Chapter 12, Problem 18E

Repeat Exercise 17 with R_w = 10 Ω, and verify your answers with an appropriate set of simulations if the operating frequency is 60 Hz.

Assume the system shown in Fig. 12.34 is balanced, R_w = 0, V_an = 208∠0° V, and a positive phase sequence applies. Calculate all phase and line currents, and all phase and line voltages, if Z_p is equal to (a) 1 kΩ; (b) 100 + j48 Ω; (c) 100 − j48 Ω.

■ FIGURE 12.34

To determine

Find the line and phase currents, line and phase voltages at the load when the load impedance $Z_p=1\text{k}\Omega$ in the Figure 12.34 using LTSpice.

Answer to Problem 18E

The line and phase currents are $\underset{\_}{I_{aA}=0.206\angle 0°\text{ A,}{I}_{bB}=0.206\angle -120°\text{ A}}$, and, $\underset{\_}{I_{cC}=0.206\angle 120°\text{ A}}$. The phase voltages are $\underset{\_}{V_{AN}=206\angle 0°\text{ V,}{V}_{BN}=206\angle -120°\text{ V}}$, and $\underset{\_}{V_{CN}=206\angle 120°\text{ V}}$, and the line voltages are $\underset{\_}{V_{AB}=356.8\angle 30°\text{ V,}{V}_{BC}=356.8\angle -90°\text{ V}}$, and $\underset{\_}{V_{CA}=356.8\angle -210°\text{ V}}$.

Explanation of Solution

Given data:

The line resistance $R_w=10\Omega$.

The load impedance $Z_p=1\text{ k}\Omega$.

The source phase voltage is $V_{an}=208\angle 0°\text{ V}$.

The simulation operation frequency is $60\text{ Hz}$.

LTspice Simulation:

Draw the given circuit diagram as shown in Figure 1, where 1, 2, and 3 are placed for node representations using Label Net.

Set the values of voltages Van, Vbn and Vcn by right clicking on the voltage component, select none in “Functions” and enter the Small signal AC analysis parameters: AC amplitude as 208 and AC phase as 0 for V1, and enter other two voltage values accordingly positive phase sequence as shown in Figure 2 for V1.

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 60 Hz shown in Figure 3.

Now, run the simulation for the designed circuit. The output for the AC analysis will displays as shown in Figure 4.

For the wye-wye connection, phase currents and line currents are equal and they are equals to $I_{aA},I_{bB}$ and $I_{cC}$ respectively. From above simulation results I(R1) or I(Z1) is equals to $I_{aA}$, I(R2) or I(Z2) is equals to $I_{bB}$, I(R3) or I(Z3) is equals to $I_{cC}$.

Then,

$\begin{array}{c}{I}_{aA}=0.205941\angle 0°\text{ A}\\ \cong 0.206\angle 0°\text{ A}\\ I_{bB}=0.205941\angle -120°\text{ A}\\ \cong 0.206\angle -120°\text{ A}\\ I_{cC}=0.205941\angle 120°\text{ A}\\ \cong 0.206\angle 120°\text{ A}\end{array}$

In above simulation results, the phase voltages at the load $V_{AN},V_{BN}$  and $V_{CN}$ are represented by V(1), V(2) and V(3).

Then, the phase voltages at the load are,

$\begin{array}{c}{V}_{AN}=205.941\angle 0°\text{ V}\\ \cong 206\angle 0°\text{ V}\\ V_{BN}=205.941\angle -120°\text{ V}\\ \cong 206\angle -120°\text{ V}\\ V_{CN}=205.941\angle 120°\text{ V}\\ \cong 206\angle 120°\text{ V}\end{array}$

The magnitude of the phase voltages is $V_p=206\text{ V}$.

Write the formula to find the line voltage $V_{AB}$ as follows.

$V_{AB}=\sqrt{3}{V}_p\angle 30°\text{ V}$

Substitute $206\text{ V}$ for $V_p$.

$\begin{array}{l}{V}_{AB}=\sqrt{3}\left(206\angle 0°\right)\angle 30°\text{ V}\\ V_{AB}=356.8\angle 30°\text{ V}\end{array}$

Write the formula to find the line voltage $V_{BC}$ as follows.

$V_{BC}=\sqrt{3}{V}_p\angle -90°\text{ V}$

Substitute $206\text{ V}$ for $V_p$.

