Question 1

Using the operator $a=1 \angle 120^{\circ}$, evaluate the following in polar form: (a) $(a+1) /$ $\left(1+a-a^{2}\right)$, (b) $\left(a^{2}+a+j\right) /\left(j a-a^{2}\right),\left(\right.$ c) $(1-a)\left(1+a^{2}\right)$, (d) $\left(a+a^{2}\right)\left(a^{2}+1\right)$.

Accepted Answer

Using the identities given in Table 8.1:
(a)
$\begin{aligned}
\frac{a+1}{1+a-a^{2}} & =\frac{1 \angle 60^{\circ}}{\underbrace{\left(1+a+a^{2}\right)}_{0}-2 a^{2}}=\frac{1 \angle 60^{\circ}}{(-2)\left(1 \angle 240^{\circ}\right)}=-\frac{1}{2} \angle-180^{\circ} \
& =+\frac{1}{2} \angle 0^{\circ}=\frac{1}{2}
\end{aligned}$
(b) $\frac{\left(a^{2}+a\right)+j}{\left(j a-a^{2}\right)}=\frac{-1+j}{a(j-a)}=\frac{\sqrt{2} \angle 135^{\circ}}{\left(1 \angle 120^{\circ}\right)\left(j+\frac{1}{2}-j \frac{\sqrt{3}}{2}\right)}$
$=\frac{\sqrt{2} \angle 15^{\circ}}{\frac{1}{2}+j\left(1-\frac{\sqrt{3}}{2}\right)}=\frac{\sqrt{2} \angle 15^{\circ}}{0.5177 \angle 15^{\circ}}=\underline{\underline{2.732 \angle 0^{\circ}}}$
(c)
$(1-a)\left(1+a^{2}\right)=\left(\sqrt{3} \angle-30^{\circ}\right)\left(1 \angle-60^{\circ}\right)=\underline{\underline{\sqrt{3} \angle-90^{\circ}}}$
(d)
$\left(a+a^{2}\right)\left(a^{2}+1\right)=(-1)\left(1 \angle-60^{\circ}\right)=\underline{\underline{1 \angle 120^{\circ}}}=a$

Question 2

Determine the symmetrical components of the following line currents: (a) $I_{a}=10 \angle 90^{\circ}$, $I_{b}=10 \angle 340^{\circ}, I_{c}=10 \angle 200^{\circ} \mathrm{A}$; (b) $I_{a}=100, I_{b}=j 100, I_{c}=0 \mathrm{~A}$.

Accepted Answer

(a)
$\begin{aligned}
{\left[\begin{array}{l}
\bar{I}_{0} \
\bar{I}_{1} \
\bar{I}_{2}
\end{array}\right] } & =\frac{1}{3}\left[\begin{array}{ccc}
1 & 1 & 1 \
1 & a & a^{2} \
1 & a^{2} & a
\end{array}\right]\left[\begin{array}{c}
10 \angle 90^{\circ} \
10 \angle 340^{\circ} \
10 \angle 200^{\circ}
\end{array}\right]=\frac{10}{3}\left[\begin{array}{c}
1 \angle 90^{\circ}+1 \angle 340^{\circ}+1 \angle 200^{\circ} \
1 \angle 90^{\circ}+1 \angle 100^{\circ}+1 \angle 80^{\circ} \
1 \angle 90^{\circ}+1 \angle 220^{\circ}+1 \angle 320^{\circ}
\end{array}\right] \
& =\frac{10}{3}\left[\begin{array}{c}
0+j 0.316 \
0+j 2.9696 \
0-j 0.2856
\end{array}\right]=\left[\begin{array}{c}
1.0533 \angle 90^{\circ} \
9.8987 \angle 90^{\circ} \
0.9520 \angle-90^{\circ}
\end{array}\right] \mathrm{A}
\end{aligned}$
(b) $\left[\begin{array}{l}\bar{I}_{0} \ \bar{I}_{1} \ \bar{I}_{2}\end{array}\right]=\frac{1}{3}\left[\begin{array}{ccc}1 & 1 & 1 \ 1 & a & a^{2} \ 1 & a^{2} & a\end{array}\right]\left[\begin{array}{c}100 \angle 0^{\circ} \ 100 \angle 90^{\circ} \ 0\end{array}\right]=\frac{100}{3}\left[\begin{array}{c}1 \angle 0^{\circ}+1 \angle 90^{\circ} \ 1 \angle 0^{\circ}+1 \angle 210^{\circ} \ 1 \angle 0^{\circ}+1 \angle 330^{\circ}\end{array}\right]$
(c)
$=\frac{100}{3}\left[\begin{array}{c}
\sqrt{2} \angle 45^{\circ} \
0.5176 \angle-75^{\circ} \
1.9319 \angle-15^{\circ}
\end{array}\right]=\left[\begin{array}{c}
47.13 \angle 45^{\circ} \
17.253 \angle-75^{\circ} \
64.4 \angle-15^{\circ}
\end{array}\right] \mathrm{A}$

Question 3

Find the phase voltages $V_{a n}, V_{b n}$, and $V_{c n}$ whose sequence components are: $V_{0}=20 \angle 80^{\circ}, V_{1}=100 \angle 0^{\circ}, V_{2}=30 \angle 180^{\circ} \mathrm{V}$.

