EE301
Exam 2 Review
Exam 2 Review
1) In a DC circuit when steady-state conditions are reached a Capacitor acts like a what?
OPEN
2) In a DC circuit when steady-state conditions are reached an Inductor acts like a what?
SHORT
3) The circuits below have been sitting for a long time; apply the steady-state conditions of the
Capacitor and the Inductor to determine the voltages at terminals a-b. (Assume the Capacitors
and Inductors are ideal).
4.2 mH
500 Ω
a
+
12 V
+
-
1 kΩ
120 nF
𝑽𝑽𝑶𝑶𝑶𝑶𝑶𝑶
-
b
𝑽𝑽𝑶𝑶𝑶𝑶𝑶𝑶 = 𝟏𝟏𝟏𝟏 𝑽𝑽
750 Ω
500 Ω
a
+
12 V
+
-
1 kΩ
4.2 mH
𝑽𝑽𝑶𝑶𝑶𝑶𝑶𝑶
-
b
𝑽𝑽𝑶𝑶𝑶𝑶𝑶𝑶 = 𝟎𝟎 𝑽𝑽
4) Initially the switch is at Position 1 for a long time and then moves to Position 2. Determine the
initial Voltage (𝑽𝑽𝒐𝒐 ) across the Capacitor.
6 kΩ
1
7 kΩ
2
40 V
+
-
10 kΩ
+
𝑽𝑽𝒄𝒄
-
2 kΩ
𝑪𝑪 = 𝟓𝟓𝝁𝝁𝝁𝝁
10𝑥103
𝑉𝑜 = (6𝑥103 )+(10𝑥103 ) (40) = 𝟏𝟏𝟓𝟓 𝑽𝑽
5) Redraw the circuit in Position 2 and determine the equivalent Resistance (𝑹𝑹𝒆𝒒 ) as seen by the
Capacitor. Also determine the time constant (𝝉) and the final Voltage (𝑽𝑽𝒇 ) across the Capacitor.
7 kΩ
2
2 kΩ
+
𝑽𝑽𝒄𝒄
-
𝑅𝑒𝑞 = (7𝑥103 ) + (2𝑥103 ) = 𝟏𝟏 kΩ
𝜏 = 𝑅𝑒𝑞 𝐶 = (9𝑥103 )(5𝑥10−6 ) = 𝟏𝟏𝟓𝟓 ms
𝑪𝑪 = 𝟓𝟓𝝁𝝁𝝁𝝁
𝑉𝑓 = 𝟎𝟎 𝑽𝑽
6) Develop the transit response equation for the Voltage across the Capacitor and plot the
response below. Make sure to include the time (in milliseconds) when steady-state is reached.
𝑡
𝑣(𝑡) = 𝑉𝑓 + �𝑉𝑜 − 𝑉𝑓 �𝑒 − 𝜏
𝑡
25 V
𝑽𝑽
5𝜏 = 5(45𝑥10−3 ) = 𝟏𝟏𝟏𝟏𝟓𝟓 ms
𝒕𝒕
−
𝑣(𝑡) = 0 + (25 − 0)𝑒 − 45 𝑚𝑠 = 𝟏𝟏𝟓𝟓𝒆 𝟏𝟏𝟓𝟓 𝟑𝟑𝟐𝟐 𝑽𝑽
225 ms
𝒕𝒕
7) Initially the switch is at Position 2 for a long time and then moves to Position 1. Determine the
initial Current (𝒊𝒊𝒐𝒐 ) through the Inductor.
6 kΩ
1
7 kΩ
2
+
-
40 V
10 kΩ
+
𝒊𝒊𝑳𝑳
-
2 kΩ
𝑳𝑳 = 𝟑𝟑𝟑𝟑𝟑𝟑
𝑖𝑜 = 𝟎𝟎 𝑨𝑨
8) Redraw the circuit in Position 1 and determine the equivalent Resistance (𝑹𝑹𝒆𝒒 ) as seen by the
Inductor. Also determine the time constant (𝝉) and the final Current (𝒊𝒊𝒇 ) through the Inductor.
