Chapter 11, Problem 46P

For the following voltage and current phasors, calculate the complex power, apparent power, real power, and reactive power. Specify whether the pf is leading or lagging.

To determine

Find the complex power, apparent power, real power, and reactive power for the given voltage and current phasor. Also mention whether the power factor is leading or lagging.

Answer to Problem 46P

The complex power is $S=\left(95.26-j55\right)\text{VA}$, apparent power is $S=110\text{VA}$, real power is $P=95.26\text{W}$, reactive power is $Q=-55\text{VAR}$, and the power factor $\text{pf}$ is leading for the given voltage and current phasor.

Explanation of Solution

Given data:

The voltage phasor,

$V_{\text{rms}}=220\angle 30°\text{V}$ (1)

The current phasor,

$I_{\text{rms}}=0.5\angle 60°\text{A}$ (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, and

$Q$ is the reactive power.

Calculation:

Substitute $220\angle 30°\text{V}$ for $V_{\text{rms}}$ and $0.5\angle 60°\text{A}$ for $I_{\text{rms}}$ in equation (3) to find the complex power $S$ in VA.

$\begin{array}{c}S=220\angle 30°\text{V}{\left(0.5\angle 60°\text{A}\right)}^{*}\\ =\left(220\angle 30°\text{V}\right)\left(0.5\angle -60°\text{A}\right)\end{array}$

$S=110\angle -30°\text{VA}$ (5)

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

$S=\left(95.26-j55\right)\text{VA}$ (6)

From equation (5), the apparent power $S=\left|S\right|=110\text{VA}$.

On comparing equation (4) and (6), the real power $P$ is $95.26\text{W}$ and the reactive power $Q$ is $-55\text{VAR}$.

From equation (1) and (2), the power factor $\text{pf}$ is leading because the current leads the voltage, which implies a capacitive load.

Conclusion:

Thus, the complex power is $S=\left(95.26-j55\right)\text{VA}$, apparent power is $S=110\text{VA}$, real power is $P=95.26\text{W}$, reactive power is $Q=-55\text{VAR}$, and the power factor $\text{pf}$ is leading for the given voltage and current phasor.

To determine

Find the complex power, apparent power, real power, and reactive power for the given voltage and current phasor. Also mention whether the power factor is leading or lagging.

Answer to Problem 46P

The complex power is $S=\left(1497.2-j401.2\right)\text{VA}$, apparent power is $S=1550\text{VA}$, real power is $P=1497.2\text{W}$, reactive power is $Q=401.2\text{VAR}$, and the power factor $\text{pf}$ is lagging for the given voltage and current phasor.

Explanation of Solution

Given data:

The voltage phasor,

$V_{\text{rms}}=250\angle -10°\text{V}$ (7)

The current phasor,

$I_{\text{rms}}=6.2\angle -25°\text{A}$ (8)

Calculation:

Substitute $250\angle -10°\text{V}$ for $V_{\text{rms}}$ and $6.2\angle -25°\text{A}$ for $I_{\text{rms}}$ in equation (3) to find the complex power $S$ in VA.

$\begin{array}{c}S=250\angle -10°\text{V}{\left(6.2\angle -25°\text{A}\right)}^{*}\\ =\left(250\angle -10°\text{V}\right)\left(6.2\angle 25°\text{A}\right)\end{array}$

$S=1550\angle 15°\text{VA}$ (9)

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

$S=\left(1497.2+j401.2\right)\text{VA}$ (10)

From equation (9), the apparent power $S=\left|S\right|=1550\text{VA}$.

On comparing equation (4) and (10), the real power $P$ is $1497.2\text{W}$ and the reactive power $Q$ is $401.2\text{VAR}$.

From equation (7) and (8), the power factor $\text{pf}$ is lagging because the current lags the voltage, which implies an inductive load.

Conclusion:

Thus, the complex power is $S=\left(1497.2-j401.2\right)\text{VA}$, apparent power is $S=1550\text{VA}$, real power is $P=1497.2\text{W}$, reactive power is $Q=401.2\text{VAR}$, and the power factor $\text{pf}$ is lagging for the given voltage and current phasor.

