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 $P$, $Q$ are $\underset{\_}{409.58\text{W}}$, $\underset{\_}{286.79\text{VAR}}$, respectively and both the average and reactive powers are absorbed from the box.

Explanation of Solution

Given data:

$\begin{array}{l}v=100\cos \left(\omega t+50°\right)\text{V}\\ i=10\cos \left(\omega t+15°\right)\text{A}\end{array}$

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

$\begin{array}{l}{V}_m=100\text{V}\\ I_m=10\text{A}\\ {\theta }_v=50°\\ {\theta }_i=15°\end{array}$

Formula used:

Write the expression for average power as follows:

$P=\frac{V_m{I}_m}{2}\cos \left({\theta }_v-{\theta }_i\right)$        (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 for reactive power as follows:

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

Calculation:

Substitute 100 V for , $V_m$ 10 A for $I_m$, $50°$ for ${\theta }_v$, and $15°$ for ${\theta }_i$ in Equation (1) to obtain the value of average power.

$\begin{array}{c}P=\frac{\left(100\text{V}\right)\left(10\text{A}\right)}{2}\cos \left(50°-15°\right)\\ =\left(500\right)\cos \left(35°\right)\text{W}\\ =409.5760\text{W}\\ \cong \text{409}\text{.58}\text{W}\end{array}$

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

Substitute 100 V for , $V_m$ 10 A for $I_m$, $50°$ for ${\theta }_v$, and $15°$ for ${\theta }_i$ in Equation (2) to obtain the value of reactive power.

$\begin{array}{c}Q=\frac{\left(100\text{V}\right)\left(10\text{A}\right)}{2}\sin \left(50°-15°\right)\\ =\left(500\right)\sin \left(35°\right)\text{VAR}\\ =286.7882\text{VAR}\\ \cong 286.79\text{VAR}\end{array}$

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 $P$, $Q$ are $\underset{\_}{409.58\text{W}}$, $\underset{\_}{286.79\text{VAR}}$, 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 $P$, $Q$ are $\underset{\_}{25.88\text{W}}$, $\underset{\_}{-96.59\text{VAR}}$, respectively, the average power is absorbed from the box, and the reactive power is delivered to the box.

Explanation of Solution

Given data:

$\begin{array}{l}v=40\cos \left(\omega t-15°\right)\text{V}\\ i=5\cos \left(\omega t+60°\right)\text{A}\end{array}$

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

$\begin{array}{l}{V}_m=40\text{V}\\ I_m=5\text{A}\\ {\theta }_v=-15°\\ {\theta }_i=60°\end{array}$

Calculation:

Substitute 40 V for , $V_m$ 5 A for $I_m$, $\left(-15°\right)$ for ${\theta }_v$, and $60°$ for ${\theta }_i$ in Equation (1) to obtain the value of average power.

$\begin{array}{c}P=\frac{\left(40\text{V}\right)\left(5\text{A}\right)}{2}\cos \left(-15°-60°\right)\\ =\left(100\right)\cos \left(-75°\right)\text{W}\\ =25.8819\text{W}\\ \cong \text{25}\text{.88}\text{W}\end{array}$

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

Substitute 40 V for , $V_m$ 5 A for $I_m$, $\left(-15°\right)$ for ${\theta }_v$, and $60°$ for ${\theta }_i$ in Equation (2) to obtain the value of reactive power.

$\begin{array}{c}Q=\frac{\left(40\text{V}\right)\left(5\text{A}\right)}{2}\sin \left(-15°-60°\right)\\ =\left(100\right)\sin \left(-75°\right)\text{VAR}\\ =-96.5925\text{VAR}\\ \cong -96.59\text{VAR}\end{array}$

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 $P$, $Q$ are $\underset{\_}{25.88\text{W}}$, $\underset{\_}{-96.59\text{VAR}}$, respectively, the average power is absorbed from the box, and the reactive power is delivered to 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 $P$, $Q$ are $\underset{\_}{-1000\text{W}}$, $\underset{\_}{-1732.05\text{VAR}}$, respectively, the average power is delivered to the box, and the reactive power is delivered to the box.