𝑽𝑽
6 kΩ
40 V
+
7 kΩ
1
+
𝒊𝒊𝑳𝑳
-
10 kΩ
-
**Using Nodal Analysis
𝑳𝑳 = 𝟑𝟑𝟑𝟑𝟑𝟑
𝑉 − 40
𝑉
𝑉
+
+
=0
3
3
6𝑥10
10𝑥10
7𝑥103
𝑉�
(6𝑥103 )(10𝑥103 )
𝑅𝑒𝑞 = (6𝑥103 )+(10𝑥103 ) + 7𝑥103 = 𝟏𝟏𝟎𝟎. 𝟑𝟑𝟓𝟓 kΩ
𝜏=
𝐿
𝑅𝑒𝑞
𝑖𝑓 =
𝑉[4.095𝑥10−4 ] = 6.67𝑥10−3
3𝑥10−3
= 10.75𝑥103 = 𝟏𝟏𝟑𝟑𝟏𝟏 ns
𝑉
7𝑥103
=
16.28
7𝑥103
40
1
1
1
+
+
�=
3
3
3
6𝑥103
10𝑥10
7𝑥10
6𝑥10
𝑉 = 16.28 𝑉
= 𝟏𝟏. 𝟑𝟑𝟑𝟑 𝟑𝟑𝑨𝑨
9) Develop the transit response equation for the Current through the Inductor and plot the
response below. Make sure to include the time (in milliseconds) when steady-state is reached.
2.33 mA
𝑰𝑰
𝑡
𝑖(𝑡) = 𝑖𝑓 + (𝑖𝑜 − 𝑖𝑓 )𝑒 − 𝜏
𝑡
𝑖(𝑡) = 2.33𝑚𝐴 + (0 − 2.33𝑚𝐴)𝑒 − 279 𝑛𝑠
1.395 µs
𝒕𝒕
𝒕𝒕
𝑖(𝑡) = 𝟏𝟏. 𝟑𝟑𝟑𝟑(𝟏𝟏 − 𝒆− 𝟏𝟏𝟑𝟑𝟏𝟏 𝟐𝟐𝟐𝟐 ) 𝟑𝟑𝑨𝑨
5𝜏 = 5(279𝑥10−9 ) = 𝟏𝟏. 𝟑𝟑𝟏𝟏𝟓𝟓 µs
10) How long will it take for the Current through the Inductor to reach 1.94 mA and what will be the
energy stored in the Inductor at this point in time?
𝑡
1.94 𝑚𝐴 = 2.33(1 − 𝑒 − 279 𝑛𝑠 ) 𝑚𝐴
𝑡
1.94 𝑚𝐴
− 1 = −𝑒 − 279 𝑛𝑠
2.33 𝑚𝐴
𝑡
−0.167 = −𝑒 − 279 𝑛𝑠
𝑙𝑛(0.167) = −
𝑡
279 𝑛𝑠
𝑡 = − 𝑙𝑛(0.167) 𝑥(279𝑥10−9 ) = 𝟎𝟎. 𝟓𝟓 µs
1
1
𝑊 = 𝐿𝑖 2 = (3𝑥10−3 )(1.94𝑥10−3 )2 = 𝟓𝟓. 𝟏𝟏𝟓𝟓 𝟐𝟐𝑱
2
2
𝑡
𝑙𝑛(0.167) = 𝑙𝑛 �𝑒 − 279 𝑛𝑠 �
11) Express the Voltage Sources as a phasor.
𝟐𝟐𝟐𝟐. 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐(𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 − 𝟔𝟔𝟔𝟔𝒐𝒐 ) V
**Make sure to convert the Amplitude
to the RMS value.
