Chapter 21, Problem 85A

(a)

To determine

The average power required to energize the capacitor.

Answer to Problem 85A

  $P=1.8\times {10}^{-2}\text{ W}$

Explanation of Solution

Given:

The energy stored in the capacitor is $W\text{ = 0}\text{.45 J}$

The time taken by the capacitor to charge is $t=25\text{ s}$

Formula used:

The power $\left(P\right)$ required to charge the capacitor can be calculated by using the equation,

  $P\text{ = }\frac{W}{t}$ m:math>(1)

Where $W$ is work and $t$ is time.

Calculation:

Substituting the numerical values in equation $\left(1\right)$ ,

  $P\text{ = }\frac{\text{0}\text{.45 J}}{25\text{ s}}=1.8\times {10}^{-2}\text{ J/S = }1.8\times {10}^{-2}\text{ W}$

Conclusion:

The average power required to energize the capacitor to $\text{0}\text{.45 J}$ in $25\text{ s}$ is $1.8\times {10}^{-2}\text{ W}$ .

(b)

To determine

The power delivered to the lamp .

Answer to Problem 85A

  $P\text{ = 4}\text{.5}\times {10}^3\text{ W}$

Explanation of Solution

Given:

The energy stored in the capacitor is $W\text{ = 0}\text{.45 J}$

Time taken by the capacitor to discharge all its energy is $t=1.0\times {10}^{-4}\text{ s}$

Formula used:

The power $\left(P\right)$ delivered by the capacitor can be calculated by using the equation,

  $P\text{ = }\frac{W}{t}$ $\left(2\right)$

Where $W$ is work and $t$ is time.

Calculation:

Substituting the numerical values in equation $\left(2\right)$ ,

  $P\text{ = }\frac{\text{0}\text{.45 J}}{1.0\times {10}^{-4}\text{ s}}=\text{ 4}\text{.5}\times {10}^3\text{ W}$

Conclusion:

The power delivered to the lamp if the capacitor plates are discharged through the probe lampis $\text{4}\text{.5}\times {10}^3\text{ W}$ .

(c)

To determine

To explain:The possibility of large value of power as obtained in part (b).

Explanation of Solution

Introduction:

The quantity of energy transferred or converted in unit time is called power. It is a scalar quantity. Capacitors are used to store the power for future use.

The power delivered to the lamp if the capacitor plates are discharged through the probe lampis $\text{4}\text{.5}\times {10}^3\text{ W}$ .

Power is indirectly related to the capacitor charging or discharging time. Smaller the time required to discharge the given amount of energy, larger will be the power dissipated by the capacitor.