Accepted Answer

$\begin{aligned}
{\left[\begin{array}{l}
\bar{V}_{a n} \
\bar{V}_{b n} \
\bar{V}_{c n}
\end{array}\right] } & =\left[\begin{array}{ccc}
1 & 1 & 1 \
1 & a^{2} & a \
1 & a & a^{2}
\end{array}\right]\left[\begin{array}{c}
20 \angle 80^{\circ} \
100 \angle 0^{\circ} \
30 \angle 180^{\circ}
\end{array}\right]=\left[\begin{array}{c}
20 \angle 80^{\circ}+100 \angle 0^{\circ}+30 \angle 180^{\circ} \
20 \angle 80^{\circ}+100 \angle 290^{\circ}+30 \angle 300^{\circ} \
20 \angle 80^{\circ}+100 \angle 120^{\circ}+30 \angle 60^{\circ}
\end{array}\right] \
& =\left[\begin{array}{c}
73.47+j 19.70 \
-31.53-j 92.89 \
-31.53+j 132.3
\end{array}\right]=\left[\begin{array}{c}
76.07 \angle 15.01^{\circ} \
98.09 \angle 251.3^{\circ} \
135.98 \angle 103.4^{\circ}
\end{array}\right] \mathrm{V}
\end{aligned}$

Question 4

One line of a three-phase generator is open circuited, while the other two are short-circuited to ground. The line currents are $I_{a}=0, I_{b}=1500 / 90^{\circ}$, and $I_{c}=$ $1500 \angle-30^{\circ}$ A. Find the symmetrical components of these currents. Also find the current into the ground.

Accepted Answer

The answer of One line of a three-phase generator is...

Question 5

Given the line-to-ground voltages $V_{a g}=280 \angle 0^{\circ}, V_{b g}=290 \angle-130^{\circ}$, and $V_{c g}=260 \angle 110^{\circ}$ volts, calculate (a) the sequence components of the line-to-ground voltages, denoted $V_{\mathrm{L} g 0}, V_{\mathrm{L} g 1}$, and $V_{\mathrm{L} g 2}$; (b) line-to-line voltages $V_{a b}, V_{b c}$, and $V_{c a}$; and (c) sequence components of the line-to-line voltages $V_{\mathrm{LL} 0}, V_{\mathrm{LL} 1}$, and $V_{\mathrm{LL} 2}$. Also, verify the following general relation: $V_{\mathrm{LL} 0}=0, V_{\mathrm{LL} 1}=\sqrt{3} V_{\mathrm{L} g 1} \angle+30^{\circ}$, and $V_{\mathrm{LL} 2}=\sqrt{3} V_{\mathrm{L} g 2} \angle-30^{\circ}$ volts.

Accepted Answer

The answer of Given the line-to-ground voltages \(V_{a g}=280 \angle...

Question 6

The currents in a $\Delta$ load are $I_{a b}=10 \angle 0^{\circ}, I_{b c}=20 \angle-90^{\circ}$, and $I_{c a}=15 \angle 90^{\circ}$ A. Calculate (a) the sequence components of the $\Delta$-load currents, denoted $I_{\Delta 0}, I_{\Delta 1}, I_{\Delta 2}$; (b) the line currents $I_{a}, I_{b}$, and $I_{c}$, which feed the $\Delta$ load; and (c) sequence components of the line currents $I_{\mathrm{L} 0}, I_{\mathrm{L} 1}$, and $I_{\mathrm{L} 2}$. Also, verify the following general relation: $I_{\mathrm{L} 0}=0$, $I_{\mathrm{L} 1}=\sqrt{3} I_{\Delta 1} \angle-30^{\circ}$, and $I_{\mathrm{L} 2}=\sqrt{3} I_{\Delta 2} \angle+30^{\circ} \mathrm{A}$.

Accepted Answer

The answer of The currents in a $\Delta$ load are...

Question 7

Given that the line-to-ground voltages of $V_{a g}=280 \angle 0^{\circ}, V_{b g}=250 \angle-110^{\circ}$, and $V_{c g}=290 \angle 130^{\circ}$ volts are applied to a balanced-Y load consisting of $(12+j 16)$ ohms per phase. The load neutral is kept open. Draw the sequence networks and calculate $I_{0}, I_{1}$, and $I_{2}$, the sequence components of the line currents. Then calculate the line currents $I_{a}, I_{b}$, and $I_{c}$.

