EE301
Final Exam Review
Final Exam Review
1) Determine the total resistance (𝑽𝑽𝑻𝑻 ) seen by the source and the currents (𝑰𝑰𝒔𝒔 ), (𝑰𝑰𝟏𝟏 ), and (𝑰𝑰𝟐𝟐 ) in the
DC circuit below.
10 Ω
𝟑𝟑𝟑𝟑 𝑽𝑽
𝑅𝑇 = 10 + �
𝐼𝑠 =
𝑰𝑰𝒔𝒔
+
-
𝑰𝑰𝟏𝟏
𝑉𝑉
15 Ω
12 Ω
−1
1
1
+
� = 𝟏𝟏𝟑𝟑. 𝟐𝟐 Ω
12 15 + 5
𝐼1 =
𝑉𝑉𝑠
30
=
= 𝟏𝟏. 𝟑𝟑𝟏𝟏𝟑𝟑 𝑽𝑽
𝑅𝑇 17.5
𝑅𝑒𝑞 = �
𝑰𝑰𝟐𝟐
𝐼2 =
−1
1
1
+
� = 𝟑𝟑. 𝟐𝟐 Ω
12 15 + 5
5Ω
𝑅𝑒𝑞
7.5
(1.714) = 𝟏𝟏. 𝟑𝟑𝟑𝟑𝟏𝟏 𝑽𝑽
(𝐼𝑠 ) =
𝑅12
12
𝑅𝑒𝑞
7.5
(1.714) = 𝟑𝟑. 𝟏𝟏𝟑𝟑𝟑𝟑 𝑽𝑽
(𝐼𝑠 ) =
𝑅15+5
15 + 5
2) Find the power delivered by the source (𝑷𝑷𝒔𝒔 ) and the power absorbed by all of the resistors
(𝑷𝑷𝟏𝟏𝟑𝟑 ), (𝑷𝑷𝟏𝟏𝟐𝟐 ), (𝑷𝑷𝟏𝟏𝟐𝟐 ), and (𝑷𝑷𝟐𝟐 ). Does the power being delivered equal to the total power being
absorbed?
𝑷𝑷𝒔𝒔 = 𝑷𝑷𝟏𝟏𝟑𝟑 + 𝑷𝑷𝟏𝟏𝟐𝟐 + 𝑷𝑷𝟏𝟏𝟐𝟐 + 𝑷𝑷𝟐𝟐
𝑃𝑠 = (𝐼𝑠 )(𝑉𝑉𝑠 ) = (1.714)(30) = 𝟐𝟐𝟏𝟏. 𝟑𝟑𝟐𝟐 𝑾𝑾
𝑃10 = (𝐼𝑠 )2 (𝑅10 ) = (1.714)2 (10) = 𝟐𝟐𝟐𝟐. 𝟑𝟑𝟔𝟔 𝑾𝑾
𝑃12 = (𝐼1 )2 (𝑅12 ) = (1.071)2 (12) = 𝟏𝟏𝟑𝟑. 𝟑𝟑𝟏𝟏 𝑾𝑾
𝑃15 =
(𝐼2 )2 (𝑅15 )
=
(0.643)2 (15)
= 𝟏𝟏. 𝟐𝟐𝟑𝟑 𝑾𝑾
𝟐𝟐𝟏𝟏. 𝟑𝟑𝟐𝟐 𝑾𝑾 = 29.38 + 13.76 + 6.20 + 2.07 = 𝟐𝟐𝟏𝟏. 𝟑𝟑𝟏𝟏 𝑾𝑾
Yes! Power delivered equals total
power absorbed.
𝑃5 = (𝐼2 )2 (𝑅5 ) = (0.643)2 (5) = 𝟐𝟐. 𝟑𝟑𝟑𝟑 𝑾𝑾
3) Use nodal analysis to find the voltage at the node (𝑽𝑽). Now use Ohm’s Law to find the voltage
across the 12 Ω resistor. Do these voltages equal each other?
𝑉𝑉 − 30 𝑉𝑉
𝑉𝑉
+
+
=0
10
12 15 + 5
𝑉𝑉 �
30
1
1
1
+
+ �=
10
10 12 20
𝑉𝑉[0.233] = 3
𝑉𝑉 =
3
= 12.876 𝑉𝑉
0.233
𝑉𝑉12 = (𝐼1 )(𝑅12 ) = (1.071)(12) = 12.85 𝑉𝑉
Yes! This demonstrates there are many ways to
solve a circuit.
4) Determine the total impedance (𝒁𝒁𝑻𝑻 ) seen by the source and the currents (𝑰𝑰𝒔𝒔 ), (𝑰𝑰𝟏𝟏 ), and (𝑰𝑰𝟐𝟐 ) in
the AC circuit below.
8Ω
𝟐𝟐𝟐𝟐∡𝟏𝟏𝟏𝟏𝒐𝒐 𝑽𝑽
𝑍𝑇 = 8 + �
+
𝑰𝑰𝒔𝒔
14 Ω
𝑰𝑰𝟏𝟏
-
12 Ω
-j10 Ω
𝑉𝑉𝑠
25∡15𝑜
=
= 𝟏𝟏. 𝟐𝟐𝟏𝟏∡𝟐𝟐𝟐𝟐. 𝟑𝟑𝟑𝟑𝒐𝒐 𝑽𝑽
𝑍𝑇 20.65∡−7.34𝑜
𝑍𝑒𝑞 = �
𝐼1 =
𝐼2 =
6Ω
j18 Ω
−1
−1
1
1
1
1
+
+
� =8+�
�
12 − 𝑗10 14 + 6 + 𝑗18
15.62∡−39.81𝑜 26.91∡41.99𝑜
𝑍𝑇 = 8 + [0.064∡39.81𝑜 + 0.037∡−41.99𝑜 ]−1 = 8 +
𝐼𝑠 =
𝑰𝑰𝟐𝟐
1
0.0784∡11.94 𝑜
= 𝟐𝟐𝟑𝟑. 𝟏𝟏𝟐𝟐∡−𝟑𝟑. 𝟑𝟑𝟑𝟑𝒐𝒐 Ω
−1
1
1
+
� = 12.76∡−11.94𝑜
12 − 𝑗10 14 + 6 + 𝑗18
𝑍𝑒𝑞
12.76∡−11.94𝑜
(𝐼𝑠 ) =
(1.21∡22.34𝑜 ) = 𝟑𝟑. 𝟐𝟐𝟔𝟔𝟔𝟔∡𝟐𝟐𝟑𝟑. 𝟐𝟐𝟏𝟏𝒐𝒐 𝑽𝑽
𝑍1
12 − 𝑗10
𝑍𝑒𝑞
12.76∡−11.94𝑜
(𝐼𝑠 ) =
(1.21∡22.34𝑜 ) = 𝟑𝟑. 𝟐𝟐𝟑𝟑𝟑𝟑∡−𝟑𝟑𝟏𝟏. 𝟐𝟐𝟐𝟐𝒐𝒐 𝑽𝑽
𝑍2
14 + 6 + 𝑗18
����⃗𝑻𝑻 ) and the Real/Reactive power of the
5) Find the complex power delivered by the source (𝑺𝑺
elements (𝑷𝑷𝟔𝟔 ), (𝑷𝑷𝟏𝟏𝟐𝟐 ), (𝑸−𝒋𝒋𝟏𝟏𝟑𝟑 ), (𝑷𝑷𝟏𝟏𝟑𝟑 ), (𝑷𝑷𝟏𝟏 ) and (𝑸𝒋𝒋𝟏𝟏𝟔𝟔 ). Does the source complex power equal to
the total Real and Reactive power of the elements?