$\begin{array}{l}{V}_{BC}=\sqrt{3}\left(206\right)\angle -90°\text{ V}\\ V_{BC}=356.8\angle -90°\text{ V}\end{array}$

Write the formula to find the line voltage $V_{CA}$ as follows.

$V_{CA}=\sqrt{3}{V}_p\angle -210°\text{ V}$

Substitute $206\text{ V}$ for $V_p$.

$\begin{array}{l}{V}_{CA}=\sqrt{3}\left(206\right)\angle -210°\text{ V}\\ V_{CA}=356.8\angle -210°\text{ V}\end{array}$

Conclusion:

Thus, the line and phase currents are $\underset{\_}{I_{aA}=0.206\angle 0°\text{ A,}{I}_{bB}=0.206\angle -120°\text{ A}}$, and, $\underset{\_}{I_{cC}=0.206\angle 120°\text{ A}}$. The phase voltages are $\underset{\_}{V_{AN}=206\angle 0°\text{ V,}{V}_{BN}=206\angle -120°\text{ V}}$, and $\underset{\_}{V_{CN}=206\angle 120°\text{ V}}$, and the line voltages are $\underset{\_}{V_{AB}=356.8\angle 30°\text{ V,}{V}_{BC}=356.8\angle -90°\text{ V}}$, and $V_{CA}=356.8\angle -210°\text{ V}$.

To determine

Find the line and phase currents, line and phase voltages at the load when the load impedance $Z_p=100+j48\Omega$ in the Figure 12.34 using LTSpice.

Answer to Problem 18E

The line and phase currents are $\underset{\_}{I_{aA}=1.733\angle -23.6°\text{ A,}{I}_{bB}=1.733\angle -143.6°\text{ A}}$, and, $\underset{\_}{I_{cC}=1.733\angle -263.6°\text{ A}}$.

The phase voltages are $\underset{\_}{V_{AN}=192.24\angle 2.06°\text{ V,}{V}_{BN}=192.24\angle -117.93°\text{ V}}$, and $\underset{\_}{V_{CN}=192.24\angle -237.93°\text{ V}}$, and the line voltages are $\underset{\_}{V_{AB}=332.96\angle 32.06°\text{ V}}$, $V_{BC}=332.96\angle -87.93°\text{ V}$ and $\underset{\_}{V_{CA}=332.96\angle -207.93°\text{ V}}$.

Explanation of Solution

Given data:

Refer to part (a).

LTspice Simulation:

The load impedance is given as,

$Z_p=100+j48\Omega$

Where load resistance is $R_p=100\Omega$, and the inductive reactance is $X_P=j48\Omega$.

Write the formula to find the inductive reactance as follows.

$\begin{array}{l}{X}_p=j\omega L\\ X_p=j2\pi fL\end{array}$

Substitute $j48\Omega$ for $X_p$, 60 Hz for f.

$\begin{array}{l}j48\Omega =j2\pi \left(60\text{ Hz}\right)L\\ L=\frac{48}{2\pi \left(60\right)}\frac{\Omega }{\text{Hz}}\\ L=0.12732\text{ H }\left\{\because \frac{\Omega }{\text{Hz}}=\text{ H}\right\}\end{array}$

Draw the given circuit diagram as shown in Figure 5 for the load impedance $Z_p=100+j48\Omega$.

Keep the same simulation settings as given in Part(a) and run the simulation, then the output for the AC analysis will displays as shown in Figure 6.

From above simulation results I(R1) or I(R4) or I(L1) is equals to $I_{aA}$, I(R2) or I(R5) or I(L2) is equals to $I_{bB}$, I(R3) or I(R6) or I(L3) is equals to $I_{cC}$.

Then,

$\begin{array}{c}{I}_{aA}=1.73309\angle -23.5739°\text{ A}\\ \cong 1.733\angle -23.6°\text{ A}\\ I_{bB}=1.73309\angle -143.574°\text{ A}\\ \cong 1.733\angle -143.6°\text{ A}\\ I_{cC}=1.73309\angle 96.4261°\text{ A or}\\ =1.733\angle -263.5739°\text{ A}\\ \cong 1.733\angle -263.6°\text{ A}\end{array}$

The phase voltages at the load $V_{AN},V_{BN}$  and $V_{CN}$ are represented by V(1), V(2) and V(3).

