Chapter 21, Problem 64QAP

To determine

(a)

The peak value of the current and average power delivered when a sinusoidal voltage of 40.0 V rms and a frequency of 100 Hz is applied to a 100 O resistor.

Answer to Problem 64QAP

When a sinusoidal voltage of 40.0 V rms and a frequency of 100 Hz is applied to a 100 O resistor, the peak current is found to be 0.57 A and the average power delivered is 16 W.

Explanation of Solution

Given:

  $\begin{array}{c}R=100\Omega \\ V_{\text{rms}}=40.0\text{ V}\\ f=\text{100 Hz}\end{array}$

Formula used:

The peak value of current $i_{\max }$ in an ac circuit is given by,

  $i_{\max }=\sqrt{2}{i}_{\text{rms}}$......(1)

Here, the rms value of current $i_{\text{rms}}$ is given by,

  $i_{\text{rms}}=\frac{V_{\text{rms}}}{R}$......(2)

The average power dissipated in the circuit is given by,

  $P_{\text{av}}=V_{\text{rms}}{i}_{\text{rms}}$......(3)

Calculation:

Calculate the rms value of current $i_{\text{rms}}$ by substituting the values of the variables in equation (2).

  $\begin{array}{c}{i}_{\text{rms}}=\frac{V_{\text{rms}}}{R}\\ =\frac{40.0\text{ V}}{100\Omega }\\ =0.40\text{ A}\end{array}$

Calculate the value of peak current using equation (1).

  $\begin{array}{c}{i}_{\max }=\sqrt{2}{i}_{\text{rms}}\\ =\sqrt{2}\left(0.40\text{ A}\right)\\ =0.57\text{ A}\end{array}$

Calculate the average power delivered using equation (3).

  $\begin{array}{c}{P}_{\text{av}}=V_{\text{rms}}{i}_{\text{rms}}\\ =\left(40.0\text{ V}\right)\left(1.2\text{ A}\right)\\ =16\text{ W}\end{array}$

Conclusion;Thus, when a sinusoidal voltage of 40.0 V rms and a frequency of 100 Hz is applied to a 100 O resistor, the peak current is found to be 0.57 A and the average power delivered is 16 W.

To determine

(b)

The peak value of the current and average power delivered when a sinusoidal voltage of 40.0 V rms and a frequency of 100 Hz is applied to a 0.200 H inductor.

Answer to Problem 64QAP

When a sinusoidal voltage of 40.0 V rms and a frequency of 100 Hz is applied to a 0.200 H inductor, the peak current is found to be 0.45A and the average power delivered is 0 W.

Explanation of Solution

Given:

  $\begin{array}{c}L=0.200\text{ H}\\ V_{\text{rms}}=40.0\text{ V}\\ f=\text{100 Hz}\end{array}$

Formula used:

The peak value of current $i_{\max }$ in an ac circuit is given by,

  $i_{\max }=\sqrt{2}{i}_{\text{rms}}$......(1)

Here, the rms value of current $i_{\text{rms}}$ is given by,

  $i_{\text{rms}}=\frac{V_{\text{rms}}}{2\pi fL}$......(4)

The average power dissipated in the circuit is given by,

  $P_{\text{av}}=V_{\text{rms}}{i}_{\text{rms}}\cos \varphi$......(5)

Here, $\varphi$ is the phase difference between current and voltage.

Calculation:

Calculate the rms value of current $i_{\text{rms}}$ by substituting the values of the variables in equation (4).

  $\begin{array}{c}{i}_{\text{rms}}=\frac{V_{\text{rms}}}{2\pi fL}\\ =\frac{40.0\text{ V}}{2\left(3.14\right)\left(100\text{ Hz}\right)\left(0.200\text{ H}\right)}\\ =0.32\text{ A}\end{array}$

Calculate the value of peak current using equation (1).

  $\begin{array}{c}{i}_{\max }=\sqrt{2}{i}_{\text{rms}}\\ =\sqrt{2}\left(0.32\text{ A}\right)\\ =0.45\text{ A}\end{array}$

Calculate the average power delivered using equation (5). The phase difference between voltage and current in a ac circuit containing an inductor is 90^o

  $\varphi =90°$

Therefore, since $\cos 90°=0$

  $\begin{array}{c}{P}_{\text{av}}=V_{\text{rms}}{i}_{\text{rms}}\cos \varphi \\ =0\end{array}$

Conclusion;Thus, when a sinusoidal voltage of 40.0 V rms and a frequency of 100 Hz is applied to a 0.200 H inductor, the peak current is found to be 0.45A and the average power delivered is 0 W.

To determine

(c)

The peak value of the current and average power delivered when a sinusoidal voltage of 40.0 V rms and a frequency of 100 Hz is applied to a 50.0 μF capacitor.

Answer to Problem 64QAP

When a sinusoidal voltage of 40.0 V rms and a frequency of 100 Hz is applied to a 50.0 μF capacitor, the peak current is found to be 1.8 A and the average power delivered is 0 W.

Explanation of Solution

Given:

  $\begin{array}{c}L=50.0\mu \text{F}=50.0\times {10}^{-6}\text{F}\\ V_{\text{rms}}=40.0\text{ V}\\ f=\text{100 Hz}\end{array}$

Formula used:

The peak value of current $i_{\max }$ in an ac circuit is given by,

  $i_{\max }=\sqrt{2}{i}_{\text{rms}}$......(1)

Here, the rms value of current $i_{\text{rms}}$ is given by,

  $i_{\text{rms}}=2\pi fC{V}_{\text{rms}}$......(6)

The average power dissipated in the circuit is given by,

  $P_{\text{av}}=V_{\text{rms}}{i}_{\text{rms}}\cos \varphi$......(3)

Here, $\varphi$ is the phase difference between current and voltage.

Calculation:

Calculate the rms value of current $i_{\text{rms}}$ by substituting the values of the variables in equation (6).

  $\begin{array}{c}{i}_{\text{rms}}=2\pi fC{V}_{\text{rms}}\\ =2\left(3.14\right)\left(100\text{ Hz}\right)\left(50.0\times {10}^{-6}\text{F}\right)\left(40.0\text{ V}\right)\\ =1.256\text{ A}\end{array}$

Calculate the value of peak current using equation (1).

  $\begin{array}{c}{i}_{\max }=\sqrt{2}{i}_{\text{rms}}\\ =\sqrt{2}\left(1.256\text{ A}\right)\\ =1.776\text{ A}\\ \text{=1}\text{.8 A}\end{array}$

Calculate the average power delivered using equation (5). The phase difference between voltage and current in a ac circuit containing a capacitor is 90^o

  $\varphi =90°$

Therefore, since $\cos 90°=0$

  $\begin{array}{c}{P}_{\text{av}}=V_{\text{rms}}{i}_{\text{rms}}\cos \varphi \\ =0\end{array}$

Conclusion;Thus, when a sinusoidal voltage of 40.0 V rms and a frequency of 100 Hz is applied to a 50.0 μF capacitor, the peak current is found to be 1.8 A and the average power delivered is 0 W.