∗
����⃗
���⃗𝑠 ��𝐼��⃗𝑠 � = (25∡15𝑜 )(1.21∡−22.34𝑜 )
𝑆𝑇 = �𝑉𝑉
����⃗𝑇 = 𝟑𝟑𝟑𝟑. 𝟐𝟐𝟐𝟐∡−𝟑𝟑. 𝟑𝟑𝟑𝟑𝒐𝒐 𝑽𝑽𝑽𝑽
𝑆
𝑃8 = (𝐼𝑠 )2 (𝑅8 ) = (1.21)2 (8) = 𝟏𝟏𝟏𝟏. 𝟑𝟑𝟏𝟏 𝑾𝑾
𝑃12 = (𝐼1 )2 (𝑅12 ) = (0.988)2 (12) = 𝟏𝟏𝟏𝟏. 𝟑𝟑𝟏𝟏 𝑾𝑾
𝑄−𝑗10 = (𝐼1 )2 (𝑋𝑐 ) = (0.988)2 (10) = −𝟐𝟐. 𝟑𝟑𝟏𝟏 𝑽𝑽𝑽𝑽𝑽𝑽
Remember that an asterisk (*) is the symbol for the complex conjugate.
𝐼 2 = (𝐼)(𝐼 ∗ ) = (𝐼∡𝜃)(𝐼∡ − 𝜃) = 𝐼 2 ∡𝜃 − 𝜃 = 𝐼 2 ∡0𝑜 = 𝐼 2
𝑃14 = (𝐼2 )2 (𝑅14 ) = (0.574)2 (14) = 𝟑𝟑. 𝟏𝟏𝟏𝟏 𝑾𝑾
𝑃6 = (𝐼2 )2 (𝑅6 ) = (0.574)2 (6) = 𝟏𝟏. 𝟐𝟐𝟔𝟔 𝑾𝑾
𝑄𝑗18 = (𝐼2 )2 (𝑋𝐿 ) = (0.574)2 (10) = 𝟐𝟐. 𝟐𝟐𝟑𝟑 𝑽𝑽𝑽𝑽𝑽𝑽
����⃗
𝑺𝑺𝑻𝑻 = (𝑷𝑷𝟔𝟔 + 𝑷𝑷𝟏𝟏𝟐𝟐 + 𝑷𝑷𝟏𝟏𝟑𝟑 + 𝑷𝑷𝟏𝟏 ) + 𝒋𝒋(𝑸−𝒋𝒋𝟏𝟏𝟑𝟑 + 𝑸𝒋𝒋𝟏𝟏𝟔𝟔 )
����⃗
𝑆
𝑇 = (11.71 + 11.71 + 4.61 + 1.98) + 𝑗(−9.76 + 5.93)
����⃗
𝑆𝑇 = (30.01) + 𝑗(−3.83) = 𝟑𝟑𝟑𝟑. 𝟐𝟐𝟐𝟐∡−𝟑𝟑. 𝟐𝟐𝟑𝟑𝒐𝒐 𝑽𝑽𝑽𝑽
6) What is the difference between Apparent Power and Complex Power?
Complex Power has a magnitude and angle while Apparent Power is just the magnitude of the
Complex Power.
7) Find the power factor (𝝁𝝁𝒑𝒑 ) for the given Complex Power. Make sure to include if it is Leading or
Lagging.
a. 𝑆⃗ = 322∡30.25𝑜 𝑉𝑉𝐴
𝐹𝑝 = 𝟑𝟑. 𝟔𝟔𝟏𝟏𝟑𝟑 𝑳𝑳𝑳𝑳𝒈𝒈𝒊𝒊𝒏𝒏𝒈
b. 𝑆⃗ = 259 − 𝑗324 𝑉𝑉𝐴
tan 𝜃 =
𝑂𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑄
=
𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑃
−324
𝜃 = tan−1 �
� = −51.36𝑜
259
𝐹𝑝 = 𝟑𝟑. 𝟏𝟏𝟐𝟐𝟑𝟑 𝑳𝑳𝒆𝒆𝑳𝑳𝑳𝑳𝒊𝒊𝒏𝒏𝒈
𝐹𝑝 = cos 𝜃 = cos(30.25) = 0.864
**Since the Complex Power Angle is Positive the Load
is Inductive and using the acronym 𝑬𝑬𝑳𝑳𝑰𝑰 current Lags
voltage, therefore it is a Lagging Power Factor.
𝑆⃗ = 𝑆∡−51.36𝑜
𝐹𝑝 = cos(−51.36) = 0.624
**Since the Complex Power Angle is
Negative the Load is Capacitive
(𝑰𝑰𝑪𝑪𝑬𝑬) so current Leads voltage.
𝟐𝟐𝟐𝟐𝟐𝟐 𝑾𝑾
𝜽𝜽
𝑺𝑺
−𝟑𝟑𝟑𝟑𝟑𝟑 𝑽𝑽𝑽𝑽𝑽𝑽
8) An AC circuit operates at a frequency of 𝟏𝟏. 𝟑𝟑𝟏𝟏𝟐𝟐 𝒌𝑻𝑻𝒛 and its impedance is 𝒁𝒁 = 𝟐𝟐𝟐𝟐 + 𝒋𝒋𝟏𝟏𝟑𝟑𝟑𝟑 Ω. Is
the impedance Inductive or Capacitive? What is the Inductive/Capacitive component value?
**Since the imaginary component is positive then the impedance is Inductive.
𝑋𝐿 = 𝜔𝐿
𝐿=
𝑋𝐿
130
=
= 𝟑𝟑. 𝟐𝟐 𝒎𝑻𝑻
𝜔 2𝜋(6.465𝑥103 )
9) An AC circuit operates at a frequency of 𝟑𝟑𝟔𝟔𝟑𝟑. 𝟏𝟏𝟑𝟑 𝒓𝑳𝑳𝑳𝑳/𝒔𝒔 and its impedance is 𝒁𝒁 = 𝟑𝟑𝟐𝟐 − 𝒋𝒋𝟏𝟏𝟑𝟑𝟐𝟐 Ω.
