Chapter 12, Problem 39E

(a) Find both wattmeter readings in Fig. 12.39 if V_A = 100∠0 ° V rms, V_B = 50∠90° V rms, Z_A = 10 − j10 Ω, Z_B = 8 + j6 Ω, and Z_C = 30 + j10 Ω. (b) Is the sum of these readings equal to the total power taken by the three loads? Verify your answer with an appropriate simulation.

■ FIGURE 12.39

To determine

Find the wattmeter readings in the circuit of Figure 12.39.

Answer to Problem 39E

The reading of wattmeter $A$ and wattmeter $B$ are $850\text{W}$ and $225\text{W}$ respectively.

Explanation of Solution

Given data:

The load impedance are $Z_A=10-j10\Omega$, $Z_B=8+j6\Omega$, and $Z_C=30+j10\Omega$.

Calculation:

The given figure with mesh currents is shown in Figure 1.

Apply Kirchhoff’s voltage law to the loop with current $I_1$.

$\left(10-j10\right)\left(I_1-I_3\right)-100=0$

$\left(10-j10\right)I_1-\left(10-j10\right)I_3=100$        (1)

Apply Kirchhoff’s voltage law to the loop with current $I_2$.

$\left(8+j6\right)\left(I_2-I_3\right)-\left(50\angle 90°\right)=0$

$\left(8+j6\right)I_2-\left(8+j6\right)I_3=50\angle 90°$        (2)

Apply Kirchhoff’s voltage law to the loop with current $I_3$.

$\begin{array}{l}\left(30+j10\right)I_3+\left(8+j6\right)\left(I_3-I_2\right)+\left(10-j10\right)\left(I_3-I_1\right)=0\\ \left(30+j10\right)I_3+\left(8+j6\right)I_3-\left(8+j6\right)I_2+\left(10-j10\right)I_3-\left(10-j10\right)I_1=0\end{array}$

$-\left(10-j10\right)I_1-\left(8+j6\right)I_2+\left(48+j6\right)I_3=0$        (3)

Rearrange equation (1).

$\begin{array}{l}\left(10-j10\right)I_1-\left(10-j10\right)I_3=100\\ \left(10-j10\right)I_3=\left(10-j10\right)I_1-100\\ I_3=\frac{\left(10-j10\right)I_1}{\left(10-j10\right)}-\frac{100}{\left(10-j10\right)}\end{array}$

$I_3=I_1-\left(5+j5\right)$        (4)

Substitute equation (4) in (3).

$\begin{array}{l}-\left(10-j10\right)I_1-\left(8+j6\right)I_2+\left(48+j6\right)\left(I_1-\left(5+j5\right)\right)=0\\ -\left(10-j10\right)I_1-\left(8+j6\right)I_2+\left(48+j6\right)I_1-\left(48+j6\right)\left(5+j5\right)=0\\ \left(38+j16\right)I_1-\left(8+j6\right)I_2=210+j270\\ \left(38+j16\right)I_1=\left(210+j270\right)+\left(8+j6\right)I_2\end{array}$

Simplify the equation as follows.

$I_1=\frac{\left(210+j270\right)}{\left(38+j16\right)}+\frac{\left(8+j6\right)}{\left(38+j16\right)}{I}_2$

$I_1=\left(7.235+j4.058\right)+\left(0.235+j0.0588\right)I_2$        (5)

Substitute equation (5) in (4).

$I_3=\left(7.235+j4.058\right)+\left(0.235+j0.0588\right)I_2-\left(5+j5\right)$

$I_3=\left(2.235-j0.942\right)+\left(0.235+j0.0588\right)I_2$        (6)

Substitute equation (6) in (2).

$\begin{array}{l}\left(8+j6\right)I_2-\left(8+j6\right)\left[\left(2.235-j0.942\right)+\left(0.235+j0.0588\right)I_2\right]=50\angle 90°\\ \left(8+j6\right)I_2-\left(23.532+j5.874\right)+\left(1.5272+j1.8804\right)I_2=50\angle 90°\\ \left(9.5272+j7.8804\right)I_2=\left(50\angle 90°\right)+\left(23.532+j5.874\right)\end{array}$

Simplify the equation as follows.

$\begin{array}{l}\left(9.5272+j7.8804\right)I_2=23.532+j55.874\\ I_2=6.503+j4.503\\ I_2=7.91\angle 34.70°\text{A}\end{array}$

Substitute $7.91\angle 34.70°\text{A}$ for $I_2$ in equation (2).

$\begin{array}{l}\left(8+j6\right)\left(7.91\angle 34.70°\text{A}\right)-\left(8+j6\right)I_3=50\angle 90°\\ \left(8+j6\right)I_3=25+j25\\ I_3=3.54\angle 8.13°\text{A}\end{array}$

Substitute $3.54\angle 8.13°\text{A}$ for $I_3$ in equation (1).

$\begin{array}{l}\left(10-j10\right)I_1-\left(10-j10\right)\left(3.54\angle 8.13°\right)=100\\ \left(10-j10\right)I_1=140-j30\\ I_1=10.12\angle 32.91°\text{A}\end{array}$

Calculate the power delivered by phase A as follows.