Then, the phase voltages at the load are,

$\begin{array}{c}{V}_{AN}=192.24\angle 2.06623°\text{ V}\\ \cong 192.24\angle 2.06°\text{ V}\\ V_{BN}=192.24\angle -117.934°\text{ V}\\ \cong 192.24\angle -117.93°\text{ V}\\ V_{CN}=192.24\angle 122.066°\text{ V or}\\ =192.24\angle -237.934°\text{ V}\\ \cong 192.24\angle -237.93°\text{ V}\end{array}$

The magnitude of the phase voltages is $V_p=192.24\text{ V}$.

Write the formula to find the line voltage $V_{AB}$ as follows.

$V_{AB}=\sqrt{3}{V}_{AN}\angle 30°\text{ V}$        (1)

Substitute $192.24\angle 2.06°\text{ V}$ for $V_p$.

$\begin{array}{l}{V}_{AB}=\sqrt{3}\left(192.24\angle 2.06°\text{ V}\right)\angle 30°\text{ V}\\ V_{AB}=332.96\angle 32.06°\text{ V}\end{array}$

Write the formula to find the line voltage $V_{BC}$ as follows.

$V_{BC}=\sqrt{3}{V}_{BN}\angle 30°\text{ V}$        (2)

Substitute $192.24\angle -117.93°\text{ V}$ for $V_{BN}$.

$\begin{array}{l}{V}_{BC}=\sqrt{3}\left(192.24\angle -117.93°\text{ V}\right)\angle 30°\text{ V}\\ V_{BC}=332.96\angle -87.93°\text{ V}\end{array}$

Write the formula to find the line voltage $V_{CA}$ as follows.

$V_{CA}=\sqrt{3}{V}_{CN}\angle 30°\text{ V}$        (3)

Substitute $192.24\angle -237.93°\text{ V}$ for $V_{CN}$.

$\begin{array}{l}{V}_{CA}=\sqrt{3}\left(192.24\angle -237.93°\text{ V}\right)\angle 30°\text{ V}\\ V_{CA}=332.96\angle -207.93°\text{ V}\end{array}$

Conclusion:

Thus, the line and phase currents are $\underset{\_}{I_{aA}=1.733\angle -23.6°\text{ A,}{I}_{bB}=1.733\angle -143.6°\text{ A}}$, and, $\underset{\_}{I_{cC}=1.733\angle -263.6°\text{ A}}$.

The phase voltages are $\underset{\_}{V_{AN}=192.24\angle 2.06°\text{ V,}{V}_{BN}=192.24\angle -117.93°\text{ V}}$, and $\underset{\_}{V_{CN}=192.24\angle -237.93°\text{ V}}$, and the line voltages are $\underset{\_}{V_{AB}=332.96\angle 32.06°\text{ V}}$, $V_{BC}=332.96\angle -87.93°\text{ V}$ and $\underset{\_}{V_{CA}=332.96\angle -207.93°\text{ V}}$.

To determine

Find the line and phase currents, line and phase voltages at the load when the load impedance $Z_p=100-j48\Omega$ in the Figure 12.34 using LTspice.

Answer to Problem 18E

The line and phase currents are $\underset{\_}{I_{aA}=1.733\angle 23.6°\text{ A,}{I}_{bB}=1.733\angle -96.4°\text{ A}}$, and, $\underset{\_}{I_{cC}=1.733\angle -216.4°\text{ A}}$.

The phase voltages are $\underset{\_}{V_{AN}=192.24\angle -2.06°\text{ V,}{V}_{BN}=192.24\angle -122.06°\text{ V}}$, and $\underset{\_}{V_{CN}=192.24\angle -242.06°\text{ V}}$, and the line voltages are $\underset{\_}{V_{AB}=332.96\angle 27.94°\text{ V}}$, $\underset{\_}{V_{BC}=332.96\angle -92.06°\text{ V}}$ and $\underset{\_}{V_{CA}=332.96\angle -212.06°\text{ V}}$.

Explanation of Solution

Given data:

Refer to part (a).

LTspice Simulation:

The load impedance is given as,

$Z_p=100-j48\Omega$

Where load resistance is $R_p=100\Omega$, and the capacitive reactance is $X_P=-j48\Omega$.

Write the formula to find the inductive reactance as follows.