Is the impedance Inductive or Capacitive? What is the Inductive/Capacitive component value?
**Since the imaginary component is negative then the impedance is Capacitive.
𝑋𝑐 =
𝐶=
1
𝜔𝐶
1
1
=
= 𝟏𝟏𝟔𝟔 𝝁𝝁𝝁𝝁
𝜔(𝑋𝑐 ) (383.14)(145)
10) The following circuit is operating at a frequency of 𝝎 = 𝟐𝟐𝟑𝟑𝟑𝟑 𝒓𝑳𝑳𝑳𝑳/𝒔𝒔.
𝑰𝑰𝒔𝒔
+
𝟏𝟏𝟏𝟏𝟏𝟏 𝑾𝑾
𝟐𝟐𝟐𝟐𝟐𝟐∡𝟒𝟒𝟒𝟒𝒐𝒐 𝑽𝑽
𝟒𝟒𝟒𝟒𝟒𝟒 𝑽𝑽𝑽𝑽𝑽𝑽
-
a. Draw the Power Triangle for the Load and label the Apparent, Real, and Reactive
Power. Also include the Complex Power angle.
�𝑆⃗� = �1602 + 4002 = 430.81 𝑉𝑉𝐴
400
𝜃 = tan−1 �
� = 68.2𝑜
160
b. Find the source current (𝑰𝑰𝒔𝒔 ).
∗
����⃗𝑠 � ����⃗
𝑆⃗ = �𝑉𝑉
𝐼𝑠 �
∗
�𝐼��⃗𝑠 � =
430.81∡68.2𝑜
220∡40𝑜
𝟔𝟔𝟔𝟔. 𝟐𝟐𝒐𝒐
𝟒𝟒𝟒𝟒𝟒𝟒 𝑽𝑽𝑽𝑽𝑽𝑽
𝟏𝟏𝟏𝟏𝟏𝟏 𝑾𝑾
𝐼𝑠 = 𝟏𝟏. 𝟐𝟐𝟏𝟏∡−𝟐𝟐𝟔𝟔. 𝟐𝟐𝒐𝒐 𝑽𝑽
= 1.96∡28.2𝑜 𝐴
c. Typically we like to reduce the source current. This can be accomplished by connecting
a Capacitor in parallel to an Inductive Load. The Capacitor component value (𝑪𝑪) is
determined by choosing a Capacitance Reactive Power (𝑸𝒄𝒄 ) equal to the Inductance
Reactive Power (𝑸𝑳𝑳 ).
Find the Capacitor component value so all of the Reactive Power at the Load is
cancelled. In other words, what component value will correct the power factor to
unity?
|𝑄𝐿 | = |𝑄𝑐 | = 400 𝑉𝑉𝐴𝑅
𝑉𝑉 2
𝑄𝑐 =
=>
𝑋𝑐
𝑋𝑐 =
1
=>
𝜔𝐶
(220)2
𝑋𝑐 =
= 121 Ω
400
𝐶=
1
= 𝟏𝟏𝟏𝟏. 𝟐𝟐𝟑𝟑 𝝁𝝁𝝁𝝁
(500)(121)
d. Find the source current (𝑰𝑰𝒔𝒔 ) when a Capacitor is connected in parallel with the Load.
Assume the Capacitance Reactive Power is −𝟑𝟑𝟑𝟑𝟑𝟑 𝑽𝑽𝑽𝑽𝑽𝑽.
+
𝑰𝑰𝒔𝒔
𝑰𝑰𝟏𝟏
𝟏𝟏𝟏𝟏𝟏𝟏 𝑾𝑾
𝟐𝟐𝟐𝟐𝟐𝟐∡𝟒𝟒𝟒𝟒𝒐𝒐 𝑽𝑽
𝐼1 = 1.96∡−28.2𝑜 𝐴
(𝐼2
=
−𝟒𝟒𝟒𝟒𝟒𝟒 𝑽𝑽𝑽𝑽𝑽𝑽
𝟒𝟒𝟒𝟒𝟒𝟒 𝑽𝑽𝑽𝑽𝑽𝑽
-
)∗
𝑰𝑰𝟐𝟐
�𝑆⃗
���⃗
𝑉𝑉𝑠
=
400∡−900
220∡40𝑜
𝐼2 = 1.82∡130𝑜 𝐴
𝐼𝑠 = 𝐼1 + 𝐼2
𝑜
= 1.82∡−130 𝐴
𝐼𝑠 = (1.96∡−28.2𝑜 ) + (1.82∡130𝑜 )
𝐼𝑠 = 𝟑𝟑. 𝟑𝟑𝟐𝟐𝟔𝟔∡𝟑𝟑𝟑𝟑. 𝟑𝟑𝟏𝟏𝒐𝒐 𝑽𝑽
11) Find the Primary (𝑰𝑰𝒑𝒑 ) and Secondary (𝑰𝑰𝒔𝒔 ) currents for the circuit shown below.
𝟏𝟏𝟏𝟏 Ω
𝟔𝟔𝟔𝟔∡𝟎𝟎𝒐𝒐 𝑽𝑽
+
𝑰𝑰𝒑𝒑
𝒋𝒋𝒋𝒋𝒋𝒋 Ω
𝟔𝟔𝟔𝟔∡𝟎𝟎𝒐𝒐 𝑽𝑽
+
-
𝑰𝑰𝒑𝒑
𝑰𝑰𝒔𝒔
-
−𝒋𝒋𝒋𝒋 Ω
𝟔𝟔Ω
**Reflecting everything to the Primary side.
𝟏𝟏𝟏𝟏 Ω
4:6
𝒋𝒋𝒋𝒋𝒋𝒋 Ω
𝒁𝒁−𝒋𝒋𝒋𝒋
𝒁𝒁𝟔𝟔
𝑎=
𝑁𝑝 4 2
= =
𝑁𝑠 6 3
2 2
𝑍−𝑗8 = 𝑎2 (𝑍𝑠 ) = � � (−𝑗8) = −𝑗3.56 Ω
3
2 2
𝑍6 = � � (6) = 2.67 Ω
3
𝐼𝑝 =
𝑉𝑉𝑠
60∡0𝑜
=
= 𝟑𝟑. 𝟐𝟐𝟏𝟏∡−𝟑𝟑𝟐𝟐. 𝟑𝟑𝟔𝟔𝒐𝒐 𝑽𝑽
𝑍𝑇 10 + 𝑗15 − 𝑗3.56 + 2.67
2
𝐼𝑠 = 𝑎�𝐼𝑝 � = (3.51∡−42.08𝑜 ) = 𝟐𝟐. 𝟑𝟑𝟑𝟑∡−𝟑𝟑𝟐𝟐. 𝟑𝟑𝟔𝟔𝒐𝒐 𝑽𝑽
3
12) Refer to the circuit in problem 11. What component values do we need if we treat the
secondary side as the Load and can replace the Load impedances (𝒁𝒁𝑳𝑳𝒐𝒐𝑳𝑳𝑳𝑳 ) with components that
will achieve Maximum Power Transfer?