+
-
21.21
√2
= 15 𝑉
∴
𝟏𝟏𝟓𝟓∡−𝟏𝟏𝟑𝟑𝒐𝒐 𝑽𝑽
12) Find the Impedance values for the given elements.
a. Let 𝒇 = 𝟏𝟏𝟎𝟎 𝟑𝟑𝒛
132.63 mH
b. Let 𝝎 = 𝟏𝟏𝟑𝟑𝟓𝟓 𝒓𝑳𝑳𝑳𝑳/𝟐𝟐
193.42 µF
c. Let 𝝎 = 𝟏𝟏𝟓𝟓𝟎𝟎𝟎𝟎 𝒓𝑳𝑳𝑳𝑳/𝟐𝟐
80.5 Ω
𝜔 = 2𝜋𝑓 = (2𝜋)(60) = 377 𝑟𝑎𝑑/𝑠
𝑍𝐿 = 𝑗𝜔𝐿 = 𝑗(377)(132.63𝑥10−3 ) = 𝒋𝒋𝟓𝟓𝟎𝟎 Ω
𝑍𝐶 = −𝑗
1
1
= −𝑗
𝜔𝐶
(235)(193.42𝑥10−6 )
𝑍𝐶 = −𝒋𝒋𝟏𝟏𝟏𝟏 Ω
𝑍𝑅 = 𝑅 = 𝟖𝟖𝟎𝟎. 𝟓𝟓 Ω
13) Redraw the circuit below. Express the source in phasor form and the elements as Impedances.
10 Ω
+
𝒐𝒐
𝟏𝟏𝟏𝟏. 𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝟖(𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑 + 𝟑𝟑𝟑𝟑 ) V
𝑉𝑠 =
19.8
√2
-
𝑰𝑰𝒔𝒔
𝑰𝑰𝟏𝟏
10 Ω
𝑽𝑽𝒔𝒔 = 𝟏𝟏𝟏𝟏∡𝟑𝟑𝟑𝟑 V
-
𝑰𝑰𝒔𝒔
1
𝑍𝐿 = 𝑗(377)(53.05𝑥10−3 ) = 𝒋𝒋𝟏𝟏𝟎𝟎 Ω
1
+
𝟓𝟓𝟓𝟓. 𝟎𝟎𝟎𝟎 𝒎𝒎𝒎𝒎
𝑍𝐶2 = −𝑗 (377)(530.5𝑥10−6 ) = −𝒋𝒋𝟓𝟓 Ω
𝑍𝐶1 = −𝑗 (377)(221𝑥10−6 ) = −𝒋𝒋𝟏𝟏𝟏𝟏 Ω
𝒐𝒐
+
𝑽𝑽𝑳𝑳
-
𝟐𝟐𝟐𝟐𝟐𝟐 𝝁𝝁𝝁𝝁 𝑰𝑰𝟐𝟐 𝟐𝟐𝟐𝟐 Ω
𝑍21 = 𝟏𝟏𝟏𝟏 Ω
∡30𝑜 = 𝟏𝟏𝟏𝟏∡𝟑𝟑𝟎𝟎𝒐𝒐 𝑽𝑽
𝑍10 = 𝟏𝟏𝟎𝟎 Ω
𝟓𝟓𝟓𝟓𝟓𝟓. 𝟓𝟓 𝝁𝝁𝝁𝝁
𝑰𝑰𝟑𝟑
𝑰𝑰𝟑𝟑
𝑰𝑰𝟏𝟏
−𝒋𝒋𝒋𝒋𝒋𝒋 Ω
𝑰𝑰𝟐𝟐 𝟐𝟐𝟐𝟐 Ω
−𝒋𝒋𝒋𝒋 Ω
+
𝑽𝑽𝑳𝑳
-
𝒋𝒋𝒋𝒋𝒋𝒋 Ω
14) Find the total Impedance as seen by the Source.
