Chapter 11, Problem 47P

For each of the following cases, find the complex power, the average power, and the reactive power:

1. (a) v(t) = 169.7 sin (377t + 45°) V,

i(t) = 5.657 sin (377t) A

1. (b) v(t) = 339.4 sin (377t + 90°) V,

i(t) = 5.657 sin (377t + 45°) A

To determine

Find the complex power, average power, and reactive power for the given instantaneous voltage and current.

Answer to Problem 47P

The complex power is $S=339.4\text{W}+j339.4\text{VAR}$, average power is $P=339.4\text{W}$, and reactive power is $Q=339.4\text{VAR}$.

Explanation of Solution

Given data:

The voltage phasor is,

$v(t)=169.7\sin \left(377t+45°\right)\text{V}$ (1)

The current phasor is,

$i(t)=5.567\sin \left(377t\right)\text{A}$ (2)

Formula used:

Write the expression to find the complex power $S$.

$S=V_{\text{rms}}{\left(I_{\text{rms}}\right)}^{*}$ (3)

Here,

$V_{\text{rms}}$ is the root mean square value of voltage, and

$I_{\text{rms}}$ is the root mean square value of current.

Write the expression for complex power $S$.

$S=P+jQ$ (4)

Here,

$P$ is the real power or average power, and

$Q$ is the reactive power.

Calculation:

From equation (1), the rms value of the voltage is

$\begin{array}{c}{V}_{\text{rms}}=\frac{169.7}{\sqrt{2}}\angle 45°\left\{\because V_m=169.7,{\theta }_v=45°\right\}\\ =120\angle 45°\text{V}\end{array}$

From equation (2), the rms value of the current is

$\begin{array}{c}{I}_{\text{rms}}=\frac{5.657}{\sqrt{2}}\angle 0°\left\{\because I_m=5.657,{\theta }_i=0°\right\}\\ =4\angle 0°\text{A}\end{array}$

Substitute $220\angle 45°\text{V}$ for $V_{\text{rms}}$ and $4\angle 0°\text{A}$ for $I_{\text{rms}}$ in equation (3) to find the complex power $S$ in VA.

$\begin{array}{c}S=220\angle 45°\text{V}{\left(4\angle 0°\text{A}\right)}^{*}\\ =\left(220\angle 45°\text{V}\right)\left(4\angle -0°\text{A}\right)\end{array}$

$S=480\angle 45°\text{VA}$ (5)

Convert equation (5) from polar form to rectangular form. Therefore,

$S=339.4\text{W}+j339.4\text{VAR}$ (6)

On comparing equation (4) and (6), the average power $P$ is $339.4\text{W}$ and the reactive power $Q$ is $339.4\text{VAR}$.

Conclusion:

Thus, the complex power is $S=339.4\text{W}+j339.4\text{VAR}$, average power is $P=339.4\text{W}$, and reactive power is $Q=339.4\text{VAR}$.

To determine

Find the complex power, average power, and reactive power for the given instantaneous voltage and current.

Answer to Problem 47P

The complex power is $S=678.8\text{W}+j678.8\text{VAR}$, average power is $P=678.8\text{W}$, and reactive power is $Q=678.8\text{VAR}$.

Explanation of Solution

Given data:

The voltage phasor,

$v\left(t\right)=339.4\sin \left(377t+90°\right)\text{V}$ (7)

The current phasor,

$i\left(t\right)=5.657\sin \left(377t+45°\right)\text{A}$ (8)

Calculation:

From equation (7), the rms value of the voltage is

$\begin{array}{c}{V}_{\text{rms}}=\frac{339.4}{\sqrt{2}}\angle 90°\left\{\because V_m=339.4,{\theta }_v=90°\right\}\\ =240\angle 90°\text{V}\end{array}$

From equation (8), the rms value of the current is

$\begin{array}{c}{I}_{\text{rms}}=\frac{5.657}{\sqrt{2}}\angle 45°\left\{\because I_m=5.657,{\theta }_i=45°\right\}\\ =4\angle 45°\text{A}\end{array}$

Substitute $240\angle 90°\text{V}$ for $V_{\text{rms}}$ and $4\angle 45°\text{A}$ for $I_{\text{rms}}$ in equation (3) to find the complex power $S$ in VA.