Accepted Answer

The answer of Given that the line-to-ground voltages of \(V_{a...

Question 8

Given that line-to-ground voltages of $V_{a g}=280 \angle 0^{\circ}, V_{b g}=250 \angle-110^{\circ}$, and $V_{c g}=290 \angle 130^{\circ}$ volts are applied to a balanced- $\Delta$ load consisting of $(12+j 16)$ ohms per phase. The load neutral is solidly grounded. Draw the sequence networks and calculate $I_{0}$, $I_{1}$, and $I_{2}$, the sequence components of the line currents. Then calculate the line currents $I_{a}, I_{b}$, and $I_{c}$.

Accepted Answer

The answer of Given that line-to-ground voltages of \(V_{a g}=280...

Question 9

Given that line-to-ground voltages of $V_{a g}=280 \angle 0^{\circ}, V_{b g}=250 \angle-110^{\circ}$, and $V_{c g}=290 \angle 130^{\circ}$ volts are applied to a balanced-Y load consisting of $(3+j 4)$ ohms per phase between the source and the load. The load neutral is solidly grounded. Draw the sequence networks and calculate $I_{0}, I_{1}$, and $I_{2}$, the sequence components of the line currents. Then calculate the line currents $I_{a}, I_{b}$, and $I_{c}$.

Accepted Answer

The answer of Given that line-to-ground voltages of \(V_{a g}=280...

Question 10

A balanced Y-connected generator with terminal voltage $V_{b c}=480 \angle 90^{\circ}$ volts is connected to a balanced- $\Delta$ load whose impedance is $20 \angle 40^{\circ}$ ohms per phase. The line impedance between the source and load is $0.5 \angle 80^{\circ}$ ohm for each phase. The generator neutral is grounded through an impedance of $j 5 \mathrm{ohms}$. The generator sequence impedances are given by $Z_{g 0}=j 7, Z_{g 1}=j 15$, and $Z_{g 2}=j 10 \mathrm{ohms}$. Draw the sequence networks for this system and determine the sequence components of the line currents.

Accepted Answer

The answer of A balanced Y-connected generator with terminal voltage...

Question 11

In a three-phase system, a synchronous generator supplies power to a 208-volt synchronous motor through a line having an impedance of $0.5 \angle 80^{\circ}$ ohm per phase. The motor draws $10 \mathrm{~kW}$ at 0.8 p.f. leading and at rated voltage. The neutrals of both the generator and motor are grounded through impedances of $j 5 \mathrm{ohms}$. The sequence impedances of both machines are $Z_{0}=j 5, Z_{1}=j 15$, and $Z_{2}=j 10$ ohms. Draw the sequence networks for this system and find the line-to-line voltage at the generator terminals. Assume balanced three-phase operation.

Accepted Answer

The answer of In a three-phase system, a synchronous generator...

Question 12

Given that line-to-ground voltages of $V_{a g}=280 \angle 0^{\circ}, V_{b g}=250 \angle-110^{\circ}$, and $V_{c g}=290 \angle 130^{\circ}$ volts are applied to a balanced-Y load consisting of $(3+j 4)$ ohms per phase between the source and the load. The load neutral is solidly grounded. Draw the sequence networks and calculate $I_{0}, I_{1}$, and $I_{2}$, the sequence components of the line currents. Then calculate the line currents $I_{a}, I_{b}$, and $I_{c}$. Also calculate the real and reactive power delivered to the three-phase load.

Accepted Answer

The answer of 
Given that line-to-ground voltages of \(V_{a g}=280...

Quiz 7: Symmetrical Components

Using the operator $a=1 \angle 120^{\circ}$ , evaluate the following in polar form: (a) $(a+1) /$ $\left(1+a-a^{2}\right)$ , (b) $\left(a^{2}+a+j\right) /\left(j a-a^{2}\right),\left(\right.$ c) $(1-a)\left(1+a^{2}\right)$ , (d) $\left(a+a^{2}\right)\left(a^{2}+1\right)$ .

Determine the symmetrical components of the following line currents: (a) $I_{a}=10 \angle 90^{\circ}$ , $I_{b}=10 \angle 340^{\circ}, I_{c}=10 \angle 200^{\circ} \mathrm{A}$ ; (b) $I_{a}=100, I_{b}=j 100, I_{c}=0 \mathrm{~A}$ .

Find the phase voltages $V_{a n}, V_{b n}$ , and $V_{c n}$ whose sequence components are: $V_{0}=20 \angle 80^{\circ}, V_{1}=100 \angle 0^{\circ}, V_{2}=30 \angle 180^{\circ} \mathrm{V}$ .

One line of a three-phase generator is open circuited, while the other two are short-circuited to ground. The line currents are $I_{a}=0, I_{b}=1500 / 90^{\circ}$ , and $I_{c}=$ $1500 \angle-30^{\circ}$ A. Find the symmetrical components of these currents. Also find the current into the ground.