𝑍𝑇 = 10 + �
𝑍𝑇 = 10 +
−1
1
1
1
+
+
� = 10 + [𝑗0.083 + 0.048 − 𝑗0.067]−1
−𝑗12 21 (−𝑗5 + 𝑗20)
1
1
= 10 +
= 10 + 19.61∡−18.43𝑜
0.048 + 𝑗0.016
0.051∡18.43𝑜
𝑍𝑇 = 10 + 18.6 − 𝑗6.2 = 28.6 − 𝑗6.2 = 𝟏𝟏𝟏𝟏. 𝟏𝟏𝟏𝟏∡−𝟏𝟏𝟏𝟏. 𝟏𝟏𝟑𝟑𝒐𝒐 Ω
15) Find (𝑰𝑰𝟐𝟐 ), (𝑰𝑰𝟏𝟏 ), (𝑰𝑰𝟏𝟏 ), and (𝑰𝑰𝟑𝟑 ). Use Current Divider Rule when finding (𝑰𝑰𝟏𝟏 ), (𝑰𝑰𝟏𝟏 ), and (𝑰𝑰𝟑𝟑 ).
𝐼𝑠 =
𝑉𝑠
14∡30𝑜
=
= 𝟎𝟎. 𝟏𝟏𝟑𝟑𝟖𝟖∡𝟏𝟏𝟏𝟏. 𝟏𝟏𝟑𝟑𝒐𝒐 𝑨𝑨
𝑍𝑇 29.26∡−12.23𝑜
𝑍𝑒𝑞 = �
−1
1
1
1
+
+
�
−𝑗12 21 (−𝑗5 + 𝑗20)
𝑍𝑒𝑞 = 19.61∡−18.43𝑜 Ω
𝐼1 =
𝐼2 =
𝐼3 =
𝑍𝑒𝑞
19.61∡−18.43𝑜
(𝐼𝑠 ) =
(0.478∡42.23𝑜 ) = 𝟎𝟎. 𝟑𝟑𝟖𝟖𝟏𝟏∡𝟏𝟏𝟏𝟏𝟑𝟑. 𝟖𝟖𝒐𝒐 𝑨𝑨
𝑍𝐶1
12∡−90𝑜
𝑍𝑒𝑞
19.61∡−18.43𝑜
(𝐼𝑠 ) =
(0.478∡42.23𝑜 ) = 𝟎𝟎. 𝟏𝟏𝟏𝟏𝟏𝟏∡𝟏𝟏𝟑𝟑. 𝟖𝟖𝒐𝒐 𝑨𝑨
𝑍21
21∡0𝑜
𝑍𝑒𝑞
19.61∡−18.43𝑜
(𝐼𝑠 ) =
(0.478∡42.23𝑜 ) = 𝟎𝟎. 𝟏𝟏𝟏𝟏𝟓𝟓∡−𝟏𝟏𝟏𝟏. 𝟏𝟏𝒐𝒐 𝑨𝑨
𝑍𝐶2 +𝐿
15∡90𝑜
16) Find the voltage across the Inductor (𝑽𝑽𝑳𝑳 ).
𝑉𝐿 = (𝐼3 )(𝑍𝐿 ) = (0.625∡−66.2𝑜 )(20∡90𝑜 ) = 𝟏𝟏𝟏𝟏. 𝟓𝟓∡𝟏𝟏𝟑𝟑. 𝟖𝟖𝒐𝒐 𝑽𝑽
17) When solving Thevenin Equivalent problems there are logical steps that hold true no matter if
the circuit is DC or AC. Answer the following with True or False.
a. ____
T If a Load is shown on the circuit, the first step is to remove the Load.
b. ____
F The Thevenin Impedance is always the impedance as seen by the Source.
c. ____
F When solving the Thevenin Impedance, we must “turn off” the sources by
replacing both current and voltage sources as SHORTS.
d. ____
F When solving the Thevenin Impedance, we must “turn off” the sources by
replacing the current source a SHORT and the voltage sources as an OPEN.
e. ____
T When solving the Thevenin Impedance, we must “turn off” the sources by
replacing the current source an OPEN and the voltage sources as a SHORT.
f. ____
T When finding the Thevenin Voltage we use the open terminals were the Thevenin
perspective is.
18) Find the Thevenin Impedance and Voltage external to 𝒁𝒁𝑳𝑳𝒐𝒐𝑳𝑳𝑳𝑳 for the circuit below. Draw the
Thevenin circuit with the Load attached.
