^{1.5}=1.125\times {10}^{-5}\text{J}\end{array}$

Now,

  $\frac{U_{ultra}}{U}=\frac{1.125\times {10}^{3}J}{1.125\times {10}^{-5}J}=1\times {10}^8$

Conclusion:

Thus, energy stored in the ultra-capacitor is $1.125\times {10}^3\text{J}$ and this energy is $1\times {10}^8$ times larger than energy stored in normal capacitor.

To determine

(c)

The comparison the energy stored in ultra-capacitor and the energy stored in the AAA battery.

Answer to Problem 92QAP

The energy stored in ultra-capacitor is $0.17$ times the energy stored in the AAA battery.

Explanation of Solution

Given:

Energy stored in the ultra-capacitor = $U_{ultra}=1.125\times {10}^3\text{J}$

Energy stored in the AAA battery = $E_{battery}=6750\text{J}$

Calculation:

  $\frac{U_{ultra}}{E_{battery}}=\frac{1.125\times {10}^{3}J}{6750J}=0.17$

Thus, from above discussion, we find that the energy stored in ultra-capacitor is $0.17$ times the energy stored in the AAA battery.

To determine

(d)

The resistance in the wire of the circuit when ultra-capacitor is discharged.

Answer to Problem 92QAP

The resistance in the wire of the circuit is $0.06\Omega$

Explanation of Solution

Given:

Time interval = $t=1.0\min =60s$

Formula used:

Charge on the ultra-capacitor at time t,

  $\begin{array}{l}q=\frac{q_o}{e}\\ q=\text{ charge stored in the capacitor at time }t\\ q_o=\text{initial maximum charge stored in the capacitor}\end{array}$

Discharging of charge in capacitor in CR circuit is defined as,

  $q=q_o{e}^{-t/CR}$

Calculation:

From above formula,

  $\begin{array}{l}q=q_o{e}^{-t/CR}\\ \frac{q_o}{e}=q_o{e}^{-t/CR}\\ e^{-1}=e^{-t/CR}\\ -1=\frac{-t}{CR}\\ R=\frac{t}{C}=\frac{60\sec }{1000\text{F}}=0.06\Omega \end{array}$

Conclusion:

So, the resistance in the wire while capacitor is discharged is $0.06\Omega$.