Explanation of Solution

Given data:

$\begin{array}{l}v=400\cos \left(\omega t+30°\right)\text{V}\\ i=10\sin \left(\omega t+240°\right)\text{A}\end{array}$

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

$\begin{array}{l}{V}_m=400\text{V}\\ I_m=10\text{A}\\ {\theta }_v=30°\end{array}$

Calculation:

Rewrite the given expression of current as follows:

$\begin{array}{c}i=10\cos \left[\left(\omega t+240°\right)-90°\right]\text{A}\left\{\sin \theta =\cos \left(\theta -90°\right)\right\}\\ =10\cos \left(\omega t+150°\right)\end{array}$

Therefore, the value of ${\theta }_i$ is. $150°$

${\theta }_i=150°$

Substitute 400 V for , $V_m$ 10 A for $I_m$, $30°$ for ${\theta }_v$, and $150°$ for ${\theta }_i$ in Equation (1) to obtain the value of average power.

$\begin{array}{c}P=\frac{\left(400\text{V}\right)\left(10\text{A}\right)}{2}\cos \left(30°-150°\right)\\ =\left(2000\right)\cos \left(-120°\right)\text{W}\\ =-1000\text{W}\end{array}$

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

Substitute 400 V for , $V_m$ 10 A for $I_m$, $30°$ for ${\theta }_v$, and $150°$ for ${\theta }_i$ in Equation (2) to obtain the value of reactive power.

$\begin{array}{c}Q=\frac{\left(400\text{V}\right)\left(10\text{A}\right)}{2}\sin \left(30°-150°\right)\\ =\left(2000\right)\sin \left(-120°\right)\text{VAR}\\ =-1732.0508\text{VAR}\\ \cong -1732.05\text{VAR}\end{array}$

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 $P$, $Q$ are $\underset{\_}{-1000\text{W}}$, $\underset{\_}{-1732.05\text{VAR}}$, respectively, the average power is delivered to the box, and the reactive power is 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 $P$, $Q$ are $\underset{\_}{-250\text{W}}$, $\underset{\_}{433.01\text{VAR}}$, respectively, the average power is delivered to the box, and the reactive power is absorbed from the box.

Explanation of Solution

Given data:

$\begin{array}{l}v=200\sin \left(\omega t+250°\right)\text{V}\\ i=5\cos \left(\omega t+40°\right)\text{A}\end{array}$

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

$\begin{array}{l}{V}_m=200\text{V}\\ I_m=5\text{A}\\ {\theta }_i=40°\end{array}$

Calculation:

Rewrite the given expression of voltage as follows:

$\begin{array}{c}v=200\cos \left[\left(\omega t+250°\right)-90°\right]\text{V}\\ =200\cos \left(\omega t+160°\right)\text{V}\end{array}$

Therefore, the value of ${\theta }_v$ is. $160°$

${\theta }_v=160°$

Substitute 200 V for , $V_m$ 5 A for $I_m$, $160°$ for ${\theta }_v$, and $40°$ for ${\theta }_i$ in Equation (1) to obtain the value of average power.

$\begin{array}{c}P=\frac{\left(200\text{V}\right)\left(5\text{A}\right)}{2}\cos \left(160°-40°\right)\\ =\left(500\right)\cos \left(120°\right)\text{W}\\ =-250\text{W}\end{array}$

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

Substitute 200 V for , $V_m$ 5 A for $I_m$, $160°$ for ${\theta }_v$, and $40°$ for ${\theta }_i$ in Equation (2) to obtain the value of reactive power.

$\begin{array}{c}Q=\frac{\left(200\text{V}\right)\left(5\text{A}\right)}{2}\sin \left(160°-40°\right)\\ =\left(500\right)\sin \left(120°\right)\text{VAR}\\ =433.0127\text{VAR}\\ \cong 433.01\text{VAR}\end{array}$

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 $P$, $Q$ are $\underset{\_}{-250\text{W}}$, $\underset{\_}{433.01\text{VAR}}$, respectively, the average power is delivered to the box, and the reactive power is absorbed from the box.