𝟐𝟐𝟐𝟐 Ω
𝒋𝒋𝒋𝒋𝒋𝒋 Ω
𝟖𝟖∡𝟑𝟑𝟑𝟑𝒐𝒐 𝑨𝑨
𝟏𝟏𝟏𝟏 Ω
-𝒋𝒋𝒋𝒋 Ω
𝒋𝒋𝒋𝒋. 𝟒𝟒 Ω
𝒁𝒁𝑳𝑳𝑳𝑳𝑳𝑳𝑳𝑳
First step is to remove the load and find 𝒁𝒁𝑶𝑶𝟑𝟑 .
Remember the current source is replaced with an
OPEN.
𝟐𝟐𝟐𝟐 Ω
𝒋𝒋𝒋𝒋𝒋𝒋 Ω
𝑍𝑇𝐻 =
+
𝟏𝟏𝟏𝟏 Ω
-𝒋𝒋𝒋𝒋 Ω
-
𝟑𝟑. 𝟐𝟐 Ω
𝒁𝒁𝑻𝑻𝑻𝑻
(16∡0𝑜 )(8∡−90𝑜 ) (16∡0𝑜 )(8∡−90𝑜 )
=
16 − 𝑗8
17.89∡−26.57𝑜
𝑍𝑇𝐻 = 7.15∡−63.43𝑜 = 𝟑𝟑. 𝟏𝟏 − 𝒋𝒋𝟏𝟏. 𝟏𝟏 Ω
Now “turn on” the source and find 𝑽𝑽𝑶𝑶𝟑𝟑 at the open terminals.
𝟐𝟐𝟐𝟐 Ω
𝒋𝒋𝒋𝒋𝒋𝒋 Ω
𝟖𝟖∡𝟑𝟑𝟑𝟑𝒐𝒐 𝑨𝑨
𝑉𝑇𝐻 is the voltage across the 16 Ω and the –j8 Ω.
We can use Nodal Analysis or combine the 16 Ω
and –j8 Ω impedances in parallel and find the
voltage across them.
+
𝟏𝟏𝟏𝟏 Ω
-𝒋𝒋𝒋𝒋 Ω
𝑽𝑽𝑻𝑻𝑻𝑻
-
𝑍𝑒𝑞 = (16)||(−𝑗8) = 7.15∡−63.43𝑜 Ω
𝑉𝑇𝐻 = (𝐼𝑠 )�𝑍𝑒𝑞 � = (8∡30𝑜 )(7.15∡−63.43𝑜 ) = 𝟓𝟓𝟑𝟑. 𝟏𝟏∡−𝟑𝟑𝟑𝟑. 𝟏𝟏𝟑𝟑𝒐𝒐 𝑽𝑽
𝟑𝟑. 𝟐𝟐 Ω −𝒋𝒋𝒋𝒋. 𝟒𝟒 Ω
𝟓𝟓𝟓𝟓. 𝟐𝟐∡−𝟑𝟑𝟑𝟑. 𝟒𝟒𝟒𝟒𝒐𝒐 𝑽𝑽
𝟑𝟑. 𝟐𝟐 Ω
+
-
𝒋𝒋𝒋𝒋. 𝟒𝟒 Ω
19) Is Maximum Power being delivered to the Load and why?
𝒁𝒁𝑶𝑶𝟑𝟑 = 𝒁𝒁∗𝑳𝑳
YES!
3.2 − 𝑗6.4 = 3.2 + 𝑗6.4
7.15∡−63.43𝑜 = 7.15∡63.43𝑜
20) The acronym used for Capacitors in AC circuits is “ICE”. Explain what is meant with this
acronym.
Current LEADS Voltage for a capacitive circuit. If we are just analyzing a Capacitor’s current and
voltage relationship we will see the current LEADS the voltage by 900 .
