# EE-403-Finals-Examination-Problem Solving ```III.
PROBLEM-SOLVING
DIRECTION: Solve the following problem manually on a clean sheet of
round the final answer to four decimal places, magnitude only and label it
with the correct units. Avoid too much ERASURE.
A. A circuit consisting of a coil with inductance 10 mH and resistance 20
ohms is connect in series with a capacitor and a generator with an rms
voltage of 120 V. Find: (10 points)
a. The value of the capacitance that will cause the circuit to be in
resonance at 15 kHz
b. The current through the coil at resonance
c. The Q of the circuit
SOLUTION
a. Solving the capacitance of the circuit
L = 10 mH; R = 20 Ohms; Vs=120 V; fr = 15 kHz
1
𝑓𝑟 =
2𝜋√𝐿𝐶
1
15𝑥103 =
2𝜋√(10𝑥10−3 )𝐶
1
√(10𝑥10−3 )𝐶 =
2𝜋(15𝑥103 )
2
(√(10𝑥10−3 )𝐶) = (1.0610𝑋10−5 )2
(1.1258𝑋10^ − 10)
(10𝑥10−3 )
𝐶 = 11.258𝑋10−9 𝑓
𝑪 = 𝟏𝟏. 𝟐𝟓𝟖 𝒏𝑭
b. Solving the current through the coil at resonance
Consider the visualization of the circuit on the figure below
𝐶=
𝑉𝑠 = 120𝑉
Converting the circuit to the domain of frequency
𝑤𝑜 = 2𝜋𝑓𝑜
𝑤𝑜 = 2𝜋(15𝑥103 )
𝑤𝑜 = 94.2478 𝑘𝐻𝑧
𝑋𝐿 = 𝑗𝑤𝐿
𝑋𝐿 = 𝑗(94.2478𝑥103 )(100𝑥10−3 )
𝑋𝐿 = 𝑗9424.7780 𝑂ℎ𝑚𝑠 𝑜𝑟 9425 𝑜ℎ𝑚𝑠
1
𝑋𝐶 =
𝑗𝑤𝐶
1
𝑋𝐶 =
3
𝑗(94.2478𝑥10 )(1.1258𝑥10−9 )
𝑋𝐶 = −𝑗9424.6999 𝑂ℎ𝑚𝑠 𝑜𝑟 − 𝑗9425 𝑜ℎ𝑚𝑠
𝑍𝑇 = 𝑅 + 𝑗𝑋𝐿 − 𝑗𝑋𝐶
𝑍𝑇 = 20 + 𝑗9425 − 𝑗9425
𝑍𝑇 = 20 𝑂ℎ𝑚𝑠
𝑉𝑠
𝐼𝑇 =
𝑍𝑇
120 𝑉
𝐼𝑇 =
20 𝑂ℎ𝑚𝑠
𝑰𝑻 = 𝟔 𝑨
c. Solving Quality Factor
1
𝑤𝑜 𝐿
𝑄=
𝑜𝑟
𝑤𝑜 𝐶𝑅
𝑅
𝑄=
(94.2478𝑥103 )(10𝑥10−3 )
1
𝑜𝑟
(94.2478𝑥103 )(11.258𝑥10−9 )(20)
20
𝑸 = 𝟒𝟕. 𝟐𝟑𝟓𝟎 𝒐𝒓 𝟒𝟕. 𝟏𝟐𝟑𝟗
B. Two wattmeters are connected to measure the input power to a balanced
three-phase load. If the wattmeter readings are 9.3 kW and 5.4 kW,
determine (5 points)
a. the total output power, and
SOLUTION
a. Total output power
𝑃𝑇 = 𝑊𝑀𝑅1 + 𝑊𝑀𝑅2
𝑃𝑇 = 9300𝑊 + 5400𝑊
𝑷𝑻 = 𝟏𝟒𝟕𝟎𝟎 𝑾 𝒐𝒓 𝟏𝟒. 𝟕𝒌𝑾
√3(𝑊𝑀𝑅1 − 𝑊𝑀𝑅2 )
𝑊𝑀𝑅1 + 𝑊𝑀𝑅2
√3(9300𝑊 − 5400𝑊)
𝑡𝑎𝑛𝜃 =
9300𝑊 + 5400𝑊
13
𝜃 = tan−1( √3)
49
𝜃 = 24.6799
𝑝𝑓 = 𝑐𝑜𝑠𝜃
𝑝𝑓 = 𝑐𝑜𝑠24.6799
𝒑𝒇 = 𝟎. 𝟗𝟎𝟖𝟕
𝑡𝑎𝑛𝜃 =
C.
D. Solution:
usp=sharing
VI.
NETWORK ANALYSIS
DIRECTION: Solve the following problem manually on a clean sheet of
paper. Box your final answer with the corresponding signature. Make sure you
round the final answer to four decimal places, polar form, and label it with the
correct units. Avoid too much ERASURE.
24. For the three phase circuit below IbB = 30∠ 60&deg; A and VBC = 220∠ 10&deg; V. Find the
Van, VAB, IAC and Z. (10 points)
SOLUTION:
`
𝑉𝑎𝑛 =
𝑉𝑏𝑐
√3∠−90&deg;
=
220∠10&deg;
√3∠−90&deg;
; 𝑉𝑏𝑐 = 𝑉𝐵𝐶
𝑽𝒂𝒏 = 𝟏𝟐𝟕∠𝟏𝟎𝟎&deg; 𝑽
Since balance system, we can relate them with the positive sequence
of the network.