To determine

Find the complex power, apparent power, real power, and reactive power for the given voltage and current phasor. Also mention whether the power factor is leading or lagging.

Answer to Problem 46P

The complex power is $S=\left(278.2+j74.54\right)\text{VA}$, apparent power is $S=288\text{VA}$, real power is $P=278.2\text{W}$, reactive power is $Q=74.54\text{VAR}$, and the power factor $\text{pf}$ is lagging for the given voltage and current phasors.

Explanation of Solution

Given data:

The voltage phasor,

$V_{\text{rms}}=120\angle 0°\text{V}$ (11)

The current phasor,

$I_{\text{rms}}=2.4\angle -15°\text{A}$ (12)

Calculation:

Substitute $120\angle 0°\text{V}$ for $V_{\text{rms}}$ and $2.4\angle -15°\text{A}$ for $I_{\text{rms}}$ in equation (3) to find the complex power $S$ in VA.

$\begin{array}{c}S=120\angle 0°\text{V}{\left(2.4\angle -15°\text{A}\right)}^{*}\\ =\left(120\angle 0°\text{V}\right)\left(2.4\angle 15°\text{A}\right)\end{array}$

$S=288\angle 15°$ (13)

Convert equation (13) from polar form to rectangular form. Therefore,

$S=\left(278.2+j74.54\right)\text{VA}$ (14)

From equation (13), the apparent power $S=\left|S\right|=288\text{VA}$.

On comparing equation (4) and (14), the real power $P$ is $278.2\text{W}$ and the reactive power $Q$ is $74.54\text{VAR}$.

From equation (11) and (12), the power factor $\text{pf}$ is lagging because the current lags the voltage, which implies an inductive load.

Conclusion:

Thus, the complex power is $S=\left(278.2+j74.54\right)\text{VA}$, apparent power is $S=288\text{VA}$, real power is $P=278.2\text{W}$, reactive power is $Q=74.54\text{VAR}$, and the power factor $\text{pf}$ is lagging for the given voltage and current phasor.

To determine

Find the complex power, apparent power, real power, and reactive power for the given voltage and current phasor. Also mention whether the power factor is leading or lagging.

Answer to Problem 46P

The complex power is $S=\left(961.7-j961.7\right)\text{VA}$, apparent power is $S=1360\text{VA}$, real power is $P=961.7\text{W}$, reactive power is $Q=-961.7\text{VAR}$, and the power factor $\text{pf}$ is leading for the given voltage and current phasors.

Explanation of Solution

Given data:

The voltage phasor,

$V_{\text{rms}}=160\angle 45°\text{V}$ (15)

The current phasor,

$I_{\text{rms}}=8.5\angle 90°\text{A}$ (16)

Calculation:

Substitute $160\angle 45°\text{V}$ for $V_{\text{rms}}$ and $8.5\angle 90°\text{A}$ for $I_{\text{rms}}$ in equation (3) to find the complex power $S$ in VA.

$\begin{array}{c}S=160\angle 45°\text{V}{\left(8.5\angle 90°\text{A}\right)}^{*}\\ =\left(160\angle 45°\text{V}\right)\left(8.5\angle -90°\text{A}\right)\end{array}$

$S=1360\angle -45°$ (17)

Convert equation (17) from polar form to rectangular form. Therefore,

$S=\left(961.7-j961.7\right)\text{VA}$ (18)

From equation (17), the apparent power $S=\left|S\right|=1360\text{VA}$.

On comparing equation (4) and (18), the real power $P$ is $961.7\text{W}$ and the reactive power $Q$ is $-961.7\text{VAR}$.

From equation (15) and (16), the power factor $\text{pf}$ is leading because the current leads the voltage, which implies a capacitive load.

Conclusion:

Thus, the complex power is $S=\left(961.7-j961.7\right)\text{VA}$, apparent power is $S=1360\text{VA}$, real power is $P=961.7\text{W}$, reactive power is $Q=-961.7\text{VAR}$, and the power factor $\text{pf}$ is leading for the given voltage and current phasor.