𝑰𝑰𝑪𝑪
𝑪𝑪
+
𝑽𝑽𝑪𝑪
-
𝑰𝑰𝑪𝑪
𝜽𝜽𝑽𝑽
𝑽𝑽𝑪𝑪
𝟎𝟎𝒐𝒐
This implies the Current Angle (𝜽𝜽𝒊𝒊 )
equals to 𝜽𝜽𝑽𝑽 + 𝟗𝟗𝟗𝟗𝒐𝒐 .
21) The acronym used for Inductors in AC circuits is “ELI”. Explain what is meant with this acronym.
Voltage LEADS Current for an inductive circuit. If we are just analyzing an Inductor’s current and
voltage relationship we will see the voltage LEADS the current by 900 .
𝑰𝑰𝑳𝑳
𝑳𝑳
+
𝑽𝑽𝑳𝑳
-
𝑽𝑽𝑳𝑳
𝑰𝑰𝑳𝑳
𝜽𝜽𝒊𝒊
𝟎𝟎𝒐𝒐
This implies the Voltage Angle (𝜽𝜽𝑽𝑽 )
equals to 𝜽𝜽𝒊𝒊 + 𝟗𝟗𝟗𝟗𝒐𝒐 .
The Graph and circuit below are for questions 22 through 25
V
𝑽𝑽𝒔𝒔
6
3
𝑽𝑽𝑹𝑹
t
0
1.25
2.5
5
3.75
6.25
7.5
-3
**Time in milliseconds
-6
𝑽𝑽𝒔𝒔
+
+
-
- 𝑽𝑽𝑹𝑹 +
𝟐𝟐 Ω
𝑽𝑽𝑳𝑳
-
𝒁𝒁𝑳𝑳𝑳𝑳𝑳𝑳𝑳𝑳
22) For the Graph on the other page answer the following questions.
a. What is the period (𝑶𝑶) of the waveforms?
𝑇 = 𝟓𝟓 𝟑𝟑𝟐𝟐
b. What is the frequency of the waveforms (𝒇)?
𝑓=
1
1
=
= 𝟏𝟏𝟎𝟎𝟎𝟎 𝟑𝟑𝒛
𝑇 5𝑥10−3
c. What is the time difference (∆𝒕𝒕) between 𝑽𝑽𝟐𝟐 𝑎𝑛𝑑 𝑽𝑽𝑹𝑹 ?
The ∆𝒕𝒕 between the peak of 𝑽𝑽𝟐𝟐 and the peak of 𝑽𝑽𝑹𝑹 is about 0.5 ms.
∆𝑡 = 𝟎𝟎. 𝟓𝟓 𝟑𝟑𝟐𝟐
d. What is the phase difference (∆𝜽𝜽) between 𝑽𝑽𝟐𝟐 𝑎𝑛𝑑 𝑽𝑽𝑹𝑹 ?
∆𝜃 =
∆𝑡
0.5𝑥10−3
(360𝑜 ) =
(360𝑜 ) = 𝟑𝟑𝟏𝟏𝒐𝒐
5𝑥10−3
𝑇
e. Write the function for 𝑽𝑽𝟐𝟐 (𝒕𝒕).
f.
𝑉𝑠 (𝑡) = 𝟏𝟏 𝐬𝐢𝐧[𝟏𝟏𝝅(𝟏𝟏𝟎𝟎𝟎𝟎)𝒕𝒕] 𝑽𝑽
Express 𝑽𝑽𝟐𝟐 (𝒕𝒕) in phasor form.
𝑉𝑠 =
6
√2
∡0𝑜 = 𝟏𝟏. 𝟏𝟏𝟏𝟏∡𝟎𝟎𝒐𝒐 𝑽𝑽
g. Write the function for 𝑽𝑽𝑹𝑹 (𝒕𝒕). (Use 𝑉𝑠 as the reference angle)
𝑉𝑅 (𝑡) = 𝟑𝟑 𝐬𝐢𝐧[𝟏𝟏𝝅(𝟏𝟏𝟎𝟎𝟎𝟎)𝒕𝒕 − 𝟑𝟑𝟏𝟏𝒐𝒐 ] 𝑽𝑽
h. Express 𝑽𝑽𝑹𝑹 (𝒕𝒕) in phasor form.