𝑍𝐿𝑜𝑎𝑑 =
(𝑍𝑇𝐻 )∗ 10 − 𝑗15
=
= 𝟐𝟐𝟐𝟐. 𝟐𝟐 − 𝒋𝒋𝟑𝟑𝟑𝟑. 𝟑𝟑𝟐𝟐 Ω
𝑎2
2 2
� �
3
13) Find the Thevenin Equivalent (𝑽𝑽𝑻𝑻𝑻𝑻) and (𝒁𝒁𝑻𝑻𝑻𝑻 ) for the circuit below.
220 Ω
j30 Ω
100 Ω
𝒁𝒁𝑳𝑳𝑳𝑳𝑳𝑳𝑳𝑳
-j50 Ω
𝟓𝟓∡𝟒𝟒𝟒𝟒𝒐𝒐 𝑨𝑨
-j20 Ω
j30 Ω
220 Ω
First remove the Load. Next turn off the source
(Current Source = Open) and find the Thevenin
Impedance from the Load perspective. Note
the 220Ω and the –j20Ω are not included
because they do not provide a complete path.
100 Ω
-j50 Ω
𝒁𝒁𝑻𝑻𝑻𝑻
-j20 Ω
𝑍𝑇𝐻 = 100 + 𝑗30 − 𝑗50 = 𝟏𝟏𝟑𝟑𝟑𝟑 − 𝒋𝒋𝟐𝟐𝟑𝟑 Ω
j30 Ω
220 Ω
Now turn on the source and find the Thevenin
Voltage at the open terminals where the Load
was. Note the j30Ω and 100Ω are on an open
branch so no current travels through them and
no voltage drop occurs.
𝟓𝟓∡𝟒𝟒𝟒𝟒𝒐𝒐 𝑨𝑨
100 Ω
+
𝑽𝑽𝑻𝑻𝑻𝑻
-
-j50 Ω
-j20 Ω
𝑉𝑉𝑇𝐻 = (𝐼𝑠 )(𝑍−𝑗50 ) = (5∡46𝑜 )(50∡−90𝑜 ) = 𝟐𝟐𝟐𝟐𝟑𝟑∡−𝟑𝟑𝟑𝟑𝒐𝒐 𝑽𝑽
a. Determine the Load (𝒁𝒁𝑳𝑳𝒐𝒐𝑳𝑳𝑳𝑳 ) that will achieve Maximum Power Transfer and draw the
Thevenin Circuit with the Load.
𝑍𝐿𝑜𝑎𝑑 = (𝑍𝑇𝐻 )∗ = 100 + 𝑗20 Ω
𝟐𝟐𝟐𝟐𝟐𝟐∡−𝟒𝟒𝟒𝟒𝒐𝒐 𝑽𝑽
𝟏𝟏𝟏𝟏𝟏𝟏 Ω
+
-
−𝒋𝒋𝒋𝒋𝒋𝒋 Ω
𝟏𝟏𝟏𝟏𝟏𝟏 Ω
𝒋𝒋𝒋𝒋𝒋𝒋 Ω
14) Find the time constants for the circuit below if:
50 Ω
120 V
+
-
1
𝒏𝒏
2
12 Ω
𝑪𝑪 = 𝟑𝟑𝟑𝟑𝟑𝟑. 𝟑𝟑𝟑𝟑𝝁𝝁𝝁𝝁
20 Ω
+
𝑽𝑽𝒄𝒄
-
30 Ω
1 kΩ
a. The switch is in the (𝒏𝒏) position for a long time and then moves to position (𝟏𝟏) until
steady-state is reached. Is the Capacitor Charging or Discharging?
50 Ω
𝑅𝑒𝑞 =
(50)(30)
= 18.75 Ω
50 + 30
𝜏𝑛→1 = (𝑅𝑒𝑞 )(𝐶) =
Charging
+
30 Ω
(18.75)(373.33𝑥10−6 )
= 𝟑𝟑 𝒎𝒔𝒔
-
b. The switch is in the (1) position for a long time and then moves to position (2) until
steady-state is reached. Is the Capacitor Charging or Discharging?
2
(32)(373.33𝑥10−6 )
12 Ω
= 𝟏𝟏𝟏𝟏. 𝟐𝟐𝟐𝟐 𝒎𝒔𝒔
+
𝑹𝑹𝒆𝒆𝒆𝒆
-
15) In the Linear Motor below (𝑽𝑽𝒔𝒔 = 5.5 𝑘𝑉𝑉), (𝜷 = 8𝑇), (𝑽𝑽𝑽𝑽 = 12 Ω), and (𝑳𝑳 = 3𝑚). At (𝑡 = 0) the
switch is closed. Assume the sliding bar is initially at rest and there is no mechanical friction.
𝑹𝑹𝑨𝑨
𝑽𝑽𝒔𝒔
+
-
𝑰𝑰𝒔𝒔
+
𝑬𝑬𝒊𝒊𝒊𝒊𝒊𝒊 𝑳𝑳
-
a. Find the initial current and Force just after the switch is closed.
**Initial 𝐸𝑖𝑛𝑑 = 0 because there is no motion.
𝐾𝑉𝑉𝐿: − 𝑉𝑉𝑠 + (𝐼𝑠 )(𝑅𝐴 ) + 𝐸𝑖𝑛𝑑 = 0
𝐼𝑠 =
𝑉𝑉𝑠 5.5𝑥103
=
= 𝟑𝟑𝟐𝟐𝟔𝟔. 𝟑𝟑𝟑𝟑 𝑽𝑽
𝑅𝐴
12
𝑹𝑹𝒆𝒆𝒆𝒆
Discharging
𝑅𝑒𝑞 = 12 + 20 = 32 Ω
𝜏1→2 = (𝑅𝑒𝑞 )(𝐶) =
1
𝐹 = 𝐼𝐿𝛽 = (458.33)(3)(8)
𝐹 = 𝟏𝟏𝟏𝟏 𝒌𝑰𝑰
20 Ω
b. Find the velocity of the bar and Force when (𝑰𝑰𝒔𝒔 = 50 𝐴).
** 𝐸𝑖𝑛𝑑 = 𝑢𝛽𝐿
𝐾𝑉𝑉𝐿: − 𝑉𝑉𝑠 + (𝐼𝑠 )(𝑅𝐴 ) + 𝑢𝛽𝐿 = 0
𝑢=
𝑉𝑉𝑠 − (𝐼𝑠 )(𝑅𝐴 ) (5.5𝑥103 ) − (50)(12)
=
= 𝟐𝟐𝟑𝟑𝟑𝟑. 𝟏𝟏𝟑𝟑 𝒎/𝒔𝒔
𝛽𝐿
(8)(3)
𝐹 = 𝐼𝐿𝛽 = (50)(3)(8) = 𝟏𝟏. 𝟐𝟐 𝒌𝑰𝑰
c. When the bar begins to move why does the total current (𝑰𝑰𝒔𝒔 ) in the circuit decrease?