$\begin{array}{c}S=240\angle 90°\text{V}{\left(4\angle 45°\text{A}\right)}^{*}\\ =\left(240\angle 90°\text{V}\right)\left(4\angle -45°\text{A}\right)\end{array}$

$S=960\angle 45°\text{VA}$ (9)

Convert equation (9) from polar form to rectangular form. Therefore,

$S=678.8\text{W}+j678.8\text{VAR}$ (10)

On comparing equation (4) and (10), the average power $P$ is $678.8\text{W}$ and the reactive power $Q$ is $678.8\text{VAR}$.

Conclusion:

Thus, the complex power is $S=678.8\text{W}+j678.8\text{VAR}$, average power is $P=678.8\text{W}$, and reactive power is $Q=678.8\text{VAR}$.

To determine

Find the complex power, average power, and reactive power for the given voltage and impedance phasor.

Answer to Problem 47P

The complex power is $S=7.637\text{kW}+j7.637\text{kVAR}$, average power is $P=7.637\text{kW}$, and reactive power is $\text{Q}=\text{7}\text{.637}\text{kVAR}$.

Explanation of Solution

Given data:

The voltage phasor,

$V_{\text{rms}}=900\angle 90°\text{V}$ (11)

The impedance phasor,

$Z=75\angle 45°\Omega$ (12)

Formula used:

Write the expression for complex power $S$.

$S=\frac{V_{\text{rms}}^{\text{2}}}{Z^{*}}$ (13)

Here,

$Z^{*}$ is the complex conjugate of the impedance.

Calculation:

From equation (11),

$V_{\text{rms}}=\left|V_{\text{rms}}\right|=900\text{V}$

Substitute $900$ for $V_{\text{rms}}$ and $75\angle 45°\Omega$ for $Z$ in equation (13) to find the complex power $S$ in VA.

$\begin{array}{c}S=\frac{{\left(900\text{V}\right)}^2}{{\left(75\angle 45°\Omega \right)}^{*}}\\ =\frac{810000\text{V}\cdot \text{V}}{75\angle -45°\Omega }\\ =\left(1080\angle 45°\right)\times {10}^3\times {10}^{-3}\text{VA}\left\{\because 1\text{A}=\frac{1\text{V}}{1\Omega }\right\}\end{array}$

$S=10.8\angle 45°\text{kVA}$ (14)

Convert equation (14) from polar form to rectangular form. Therefore,

$S=7.637\text{kW}+j7.637\text{kVAR}$ (15)

On comparing equation (4) and (15), the average power $P$ is $7.637\text{kW}$ and the reactive power $Q$ is $\text{7}\text{.637}\text{kVAR}$.

Conclusion:

Thus, the complex power is $S=7.637\text{kW}+j7.637\text{kVAR}$, average power is $P=7.637\text{kW}$, and reactive power is $\text{Q}=\text{7}\text{.637}\text{kVAR}$.

To determine

Find the complex power, average power, and reactive power for the given voltage and impedance phasor.

Answer to Problem 47P

The complex power is $S=250\text{kW}+j\text{433}\text{kVAR}$, average power is $P=250\text{kW}$, and reactive power is $Q=433\text{kVAR}$.

Explanation of Solution

Given data:

The current phasor,

$I_{\text{rms}}=100\angle 60°\text{A}$ (16)

The impedance phasor,

$Z=50\angle 60°\Omega$ (17)

Write the expression to find the complex power $S$.

$S=I_{\text{rms}}^{\text{2}}Z$ (18)

Calculation:

From equation (16),

$I_{\text{rms}}=|I_{\text{rms}}|=100\text{A}$

Substitute $50\angle 60°\Omega$ for $Z$ and $100\text{A}$ for $I_{\text{rms}}$ in equation (18) to find the complex power $S$ in VA.

$\begin{array}{c}S={\left(100\text{A}\right)}^2\left(50\angle 60°\Omega \right)\\ =\left(500000\angle 60°\right)\times {10}^3\times {10}^{-3}{\text{A}}^2\cdot \Omega \end{array}$

$S=500\angle 60°\text{kVA}\left\{\because 1\text{V}=1\text{A}\times 1\Omega \right\}$ (19)

Convert equation (19) from polar form to rectangular form. Therefore,

$S=250\text{kW}+j\text{433}\text{kVAR}$ (20)

On comparing equation (4) and (20), the average power $P$ is $250\text{kW}$ and the reactive power $Q$ is $433\text{VAR}$.

Conclusion:

Thus, the complex power is $S=250\text{kW}+j\text{433}\text{kVAR}$, average power is $P=250\text{kW}$, and reactive power is $Q=433\text{kVAR}$.