$\begin{array}{l}{X}_p=\frac{1}{j\omega C}\\ X_p=\frac{1}{j2\pi fC}\end{array}$

Substitute $-j48\Omega$ for $X_p$, 60 Hz for f.

$\begin{array}{l}-j48\Omega =\frac{1}{j2\pi \left(60\text{ Hz}\right)C}\\ C=\frac{1}{2\pi \left(60\right)48}\frac{1}{\Omega \cdot \text{Hz}}\\ C=0.00005526\text{ F }\left\{\because \frac{1}{\Omega \cdot \text{Hz}}=\text{ F}\right\}\end{array}$

Draw the given circuit diagram as shown in Figure 5 for the load impedance $Z_p=100-j48\Omega$.

Keep the same simulation settings as given in part(a) and run the simulation, then the output for the AC analysis will displays as shown in Figure 8.

From above simulation results, since the phase currents and line currents are equal, I(R1) or I(R4) or I(C1) is equals to $I_{aA}$, I(R2) or I(R5) or I(C2) is equals to $I_{bB}$, I(R3) or I(R6) or I(C3) is equals to $I_{cC}$.

Then,

$\begin{array}{c}{I}_{aA}=1.73308\angle 23.5755°\text{ A}\\ \cong 1.733\angle 23.6°\text{ A}\\ I_{bB}=1.73308\angle -96.4245°\text{ A}\\ \cong 1.733\angle -96.4°\text{ A}\\ I_{cC}=1.73308\angle 143.576°\text{ A or}\\ =1.733\angle -216.424°\text{ A}\\ \cong 1.733\angle -216.4°\text{ A}\end{array}$

The phase voltages at the load $V_{AN},V_{BN}$  and $V_{CN}$ are represented by V(1), V(2) and V(3).

Then, the phase voltages at the load are,

$\begin{array}{c}{V}_{AN}=192.241\angle -2.06635°\text{ V}\\ \cong 192.24\angle -2.06°\text{ V}\\ V_{BN}=192.241\angle -122.066°\text{ V}\\ \cong 192.24\angle -122.06°\text{ V}\\ V_{CN}=192.241\angle 117.934°\text{ V or}\\ =192.24\angle -242.066°\text{ V}\\ \cong 192.24\angle -242.06°\text{ V}\end{array}$

Use the same formula given in equation (1) to find the line voltage $V_{AB}$, and substitute $192.24\angle -2.06°\text{ V}$ for $V_{AN}$ in equation (1).

$\begin{array}{l}{V}_{AB}=\sqrt{3}\left(192.24\angle -2.06°\text{ V}\right)\angle 30°\text{ V}\\ V_{AB}=332.96\angle 27.94°\text{ V}\end{array}$

Use the same formula given in equation (2) to find the line voltage $V_{BC}$, and substitute $192.24\angle -122.06°\text{ V}$ for $V_{BN}$ in equation (2).

$\begin{array}{l}{V}_{BC}=\sqrt{3}\left(192.24\angle -122.06°\text{ V}\right)\angle 30°\text{ V}\\ V_{BC}=332.96\angle -92.06°\text{ V}\end{array}$

Use the same formula given in equation (3) to find the line voltage $V_{CA}$ and substitute $192.24\angle -242.06°\text{ V}$ for $V_{CN}$ in equation (3).

$\begin{array}{l}{V}_{CA}=\sqrt{3}\left(192.24\angle -242.06°\text{ V}\right)\angle 30°\text{ V}\\ V_{CA}=332.96\angle -212.06°\text{ V}\end{array}$

Conclusion:

Thus, the line and phase currents are $\underset{\_}{I_{aA}=1.733\angle 23.6°\text{ A,}{I}_{bB}=1.733\angle -96.4°\text{ A}}$, and, $\underset{\_}{I_{cC}=1.733\angle -216.4°\text{ A}}$.

The phase voltages are $\underset{\_}{V_{AN}=192.24\angle -2.06°\text{ V,}{V}_{BN}=192.24\angle -122.06°\text{ V}}$, and $\underset{\_}{V_{CN}=192.24\angle -242.06°\text{ V}}$, and the line voltages are $\underset{\_}{V_{AB}=332.96\angle 27.94°\text{ V}}$, $\underset{\_}{V_{BC}=332.96\angle -92.06°\text{ V}}$ and $\underset{\_}{V_{CA}=332.96\angle -212.06°\text{ V}}$.