The current decreases because the induced voltage (𝑬𝑬𝒊𝒊𝒏𝒏𝑳𝑳 ) across the sliding bar will
increase creating a current opposing the source current (𝑰𝑰𝒔𝒔 ).
16) Match the following definitions.
___
I Personnel Safety
A. Converts Electrical Energy into Mechanical Energy.
___
D Faraday’s Law
B.
Converts Mechanical Energy into Electrical Energy.
___
G Bus
C.
Magnetic field created by a current carrying wire interacts with
an existing magnetic field to exert a developed force on the wire.
___
E Commutator
H Equipment Reliability
___
___
A Motor
D. Movement of a conductor in a magnetic field that will induce a
voltage.
E.
A segmented device commonly found in DC motors that are used
with brushes to reverse the direction of the applied current on
the rotor.
F.
A device commonly found in AC generators that are used with
brushes to pass a DC current to create an electromagnet on the
rotor.
___
J Circuit Breakers
___
F Slip Ring
___
C Lorentz Force Law
___
B Generator
G. A heavy gauge conductor that connects multiple circuits or loads
to a common voltage supply.
H. The main reason an ungrounded system is used on Navy Ships.
I.
A critical downside of using an ungrounded system on Navy
Ships.
J.
A device designed to trip when overcurrent or high currents are
reached.
17) A DC Motor was tested under two operating speeds. Find the armature resistance (𝑽𝑽𝑳𝑳 ), motor
constant (𝑲𝒅𝒅 ), and the Mechanical Torque Loss (𝑻𝑻𝑳𝑳𝒐𝒐𝒔𝒔𝒔𝒔).
Test 1: Applied (𝟔𝟔𝟑𝟑 𝑽𝑽𝑫𝑫𝑪𝑪 ) and measured (𝑰𝑰𝑳𝑳 = 𝟑𝟑. 𝟐𝟐 𝑽𝑽) at (𝟏𝟏𝟐𝟐𝟏𝟏𝟑𝟑 𝒓𝒑𝒑𝒎) with no load
Test 2: Applied (𝟔𝟔𝟑𝟑 𝑽𝑽𝑫𝑫𝑪𝑪 ) and measured (𝑰𝑰𝑳𝑳 = 𝟏𝟏𝟔𝟔 𝑽𝑽) at (𝟏𝟏𝟑𝟑𝟐𝟐𝟑𝟑 𝒓𝒑𝒑𝒎) with a load
𝑹𝑹𝒂𝒂
𝑽𝑽𝑫𝑫𝑫𝑫
+
-
𝑰𝑰𝒂𝒂
Applying KVL to the circuit to the right we get:
𝑬𝑬𝒂𝒂
−𝑉𝑉𝐷𝐶 + (𝐼𝑎 )(𝑅𝑎 ) + 𝐸𝑎 = 0
+
-
𝐸𝑎 = (𝐾𝑣 )𝜔
−𝑉𝑉𝐷𝐶 + (𝐼𝑎 )(𝑅𝑎 ) + (𝐾𝑣 )𝜔 = 0
𝑷𝑷𝑶𝑶𝑶𝑶𝑶𝑶
𝑷𝑷𝒅𝒅𝒅𝒅𝒅𝒅
𝑷𝑷𝑰𝑰𝑰𝑰
𝑷𝑷𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴 𝑳𝑳𝑳𝑳𝑳𝑳𝑳𝑳
𝑷𝑷𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬 𝑳𝑳𝑳𝑳𝑳𝑳𝑳𝑳
Test 1: −87 + (7.5)𝑅𝑎 + (𝐾𝑣 )(200.01) = 0
Test 2: −87 + (68)𝑅𝑎 + (𝐾𝑣 )(180.12) = 0
68
Multiply the Test 1 Equation by �− � and add
7.5
the two equation together to solve for 𝑲𝒅𝒅 .
+
𝑅𝑃𝑀
1910
𝜔1 = 2𝜋 �
� = 2𝜋 �
� = 200.01 𝑟𝑎𝑑/𝑠
60
60
𝑅𝑃𝑀
1720
𝜔2 = 2𝜋 �
� = 2𝜋 �
� = 180.12 𝑟𝑎𝑑/𝑠
60
60
788.8 − (68)𝑅𝑎 + (𝐾𝑣 )(−1813.42) = 0
−87 + (68)𝑅𝑎 + (𝐾𝑣 )(180.12)
=0
701.8 + (0)𝑅𝑎 + (𝐾𝑣 )(−1633.3) = 0
𝐾𝑣 =
701.8
= 𝟑𝟑. 𝟑𝟑𝟐𝟐𝟐𝟐𝟑𝟑 𝑽𝑽 − 𝒔𝒔
1633.3
Now substitute 𝐾𝑣 into Test 1 Equation to solve for 𝑽𝑽𝑳𝑳 .
−87 + (7.5)𝑅𝑎 + (0.4297)(200.01) = 0
𝑅𝑎 =
87 − (0.4297)(200.01)
= 𝟑𝟑. 𝟏𝟏𝟑𝟑𝟏𝟏 Ω
7.5
Since Test 1 is unloaded than we can
set 𝑇𝐿𝑜𝑎𝑑 = 0.
𝑇𝑑𝑒𝑣 = (𝐾𝑣 )(𝐼𝑎 ) = 𝑇𝐿𝑜𝑠𝑠 + 𝑇𝐿𝑜𝑎𝑑
𝑇𝐿𝑜𝑠𝑠 = (0.4297)(7.5) = 𝟑𝟑. 𝟐𝟐𝟐𝟐𝟑𝟑 𝑰𝑰 − 𝒎
a. Find (𝑷𝑷𝑰𝑰𝑰𝑰 ), (𝑷𝑷𝑬𝑬𝑬𝑬𝒆𝒆𝒄𝒄 𝑳𝑳𝒐𝒐𝒔𝒔𝒔𝒔), (𝑷𝑷𝑳𝑳𝒆𝒆𝒅𝒅 ), (𝑷𝑷𝑴𝑴𝒆𝒆𝒄𝒄𝑴𝑴 𝑳𝑳𝒐𝒐𝒔𝒔𝒔𝒔 ), (𝑷𝑷𝑶𝑶𝑶𝑶𝑻𝑻 ), and the efficiency (ƞ) of the Motor
using the results from Test 2. Assume 𝑇𝐿𝑜𝑠𝑠 is constant.
