Chapter 10, Problem 1P

(a)

To determine

Find the values of average power $P$, reactive power $Q$, and state whether the powers deliver to or absorb from the box in the given circuit.

Answer to Problem 1P

The values of average power $P$, reactive power $Q$ are $129.41\text{W}$, $482.96\text{VAR}$, respectively and both the average and reactive powers are absorbed from the box.

Explanation of Solution

Given data:

Refer to given figure in the textbook.

The voltage and current expressions are,

$\begin{array}{l}v=250\cos \left(\omega t+45°\right)\text{V}\\ i=4\sin \left(\omega t+60°\right)\text{A}\end{array}$

From the given expressions, the required parameters are written as follows:

$\begin{array}{l}{V}_m=250\text{V}\\ I_m=4\text{A}\\ {\theta }_v=45°\\ {\theta }_i=60°\end{array}$

Formula used:

Write the expression to calculate the average power.

$P=\frac{V_m{I}_m}{2}\cos \left({\theta }_v-{\theta }_i\right)$        (1)

Here,

$V_m$ is the maximum voltage,

$I_m$ is the maximum current,

${\theta }_v$ is the phase angle of the voltage, and

${\theta }_i$ is the phase angle of the current.

Write the expression to calculate the reactive power.

$Q=\frac{V_m{I}_m}{2}\sin \left({\theta }_v-{\theta }_i\right)$        (2)

Calculation:

Rewrite the given current expression of current as follows:

$\begin{array}{c}i=4\sin \left(\omega t+60°\right)\text{A}\\ =4\cos \left[\left(\omega t+60°\right)-90°\right]\text{A}\left\{\because \sin \theta =\cos \left(\theta -90°\right)\right\}\\ =4\cos \left(\omega t-30°\right)\end{array}$

Therefore, the value of ${\theta }_i$ is $-30°$.

${\theta }_i=-30°$

Substitute $250\text{V}$ for $V_m$, $4\text{A}$ for $I_m$, $45°$ for ${\theta }_v$, and $-30°$ for ${\theta }_i$ in Equation (1) to obtain the value of $P$.

$\begin{array}{c}P=\frac{\left(250\text{V}\right)\left(4\text{A}\right)}{2}\cos \left(45°-\left(-30°\right)\right)\\ =\frac{1000}{2}\cos \left(45°+30°\right)\text{W}\\ =500\cos \left(75°\right)\text{W}\\ =129.41\text{W}\end{array}$

As the average power is obtained with positive sign, the average power is absorbed from the terminals of the box in the given circuit.

Substitute $250\text{V}$ for $V_m$, $4\text{A}$ for $I_m$, $45°$ for ${\theta }_v$, and $-30°$ for ${\theta }_i$ in Equation (2) to obtain the value of $Q$.

$\begin{array}{c}Q=\frac{\left(250\text{V}\right)\left(4\text{A}\right)}{2}\sin \left(45°-\left(-30°\right)\right)\\ =\frac{1000}{2}\sin \left(45°+30°\right)\text{VAR}\\ =500\sin \left(75°\right)\text{VAR}\\ =482.96\text{VAR}\end{array}$

As the reactive power is obtained with positive sign, the magnetizing VARs are absorbed from the terminals of the box in the given circuit.

Conclusion:

Thus, the values of average power $P$, reactive power $Q$ are $129.41\text{W}$, $482.96\text{VAR}$, respectively and both the average and reactive powers are absorbed from the box.

(b)

To determine

Find the values of average power $P$, reactive power $Q$, and state whether the powers deliver to or absorb from the box in the given circuit.

Answer to Problem 1P

The values of average power $P$, reactive power $Q$ are $31.82\text{W}$, $31.82\text{VAR}$, respectively and both the average and reactive powers are absorbed from the box.

Explanation of Solution

Given data:

The voltage and current expressions are,

$\begin{array}{l}v=18\cos \left(\omega t-30°\right)\text{V}\\ i=5\cos \left(\omega t-75°\right)\text{A}\end{array}$

From the given expressions, the required parameters are written as follows:

$\begin{array}{l}{V}_m=18\text{V}\\ I_m=5\text{A}\\ {\theta }_v=-30°\\ {\theta }_i=-75°\end{array}$

Calculation:

Substitute $18\text{V}$ for $V_m$, $5\text{A}$ for $I_m$, $-30°$ for ${\theta }_v$, and $-75°$ for ${\theta }_i$ in Equation (1) to obtain the value of $P$.

$\begin{array}{c}P=\frac{\left(18\text{V}\right)\left(5\text{A}\right)}{2}\cos \left(-30°-\left(-75°\right)\right)\\ =\frac{90}{2}\cos \left(-30°+75°\right)\text{W}\\ =45\cos \left(45°\right)\text{W}\\ =31.82\text{W}\end{array}$

As the average power is obtained with positive sign, the average power is absorbed from the terminals of the box in the given circuit.

Substitute $18\text{V}$ for $V_m$, $5\text{A}$ for $I_m$, $-30°$ for ${\theta }_v$, and $-75°$ for ${\theta }_i$ in Equation (2) to obtain the value of $Q$.

$\begin{array}{c}Q=\frac{\left(18\text{V}\right)\left(5\text{A}\right)}{2}\sin \left(-30°-\left(-75°\right)\right)\\ =\frac{90}{2}\sin \left(-30°+75°\right)\text{VAR}\\ =45\sin \left(45°\right)\text{VAR}\\ =31.82\text{VAR}\end{array}$

As the reactive power is obtained with positive sign, the magnetizing VARs are absorbed from the terminals of the box in the given circuit.

