Chapter 21, Problem 87A

a.

To determine

The energy stored in the capacitors.

Answer to Problem 87A

The energy stored in the capacitors is $\text{W=}3.1\times {10}^6\text{ J}$ .

Given:

Capacitance, $\text{C}=0.061F$

Potential difference, $\Delta V=10kV$

Formula Used:

The energy stored in a capacitor is given by, $W=\frac{1}{2}\text{C}\Delta V^2$

Here, C is the capacitance and $\Delta V$ is the potential difference.

Calculation:

  $W=\frac{1}{2}\text{C}\Delta V^2$

Substituting the values

  $W=\frac{1}{2}\left(0.061\text{ F}\right){\left(10\text{ kV}\right)}^2$

  $\text{W=}3.1\times {10}^6\text{ J}$

Conclusion:

The energy stored in the capacitors is $\text{W=}3.1\times {10}^6\text{ J}$ .

b.

The power produced by the capacitors.

The power produced by the capacitors is $P=3.1\times {10}^{14}\text{ W}$ .

Given:

The energy stored in the capacitors is: $W=3.1\times {10}^6\text{ J}$

Discharge time, $t=\text{1}\times {10}^{-8}\text{ s}$

Formula Used:

  $P=\frac{W}{t}$

P is Power.

W is the energy stored.

t is the discharge time.

Calculation:

  $\text{P=}\frac{\text{W}}{\text{t}}$

Substituting the values

  $P=\frac{\text{3}{\text{.1×10}}^{\text{6}}\text{ W}}{{\text{1×10}}^{\text{-8}}\text{ s}}$

  $P=\text{3}\text{.1}\times {10}^{14}\text{ W}$

Conclusion:

The power produced by the capacitors is $P=\text{3}\text{.1}\times {10}^{14}\text{ W}$ .

c.

The time required to charge the capacitors.

The time required to charge the capacitors $t=\text{3}\text{.1}\times {10}^3\text{ s}$ .

Given:

The power of generator $P=\text{1}\times {10}^3\text{ W}$

The energy stored in the capacitors is $W=3.1\times {10}^6\text{ J}$

Formula Used:

  $P=\frac{W}{t}$

P is Power.

W is the energy stored.

t is the discharge time.

Calculation:

  $t=\frac{W}{P}$

Substituting the values

  $t=\frac{\text{3}{\text{.1×10}}^{\text{6}}\text{ J}}{{\text{1×10}}^{\text{3}}\text{ W}}$

  $t=\text{3}\text{.1}\times {10}^3\text{ s}$

Conclusion:

The time required to charge the capacitors $t=\text{3}\text{.1}\times {10}^3\text{ s}$ .

Explanation of Solution

Given:

Capacitance, $\text{C}=0.061F$

Potential difference, $\Delta V=10kV$

Formula Used:

The energy stored in a capacitor is given by, $W=\frac{1}{2}\text{C}\Delta V^2$

Here, C is the capacitance and $\Delta V$ is the potential difference.

Calculation:

  $W=\frac{1}{2}\text{C}\Delta V^2$

Substituting the values

  $W=\frac{1}{2}\left(0.061\text{ F}\right){\left(10\text{ kV}\right)}^2$

  $\text{W=}3.1\times {10}^6\text{ J}$

Conclusion:

The energy stored in the capacitors is $\text{W=}3.1\times {10}^6\text{ J}$ .

b.

To determine

The power produced by the capacitors.

Answer to Problem 87A

The power produced by the capacitors is $P=3.1\times {10}^{14}\text{ W}$ .

Given:

The energy stored in the capacitors is: $W=3.1\times {10}^6\text{ J}$

Discharge time, $t=\text{1}\times {10}^{-8}\text{ s}$

Formula Used:

  $P=\frac{W}{t}$

P is Power.

W is the energy stored.

t is the discharge time.

Calculation:

  $\text{P=}\frac{\text{W}}{\text{t}}$

Substituting the values

  $P=\frac{\text{3}{\text{.1×10}}^{\text{6}}\text{ W}}{{\text{1×10}}^{\text{-8}}\text{ s}}$

  $P=\text{3}\text{.1}\times {10}^{14}\text{ W}$

Conclusion:

The power produced by the capacitors is $P=\text{3}\text{.1}\times {10}^{14}\text{ W}$ .

c.

The time required to charge the capacitors.

The time required to charge the capacitors $t=\text{3}\text{.1}\times {10}^3\text{ s}$ .

Given:

The power of generator $P=\text{1}\times {10}^3\text{ W}$

The energy stored in the capacitors is $W=3.1\times {10}^6\text{ J}$

Formula Used:

  $P=\frac{W}{t}$

P is Power.

W is the energy stored.

t is the discharge time.

Calculation:

  $t=\frac{W}{P}$

Substituting the values

  $t=\frac{\text{3}{\text{.1×10}}^{\text{6}}\text{ J}}{{\text{1×10}}^{\text{3}}\text{ W}}$

  $t=\text{3}\text{.1}\times {10}^3\text{ s}$

Conclusion:

The time required to charge the capacitors $t=\text{3}\text{.1}\times {10}^3\text{ s}$ .

Explanation of Solution

Given:

The energy stored in the capacitors is: $W=3.1\times {10}^6\text{ J}$

Discharge time, $t=\text{1}\times {10}^{-8}\text{ s}$

Formula Used:

  $P=\frac{W}{t}$

P is Power.

W is the energy stored.

t is the discharge time.

Calculation:

  $\text{P=}\frac{\text{W}}{\text{t}}$

Substituting the values

  $P=\frac{\text{3}{\text{.1×10}}^{\text{6}}\text{ W}}{{\text{1×10}}^{\text{-8}}\text{ s}}$

  $P=\text{3}\text{.1}\times {10}^{14}\text{ W}$

Conclusion:

The power produced by the capacitors is $P=\text{3}\text{.1}\times {10}^{14}\text{ W}$ .

c.

To determine

The time required to charge the capacitors.

Answer to Problem 87A

The time required to charge the capacitors $t=\text{3}\text{.1}\times {10}^3\text{ s}$ .

Given:

The power of generator $P=\text{1}\times {10}^3\text{ W}$

The energy stored in the capacitors is $W=3.1\times {10}^6\text{ J}$

Formula Used:

  $P=\frac{W}{t}$

P is Power.

W is the energy stored.

t is the discharge time.

Calculation:

  $t=\frac{W}{P}$

Substituting the values

  $t=\frac{\text{3}{\text{.1×10}}^{\text{6}}\text{ J}}{{\text{1×10}}^{\text{3}}\text{ W}}$

  $t=\text{3}\text{.1}\times {10}^3\text{ s}$

Conclusion:

The time required to charge the capacitors $t=\text{3}\text{.1}\times {10}^3\text{ s}$ .

Explanation of Solution

Given:

The power of generator $P=\text{1}\times {10}^3\text{ W}$

The energy stored in the capacitors is $W=3.1\times {10}^6\text{ J}$

Formula Used:

  $P=\frac{W}{t}$

P is Power.

W is the energy stored.

t is the discharge time.

Calculation:

  $t=\frac{W}{P}$

Substituting the values

  $t=\frac{\text{3}{\text{.1×10}}^{\text{6}}\text{ J}}{{\text{1×10}}^{\text{3}}\text{ W}}$

  $t=\text{3}\text{.1}\times {10}^3\text{ s}$

Conclusion:

The time required to charge the capacitors $t=\text{3}\text{.1}\times {10}^3\text{ s}$ .