𝑃𝐼𝑁 = (𝐼𝑎 )(𝑉𝑉𝐷𝐶 ) = (68)(87) = 𝟐𝟐𝟐𝟐𝟏𝟏𝟏𝟏 𝑾𝑾
𝑃𝐸𝑙𝑒𝑐 𝐿𝑜𝑠𝑠 = (𝐼𝑎 )2 (𝑅𝑎 ) = (68)2 (0.141) = 𝟏𝟏𝟐𝟐𝟏𝟏. 𝟐𝟐𝟐𝟐 𝑾𝑾
𝑃𝑑𝑒𝑣 = (𝐼𝑎 )(𝐾𝑣 )(𝜔) = (68)(0.4297)(180.12) = 𝟐𝟐𝟐𝟐𝟏𝟏𝟑𝟑. 𝟑𝟑𝟑𝟑 𝑾𝑾
𝐴𝑙𝑠𝑜 𝑃𝑑𝑒𝑣 = 𝑃𝐼𝑁 − 𝑃𝐸𝑙𝑒𝑐 𝐿𝑜𝑠𝑠 = 5916 − 651.99 = 𝟐𝟐𝟐𝟐𝟏𝟏𝟑𝟑. 𝟑𝟑𝟏𝟏 𝑾𝑾
𝑃𝑀𝑒𝑐ℎ 𝐿𝑜𝑠𝑠 = (𝑇𝐿𝑜𝑠𝑠 )(𝜔) = (3.223)(180.12) = 𝟐𝟐𝟔𝟔𝟑𝟑. 𝟐𝟐𝟑𝟑 𝑾𝑾
𝑃𝑂𝑈𝑇 = 𝑃𝐼𝑁 − 𝑃𝐸𝑙𝑒𝑐 𝐿𝑜𝑠𝑠 − 𝑃𝑀𝑒𝑐ℎ 𝐿𝑜𝑠𝑠 = 5916 − 651.99 − 580.53 = 𝟑𝟑𝟏𝟏𝟔𝟔𝟑𝟑. 𝟏𝟏𝟔𝟔 𝑾𝑾
ƞ=
𝑃𝑂𝑈𝑇
4683.68
(100) =
(100) = 𝟑𝟑𝟐𝟐. 𝟏𝟏𝟑𝟑 %
𝑃𝐼𝑁
5916
18) In the balanced 3-phase circuit find the Phase Voltage (𝑬𝑬𝑽𝑽𝑰𝑰 ), the Line Current (𝑰𝑰𝒃𝒃 ), and the Total
Complex Power at the Load (𝑺𝑺𝑻𝑻 ). Assume positive sequence and 𝑬𝑬𝑽𝑽𝑩𝑩 = 𝟑𝟑𝟑𝟑𝟏𝟏. 𝟑𝟑𝟏𝟏∡𝟑𝟑𝟑𝟑𝒐𝒐 𝑽𝑽.
𝑨𝑨
−𝒋𝒋𝒋𝒋 Ω 𝟏𝟏𝟏𝟏 Ω
𝑰𝑰𝒂𝒂
+
𝑬𝑬𝑨𝑨
𝒂𝒂
𝒋𝒋𝒋𝒋𝒋𝒋 Ω
-
𝒋𝒋𝒋𝒋𝒋𝒋 Ω
𝑪𝑪
𝟏𝟏𝟏𝟏 Ω
𝑩𝑩
𝑰𝑰𝒃𝒃
𝑰𝑰𝒄𝒄
−𝒋𝒋𝒋𝒋 Ω 𝟏𝟏𝟏𝟏 Ω
−𝒋𝒋𝒋𝒋 Ω 𝟏𝟏𝟏𝟏 Ω
𝟏𝟏𝟏𝟏 Ω
𝒄𝒄
𝒋𝒋𝒋𝒋𝒋𝒋 Ω
𝟏𝟏𝟏𝟏 Ω
𝒃𝒃
Typically the objective in 3-phase problems is to redraw the circuit as a Single Phase with
reference to the neutral line. Phase 𝑽𝑽 is commonly chosen as the phase to work with and this
means we need to know the Line to Neutral Voltage (𝑬𝑬𝑽𝑽𝑰𝑰 ) and make sure our Load is in the
Y-formation.
𝐸𝐴𝐵 = 𝐸𝐴𝑁 (√3 ∡30𝑜 )
𝐸𝐴𝑁 =
𝐸𝐴𝐵
√3 ∡30𝑜
=
346.41∡30𝑜
√3 ∡30𝑜
= 𝟐𝟐𝟑𝟑𝟑𝟑∡𝟑𝟑𝒐𝒐 𝑽𝑽
Single Phase Circuit
𝑨𝑨
−𝒋𝒋𝒋𝒋 Ω 𝟏𝟏𝟏𝟏 Ω
𝑰𝑰𝒂𝒂
+
𝒂𝒂
𝑬𝑬𝑨𝑨𝑨𝑨
𝟏𝟏𝟏𝟏 Ω
-
𝑵𝑵
𝐼𝑎 =
𝐸𝐴𝑁
200∡0𝑜
=
= 8∡−16.26𝑜 𝐴
𝑍𝑇
−𝑗8 + 14 + 10 + 𝑗15
𝒏𝒏
𝒋𝒋𝒋𝒋𝒋𝒋 Ω
𝐼𝑏 = 𝐼𝑎 (∡−120𝑜 ) = 8∡(−16.26𝑜 − 120𝑜 ) = 𝟔𝟔∡ − 𝟏𝟏𝟑𝟑𝟏𝟏. 𝟐𝟐𝟏𝟏𝒐𝒐 𝑽𝑽
����⃗∅ = (𝐼)2 (𝑍) = (8)2 (10 + 𝑗15) = 640 + 𝑗960 𝑉𝑉𝐴
𝑆
����⃗
����⃗∅ = 1920 + 𝑗2880 = 𝟑𝟑𝟑𝟑𝟏𝟏𝟏𝟏. 𝟑𝟑𝟑𝟑∡𝟐𝟐𝟏𝟏. 𝟑𝟑𝟏𝟏𝒐𝒐 𝑽𝑽𝑽𝑽
𝑆𝑇 = (3)𝑆
19) In the balanced 3-phase circuit find the Phase Impedance (𝒁𝒁∆ ) and the Phase Current (𝑰𝑰𝑳𝑳𝒃𝒃 ). Also
����⃗𝑻𝑻 � Power delivered by the
find the Total Real (𝑷𝑷𝑻𝑻 ), Total Reactive (𝑸𝑻𝑻 ), and Total Apparent �𝑺𝑺
����⃗∅ = 𝟑𝟑𝟔𝟔𝟑𝟑 − 𝒋𝒋𝟑𝟑𝟏𝟏𝟑𝟑 𝑽𝑽𝑽𝑽),
Generator. The per-phase Complex Power at the Load is (𝑺𝑺
𝒐𝒐
𝒐𝒐
(𝑬𝑬𝑽𝑽𝑰𝑰 = 𝟔𝟔𝟑𝟑∡𝟐𝟐𝟑𝟑 𝑽𝑽), and (𝑰𝑰𝒄𝒄 = 𝟏𝟏𝟐𝟐∡𝟏𝟏𝟑𝟑𝟑𝟑 𝑽𝑽). Assume positive sequence.