Conclusion:

Thus, the values of average power $P$, reactive power $Q$ are $31.82\text{W}$, $31.82\text{VAR}$, respectively and both the average and reactive powers are absorbed from the box.

(c)

To determine

Find the values of average power $P$, reactive power $Q$, and state whether the powers deliver to or absorb from the box in the given circuit.

Answer to Problem 1P

The values of average power $P$, reactive power $Q$ are $-63.39\text{W}$, $-135.95\text{VAR}$, respectively and both the average and reactive powers are delivered to the box.

Explanation of Solution

Given data:

The voltage and current expressions are,

$\begin{array}{l}v=150\sin \left(\omega t+25°\right)\text{V}\\ i=2\cos \left(\omega t+50°\right)\text{A}\end{array}$

From the given expressions, the required parameters are written as follows:

$\begin{array}{l}{V}_m=150\text{V}\\ I_m=2\text{A}\\ {\theta }_v=25°\\ {\theta }_i=50°\end{array}$

Calculation:

Rewrite the given voltage expression of current as follows:

$\begin{array}{c}v=150\sin \left(\omega t+25°\right)\text{A}\\ =150\cos \left[\left(\omega t+25°\right)-90°\right]\text{A}\left\{\because \sin \theta =\cos \left(\theta -90°\right)\right\}\\ =150\cos \left(\omega t-65°\right)\end{array}$

Therefore, the value of ${\theta }_v$ is $-65°$.

${\theta }_v=-65°$

Substitute $150\text{V}$ for $V_m$, $2\text{A}$ for $I_m$, $-65°$ for ${\theta }_v$, and $50°$ for ${\theta }_i$ in Equation (1) to obtain the value of $P$.

$\begin{array}{c}P=\frac{\left(150\text{V}\right)\left(2\text{A}\right)}{2}\cos \left(-65°-50°\right)\\ =\frac{300}{2}\cos \left(-115°\right)\text{W}\\ =150\cos \left(-115°\right)\text{W}\\ =-63.39\text{W}\end{array}$

As the average power is obtained with negative sign, the average power is delivered to the terminals of the box in the given circuit.

Substitute $150\text{V}$ for $V_m$, $2\text{A}$ for $I_m$, $-65°$ for ${\theta }_v$, and $50°$ for ${\theta }_i$ in Equation (2) to obtain the value of $Q$.

$\begin{array}{c}Q=\frac{\left(150\text{V}\right)\left(2\text{A}\right)}{2}\sin \left(-65°-50°\right)\\ =\frac{300}{2}\sin \left(-115°\right)\text{VAR}\\ =150\sin \left(-115°\right)\text{VAR}\\ =-135.95\text{VAR}\end{array}$

As the reactive power is obtained with negative sign, the magnetizing VARs are delivered to the terminals of the box in the given circuit.

Conclusion:

Thus, the values of average power $P$, reactive power $Q$ are $-63.39\text{W}$, $-135.95\text{VAR}$, respectively and both the average and reactive powers are delivered to the box.

(d)

To determine

Find the values of average power $P$, reactive power $Q$, and state whether the powers deliver to or absorb from the box in the given circuit.

Answer to Problem 1P

The values of average power $P$, reactive power $Q$ are $257.12\text{W}$, $-306.42\text{VAR}$, respectively and the average power is absorbed from the box and the reactive power is delivered to the box.

Explanation of Solution

Given data:

The voltage and current expressions are,

$\begin{array}{l}v=80\cos \left(\omega t+120°\right)\text{V}\\ i=10\cos \left(\omega t+170°\right)\text{A}\end{array}$

From the given expressions, the required parameters are written as follows:

$\begin{array}{l}{V}_m=80\text{V}\\ I_m=10\text{A}\\ {\theta }_v=120°\\ {\theta }_i=170°\end{array}$

Calculation:

Substitute $80\text{V}$ for $V_m$, $10\text{A}$ for $I_m$, $120°$ for ${\theta }_v$, and $170°$ for ${\theta }_i$ in Equation (1) to obtain the value of $P$.

$\begin{array}{c}P=\frac{\left(80\text{V}\right)\left(10\text{A}\right)}{2}\cos \left(120°-170°\right)\\ =\frac{800}{2}\cos \left(-50°\right)\text{W}\\ =400\cos \left(-50°\right)\text{W}\\ =257.12\text{W}\end{array}$

As the average power is obtained with positive sign, the average power is absorbed from the terminals of the box in the given circuit.

Substitute $80\text{V}$ for $V_m$, $10\text{A}$ for $I_m$, $120°$ for ${\theta }_v$, and $170°$ for ${\theta }_i$ in Equation (2) to obtain the value of $Q$.

$\begin{array}{c}Q=\frac{\left(80\text{V}\right)\left(10\text{A}\right)}{2}\sin \left(120°-170°\right)\\ =\frac{800}{2}\sin \left(-50°\right)\text{VAR}\\ =400\sin \left(-50°\right)\text{VAR}\\ =-306.42\text{VAR}\end{array}$

As the reactive power is obtained with negative sign, the magnetizing VARs are delivered to the terminals of the box in the given circuit.

Conclusion:

Thus, the values of average power $P$, reactive power $Q$ are $257.12\text{W}$, $-306.42\text{VAR}$, respectively and the average power is absorbed from the box and the reactive power is delivered to the box.