$\begin{array}{c}{P}_A=\left|V_A\right|\left|I_1\right|\cos \left({\theta }_{V_A}-{\theta }_{I_1}\right)\\ =\left(100\right)\left(10.12\right)\cos \left(0-32.91\right)\\ =849.599\text{W}\\ \text{=850}\text{W}\end{array}$

Calculate the power delivered by phase B as follows.

$\begin{array}{c}{P}_B=\left|V_B\right|\left|I_2\right|\cos \left({\theta }_{V_B}-{\theta }_{I_2}\right)\\ =\left(50\right)\left(7.91\right)\cos \left(90-34.70\right)\\ =225\text{W}\end{array}$

Conclusion:

Thus, the reading of wattmeter $A$ and wattmeter $B$ are $850\text{W}$ and $225\text{W}$ respectively.

To determine

Verify that the sum of the wattmeter reading is equal to the total power taken by the three loads using LTspice.

Explanation of Solution

Formula used:

Write the expression for inductive reactance.

$X_L=j\omega L$        (7)

Here,

$\omega$ is the angular frequency, and

$L$ is the value of the inductor.

Write the expression for capacitive reactance.

$X_C=-\frac{j}{\omega C}$        (8)

Here,

$C$ is the value of the inductor.

Calculation:

Let us assume that the angular frequency $\omega =1\frac{\text{rad}}{\text{s}}$.

Substitute $j10\Omega$ for $X_L$ and $1\frac{\text{rad}}{\text{s}}$ for $\omega$ in equation (7) to find $L$.

$\begin{array}{c}j10\Omega =j\left(1\frac{\text{rad}}{\text{s}}\right)L\\ L=\frac{j10\Omega }{j\left(1\frac{\text{rad}}{\text{s}}\right)}\\ L=10\text{H}\left\{\because 1\text{H}=1\text{Ω}\cdot \text{s}\right\}\end{array}$

Substitute $j6\Omega$ for $X_L$ and $1\frac{\text{rad}}{\text{s}}$ for $\omega$ in equation (7) to find $L$.

$\begin{array}{c}j6\Omega =j\left(1\frac{\text{rad}}{\text{s}}\right)L\\ L=\frac{j6\Omega }{j\left(1\frac{\text{rad}}{\text{s}}\right)}\\ L=6\text{H}\left\{\because 1\text{H}=1\text{Ω}\cdot \text{s}\right\}\end{array}$

Substitute $-j10\Omega$ for $X_C$ and $1\frac{\text{rad}}{\text{s}}$ for $\omega$ in equation (6) to find $C$.

$\begin{array}{c}-j10\Omega =-\frac{j}{\left(1\frac{\text{rad}}{\text{s}}\right)C}\\ C=\frac{j}{\left(1\frac{\text{rad}}{\text{s}}\right)\left(-j10\Omega \right)}\left\{\because 1\text{F}=1\frac{\text{s}}{\text{Ω}}\right\}\\ C=0.1\text{F}\end{array}$

LTspice Simulation:

Draw the given figure in LTspice as shown in Figure 2. V1 zero and V2 zero connected in the circuit to find the current flows through it.

Set the values of voltages VA by right clicking on the voltage component, select none in “Functions” and enter the Small signal AC analysis parameters: AC amplitude as 100 for V1 as shown in Figure 3 for VA.

Now save the circuit, and open the “Edit Simulation command” choose AC analysis and select the sweep type as Decade, Number of points per decade 1, Start frequency and Stop frequency as 0.159155 Hz.

Now, run the simulation for the designed circuit. The output for the AC analysis will displays as shown in Figure 5.

From above simulation results, the current flows through the load resistance are given below.

$\begin{array}{c}{I}_A=7.07107\angle 45°\text{ A}\\ I_B=4.9996\angle 53.1335°\text{ A}\\ I_C=3.53543\angle 8.13067°\text{ A}\end{array}$

Calculate the power delivered to load $R_A$ as follows.

$\begin{array}{c}{P}_A={\left|I_A\right|}^2{R}_A\\ ={\left(7.07107\right)}^2\left(10\right)\\ =500\text{W}\end{array}$

Calculate the power delivered to load $R_B$ as follows.

$\begin{array}{c}{P}_B={\left|I_B\right|}^2{R}_B\\ ={\left(4.9996\right)}^2\left(8\right)\\ =200\text{W}\end{array}$

Calculate the power delivered to load $R_C$ as follows.

$\begin{array}{c}{P}_C={\left|I_C\right|}^2{R}_C\\ ={\left(3.53543\right)}^2\left(30\right)\\ =375\text{W}\end{array}$

The total power delivered to the load is $\left(P_A+P_B+P_C=500+200+375=1075\text{W}\right)$.

The sum of the wattmeter reading is $\left(850\text{W+225}\text{W=1075}\text{W}\right)$. Thus, sum of the wattmeter reading is equal to the power taken by the three loads.

Conclusion:

Thus, it is verified that sum of the wattmeter reading is equal to the total power taken by the three loads using LTspice.