𝑨𝑨
+
𝟓𝟓 Ω
𝑰𝑰𝒂𝒂
𝒋𝒋𝒋𝒋𝒋𝒋 Ω
𝒂𝒂
𝑬𝑬𝑨𝑨𝑨𝑨
-
𝑪𝑪
𝑰𝑰𝒄𝒄
𝑩𝑩
𝑰𝑰𝒃𝒃
𝟓𝟓 Ω
𝟓𝟓 Ω
𝒋𝒋𝒋𝒋𝒋𝒋 Ω
𝒋𝒋𝒋𝒋𝒋𝒋 Ω
𝒄𝒄
𝒁𝒁∆
𝒃𝒃
Transform the (∆ − 𝐿𝑜𝑎𝑑) to a (𝑌 − 𝐿𝑜𝑎𝑑) so you can construct a Single Phase Circuit.
𝒂𝒂
𝒂𝒂
𝒄𝒄
𝒁𝒁∆
𝑍𝑌 =
𝑍∆
3
𝒁𝒁𝒀𝒀
𝒃𝒃
𝒄𝒄
𝒃𝒃
Single Phase Circuit
𝑨𝑨
𝟓𝟓 Ω
𝑰𝑰𝒂𝒂
+
𝒋𝒋𝒋𝒋𝒋𝒋 Ω
𝒁𝒁𝒀𝒀
𝑬𝑬𝑨𝑨𝑨𝑨
-
𝒏𝒏
𝑵𝑵
Complex Power delivered by the Generator.
𝐼𝑐 = 𝐼𝑎 (∡+120𝑜 )
𝐼𝑎 =
𝐼𝑐
12∡130𝑜
=
= 12∡10𝑜 𝐴
∡1200
∡120𝑜
𝑜
𝐼𝑎 = 𝐼𝑎𝑏 (√3 ∡−30 )
𝐼𝑎𝑏 =
𝐼𝑎
√3 ∡−30𝑜
=
12∡10𝑜
√3 ∡−30𝑜
= 𝟏𝟏. 𝟐𝟐𝟑𝟑∡𝟑𝟑𝟑𝟑𝒐𝒐 𝑽𝑽
Single Phase Complex Power at the Load is
����⃗
𝑆∅ = 480 − 𝑗310 𝑉𝑉𝐴
����⃗∅ =
𝑆
𝑍𝑌 =
𝒂𝒂
(𝐼𝑎 )2 (𝑍𝑌 )
����⃗
𝑆∅
480 − 𝑗310
=
= 3.33 − 𝑗2.15 Ω
2
(𝐼𝑎 )
(12)2
𝑍∆ = (3)𝑍𝑌 = (3)(3.33 − 𝑗2.15) = 𝟏𝟏𝟑𝟑 − 𝒋𝒋𝟏𝟏. 𝟑𝟑𝟐𝟐 Ω
����⃗
𝑆∅ = (𝐸𝐴𝑁 )(𝐼𝑎 )∗
����⃗∅ = (80∡20𝑜 )(12∡−10𝑜 ) = 960∡10𝑜 𝑉𝑉𝐴
𝑆
����⃗
����⃗∅ = 2880∡10𝑜 𝑉𝑉𝐴
𝑆𝑇 = (3)𝑆
����⃗
𝑆
𝑇 = 2836.25 + 𝑗500.11 𝑉𝑉𝐴
𝑃𝑇 = 𝟐𝟐𝟔𝟔𝟑𝟑𝟏𝟏. 𝟐𝟐𝟐𝟐 𝑾𝑾
𝑄𝑇 = 𝟐𝟐𝟑𝟑𝟑𝟑. 𝟏𝟏𝟏𝟏 𝑽𝑽𝑽𝑽𝑽𝑽
����⃗
�𝑆
𝑇 � = 𝟐𝟐𝟔𝟔𝟔𝟔𝟑𝟑 𝑽𝑽𝑽𝑽
In previous 3-phase problems we have been only interested in the Line or Phase variables that
the Generator (source) has produced to solve for the 3-phase system characteristics. In AC
Generator problems we will step back to a more detail look of the internal parameters of the
Generator. Remember a Generator converts Mechanical Energy from a prime mover (an axle)
into Electrical Energy. An axle will rotate with an established constant magnetic field on the
rotor that will induce a voltage (𝑬𝑬𝒊𝒊𝒏𝒏𝑳𝑳 ) on the stator windings. Before the voltage potential
reaches the outer terminals of the Generator (Line Voltages) the induced voltage will suffer
internal losses due to resistance of the stator windings (𝑽𝑽𝑺𝑺 ) and mutual inductance of the stator
winding (𝑿𝑿𝑺𝑺 ). The circuit diagram on the left is the Single Phase Circuit of the Generator.
𝑰𝑰𝒂𝒂
+
𝑿𝑿𝑺𝑺
𝑹𝑹𝑺𝑺
+
+
𝑷𝑷𝑰𝑰𝑰𝑰
𝑷𝑷𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬 𝑳𝑳𝑳𝑳𝑳𝑳𝑳𝑳
𝑰𝑰𝒂𝒂
𝑬𝑬𝑨𝑨𝑨𝑨
-
-
𝑵𝑵
𝑷𝑷𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴 𝑳𝑳𝑳𝑳𝑳𝑳𝑳𝑳
𝑨𝑨
𝑬𝑬𝑨𝑨𝑨𝑨
𝑬𝑬𝒊𝒊𝒊𝒊𝒊𝒊
-
𝑨𝑨
𝑵𝑵
𝑷𝑷𝑶𝑶𝑶𝑶𝑶𝑶
𝑪𝑪
Load
𝑩𝑩
𝑰𝑰𝒃𝒃
𝑰𝑰𝒄𝒄
20) A 3-phase, Y-connected, 8-pole, 2.88 KV, 50 Hz synchronous generator is rated to deliver
4.5 MVA at a power factor of 0.8829 Lagging. The per-phase stator resistance is 0.09Ω and the
synchronous reactance is negligible. The AC Generator is operating at the rated Load with
Mechanical Losses at 410 kW.
To simplify the problem first list all the given variables.