𝑉𝑅 =
3
√2
∡−36𝑜 = 𝟏𝟏. 𝟏𝟏𝟏𝟏∡−𝟑𝟑𝟏𝟏𝒐𝒐 𝑽𝑽
23) Since we know 𝑽𝑽𝟐𝟐 𝑎𝑛𝑑 𝑽𝑽𝑹𝑹 , find the voltage across the Load (𝑽𝑽𝑳𝑳 ). Refer to the circuit at the
bottom of the other page. (Hint: use KVL)
−𝑉𝑠 + 𝑉𝐿 + 𝑉𝑅 = 0
𝑉𝐿 = 𝑉𝑠 − 𝑉𝑅 = 4.24∡0𝑜 − 2.12∡−36𝑜 = 4.24 − (1.72 − 𝑗1.25) = 2.52 + 𝑗1.25 𝑉
𝑉𝐿 = 𝟏𝟏. 𝟖𝟖𝟏𝟏∡𝟏𝟏𝟏𝟏. 𝟑𝟑𝟖𝟖𝒐𝒐 𝑽𝑽
24) Find 𝒁𝒁𝑳𝑳𝒐𝒐𝑳𝑳𝑳𝑳; Is the Load Capacitive or Inductive?
𝑉𝑅 2.12∡−36𝑜
=
= 1.06∡−36𝑜 𝐴
𝐼𝑆 = 𝐼𝑅 =
𝑅
2
𝑍𝐿𝑜𝑎𝑑
𝑜
𝑉𝐿 2.81∡26.38
=
=
= 2.65∡62.38𝑜 = 𝟏𝟏. 𝟏𝟏𝟏𝟏 + 𝒋𝒋𝟏𝟏. 𝟑𝟑𝟓𝟓 Ω
𝐼𝑠
1.06∡−36𝑜
𝑽𝑽𝒔𝒔
Since the imaginary component of 𝑍𝐿𝑜𝑎𝑑 is positive then the Load is Inductive.
25) Draw the Load equivalent circuit below.
𝑽𝑽𝒔𝒔
+
+
-
- 𝑽𝑽𝑹𝑹 +
𝟐𝟐 Ω
𝑽𝑽𝑳𝑳
-
𝟏𝟏. 𝟐𝟐𝟐𝟐 Ω
𝒋𝒋𝒋𝒋. 𝟑𝟑𝟑𝟑 Ω
𝑰𝑰𝒔𝒔
+
+
𝑰𝑰𝑹𝑹
- 𝑽𝑽𝑹𝑹 +
𝟐𝟐 Ω
𝑽𝑽𝑳𝑳
-
26) 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
𝟐𝟐𝟐𝟐𝟐𝟐∡𝟒𝟒𝟒𝟒𝒐𝒐 𝑽𝑽
𝐼1 = 1.96∡−28.2𝑜 𝐴
(𝐼2
=
𝟏𝟏𝟏𝟏𝟏𝟏 𝑾𝑾
−𝟒𝟒𝟒𝟒𝟒𝟒 𝑽𝑽𝑽𝑽𝑽𝑽
𝟒𝟒𝟒𝟒𝟒𝟒 𝑽𝑽𝑽𝑽𝑽𝑽
-
)∗
𝑰𝑰𝟐𝟐
�𝑆⃗
���⃗
𝑉𝑉𝑠
=
400∡−900
220∡40𝑜
𝐼2 = 1.82∡130𝑜 𝐴
𝑜
= 1.82∡−130 𝐴
𝐼𝑠 = 𝐼1 + 𝐼2
𝐼𝑠 = (1.96∡−28.2𝑜 ) + (1.82∡130𝑜 )
𝐼𝑠 = 𝟑𝟑. 𝟑𝟑𝟐𝟐𝟔𝟔∡𝟑𝟑𝟑𝟑. 𝟑𝟑𝟏𝟏𝒐𝒐 𝑽𝑽