3 − 𝑝ℎ𝑎𝑠𝑒 𝑠𝑦𝑡𝑒𝑚
𝑅𝑎𝑡𝑒𝑑 𝑝𝑜𝑤𝑒𝑟 = 3 − 𝑝ℎ𝑎𝑠𝑒 𝐴𝑝𝑝𝑎𝑟𝑒𝑛𝑡 𝑃𝑜𝑤𝑒𝑟
8 − 𝑝𝑜𝑙𝑒
𝑝𝑜𝑤𝑒𝑟 𝑓𝑎𝑐𝑡𝑜𝑟 = 𝑎𝑛𝑔𝑙𝑒 𝑜𝑓 𝐶𝑜𝑚𝑝𝑙𝑒𝑥 𝑃𝑜𝑤𝑒𝑟
𝑌 − 𝑐𝑜𝑛𝑛𝑒𝑐𝑡𝑒𝑑
𝑓 = 50 𝐻𝑧
𝑅𝑎𝑡𝑒𝑑 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 = 𝐿𝑖𝑛𝑒 𝑡𝑜 𝐿𝑖𝑛𝑒 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 = 2.88 𝑘𝑉𝑉
𝑃𝑀𝑒𝑐ℎ 𝐿𝑜𝑠𝑠 = 410 𝑘𝑊
����⃗
�𝑆
𝑇 � = 4.5 𝑀𝑉𝑉𝐴
𝜃 = cos−1(0.8829) = 28𝑜 *Lagging = positive angle
𝑅𝑆 = 0.09 Ω
𝑋𝑆 = 0 Ω
a. At what speed does the shaft rotate (𝒓𝒑𝒑𝒎)?
𝑁𝑟𝑝𝑚 =
120(𝑓) 120(50)
=
= 𝟑𝟑𝟐𝟐𝟑𝟑 𝒓𝒑𝒑𝒎
𝑃𝑜𝑙𝑒𝑠
8
b. Find the Line Current (𝑰𝑰𝑳𝑳 ).
Single Phase Circuit
+
-
𝑰𝑰𝒂𝒂
𝑿𝑿𝑺𝑺
𝑹𝑹𝑺𝑺
+
𝑬𝑬𝑨𝑨𝑨𝑨
𝑬𝑬𝒊𝒊𝒊𝒊𝒊𝒊
����⃗
𝑺𝑺∅
Rated Single Phase
Complex Power
-
𝑵𝑵
����⃗
𝑆𝑇 = 4.5∡28𝑜 𝑀𝑉𝑉𝐴
����⃗∅ =
𝑆
𝑨𝑨
����⃗
𝑆𝑇 4.5∡28𝑜 𝑀𝑉𝑉𝐴
=
= 1.5∡28𝑜 𝑀𝑉𝑉𝐴
3
3
𝑜
𝑘𝑉𝑉
𝐸𝐴𝐵 = 2.88∡𝜃𝐴𝐵
𝐸𝐴𝑁 =
𝐸𝐴𝐵
√3 ∡30𝑜
����⃗∅ = (𝐸𝐴𝑁 )(𝐼𝑎 )∗
𝑆
(𝐼𝑎 )∗ =
=
𝑜
2.88∡𝜃𝐴𝐵
𝑘𝑉𝑉
√3 ∡30𝑜
= 1.663∡0𝑜 𝑘𝑉𝑉
����⃗
𝑆∅
1.5∡28𝑜 𝑀𝑉𝑉𝐴
=
= 901.98∡28𝑜 𝐴
𝐸𝐴𝑁
1.663∡0𝑜 𝑘𝑉𝑉
Note the given rated Line Voltage (𝑬𝑬𝑽𝑽𝑩𝑩 )
does not include an angle so we can
choose any angle. It is common for the
ease of calculations to set (𝑬𝑬𝑽𝑽𝑰𝑰 ) as the
reference angle (𝜽𝜽𝑽𝑽𝑰𝑰 = 𝟑𝟑𝒐𝒐) or you can
think of it as setting (𝜽𝜽𝑽𝑽𝑩𝑩 = 𝟑𝟑𝟑𝟑𝒐𝒐 ).
𝐼𝑎 = 𝟐𝟐𝟑𝟑𝟏𝟏. 𝟐𝟐𝟔𝟔∡−𝟐𝟐𝟔𝟔𝒐𝒐 𝑽𝑽
c. Find (𝑷𝑷𝑶𝑶𝑶𝑶𝑻𝑻 ), (𝑷𝑷𝑬𝑬𝑬𝑬𝒆𝒆𝒄𝒄 𝑳𝑳𝒐𝒐𝒔𝒔𝒔𝒔), and (𝑷𝑷𝑰𝑰𝑰𝑰 ) of the Generator.
𝑷𝑷𝑶𝑶𝑶𝑶𝑻𝑻 is the real component of the rated Total Complex Power.
𝑃𝑂𝑈𝑇 = (4.5𝑥106 ) cos(28𝑜 ) = 𝟑𝟑. 𝟐𝟐𝟑𝟑𝟑𝟑 𝑴𝑴𝑾𝑾
𝑷𝑷𝑬𝑬𝑬𝑬𝒆𝒆𝒄𝒄 𝑳𝑳𝒐𝒐𝒔𝒔𝒔𝒔 is caused only by the stator resistance. Remember the given 𝑽𝑽𝑺𝑺 and the current we
found (𝑰𝑰𝑳𝑳 ) is for single phase, so the 𝑷𝑷𝑬𝑬𝑬𝑬𝒆𝒆𝒄𝒄 𝑳𝑳𝒐𝒐𝒔𝒔𝒔𝒔 needs to be multiplied by 3.
𝑃𝐸𝑙𝑒𝑐 𝐿𝑜𝑠𝑠 = 3(𝐼𝑎 )2 (𝑅𝑆 ) = 3(901.98)2 (0.09) = 𝟐𝟐𝟏𝟏𝟐𝟐. 𝟏𝟏𝟏𝟏 𝒌𝑾𝑾
𝑃𝐼𝑁 = 𝑃𝑂𝑈𝑇 + 𝑃𝑀𝑒𝑐ℎ 𝐿𝑜𝑠𝑠 + 𝑃𝐸𝑙𝑒𝑐 𝐿𝑜𝑠𝑠 = 3.973 + 0.41 + 0.21966 = 𝟑𝟑. 𝟏𝟏𝟑𝟑𝟑𝟑 𝑴𝑴𝑾𝑾
d. What is the efficiency (ƞ) of the generator?
ƞ=
𝑃𝑂𝑈𝑇
3.973 𝑀𝑊
(100) =
(100) = 𝟔𝟔𝟏𝟏. 𝟑𝟑𝟏𝟏%
𝑃𝐼𝑁
4.603 𝑀𝑊
e. What is the prime mover torque?
The prime mover will induce a synchronous Voltage which delivers the Power (𝑷𝑷𝑰𝑰𝑰𝑰 ) into
the Generator system.
𝜔 = 2𝜋 �
𝑅𝑃𝑀
750
� = 2𝜋 �
� = 78.54 𝑟𝑎𝑑/𝑠
60
60
𝑃𝐼𝑁 = (𝑇𝑃𝑀 )(𝜔)
𝑇𝑃𝑀 =
𝑃𝐼𝑁 4.603𝑥106
=
= 𝟐𝟐𝟔𝟔. 𝟏𝟏𝟏𝟏 𝒌𝑰𝑰 − 𝒎
𝜔
78.54