𝑉𝐴𝐵 = 𝑉𝐵𝐶 ∠𝜃 + 120 = 220∠10 + 120 = 220∠130&deg; 𝑉
𝑽𝑨𝑩 = 𝟐𝟐𝟎∠𝟏𝟑𝟎&deg; 𝑽
𝑉𝐴𝐶 = 𝑉𝐴𝐶 ∠𝜃 − 120 = 220∠10 − 120 = 220∠ − 110&deg; 𝑉
If the value of IbB is 30∠60&deg; A , then by the identity of the balance
system
𝐼𝑎𝐴 = 𝐼𝑏𝐵 ∠𝜃 + 120 = 30∠60 + 120 = 30∠180&deg; 𝐴
𝐼𝑐𝐶 = 𝑉𝐵𝐶 ∠𝜃 − 120 = 30∠60 − 120 = 30∠ − 60&deg; 𝐴
Converting phase current to line current.
𝐼𝐴𝐵 =
𝐼𝑎𝐴
√3∠−30&deg;
=
30∠180&deg;
√3∠−30&deg;
= 17.3205∠ − 150&deg; 𝐴 𝑜𝑟
𝐼𝐴𝐵 = 17.3205∠210&deg; 𝐴
𝐼𝐵𝐶 =
𝐼𝑏𝐵
√3∠−30&deg;
=
30∠60&deg;
√3∠−30&deg;
= 17.3205∠90&deg; 𝐴 𝑜𝑟
𝐼𝐵𝐶 = 17.3205∠ − 270&deg; 𝐴
𝐼𝐶𝐴 =
𝐼𝑐𝐶
√3∠−30&deg;
=
30∠−60&deg;
√3∠−30&deg;
= 17.3205∠ − 30&deg; 𝐴 𝑜𝑟
𝐼𝐶𝐴 = 17.3205∠330&deg; 𝐴
𝐼𝐴𝐶 = −𝐼𝐶𝐴 = −1(17.3205∠ − 30&deg; 𝐴)
𝑰𝑨𝑪 = (𝟏𝟕. 𝟑𝟐𝟎𝟓∠𝟏𝟓𝟎&deg; 𝑨) = (𝟏𝟕. 𝟑𝟐𝟎𝟓∠ − 𝟐𝟏𝟎&deg; 𝑨)
Use the gained value to obtain the impedance of any phase since
they are a balance system.
𝑉𝐵𝐶
𝑍=
𝐼𝐵𝐶
220∠10&deg; 𝑉
𝑍=
17.3205∠90&deg; 𝐴
𝒁 = 𝟏𝟐. 𝟕𝟎𝟏𝟕∠ − 𝟖𝟎&deg; 𝑨
25. Find the Vx in the network below: (10 points)
SOLUTION:
For a much more straightforward approach, convert the third loop
by source transformation, and the circuit will look like the figure
below.
𝑉𝑛𝑒𝑤 = 𝐼(𝑍) = 2∠0&deg;𝐴(−𝑗1𝑂ℎ𝑚𝑠) = −𝑗2𝑉
2Ω
8∠𝟑𝟎&deg;
I1
2Ω
j4
j4
-j1
I2
-j2V
Create two equations for the created loops I1 and I2.
For Loop 1
−8∠30&deg; + (2 + 𝑗4)𝐼1 − 𝑗𝐼2 = 0
(2 + 𝑗4)𝐼1 − 𝑗𝐼2 = 8∠30&deg; → 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 1
For Loop 2
(2 + 𝑗4 − 𝑗)𝐼2 − 𝑗𝐼1 = 0 → 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 2
Writing equation 2 in terms of I1
(2 + 𝑗4 − 𝑗)𝐼2 − 𝑗𝐼1 = 0
(2 + 𝑗4 − 𝑗)𝐼2
= 𝐼1
𝑗
(3 − 𝑗2)𝐼2 − 2 = 𝐼1 → 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 3
Substitute equation 3 to equation 1 to gain I2
(2 + 𝑗4)𝐼1 − 𝑗𝐼2 = 8∠30&deg;
(2 + 𝑗4)[(3 − 𝑗2)𝐼2 − 2] − 𝑗𝐼2 = 8∠30&deg;
(14 + 𝑗8)𝐼2 − 4 − 𝑗8 − 𝑗𝐼2 = 8∠30&deg;
(14 + 𝑗7)𝐼2 − 4 − 𝑗8 = 8∠30&deg;
8∠30&deg; + 4 + 𝑗8
𝐼2 =
14 + 𝑗7
𝐼2 = 1.0369∠21.1113&deg; 𝐴
Solving Vx
𝑉𝑥 = 2𝐼2
𝑉𝑥 = 2𝑜ℎ𝑚𝑠(1.0369∠21.1113&deg; 𝐴)
𝑽𝒙 = 𝟐. 𝟎𝟕𝟑𝟖∠𝟐𝟏. 𝟏𝟏𝟏𝟑&deg; 𝑽
26. A three-phase supply, with the line voltage 240 V rms positively phased, has an
unbalanced delta-connected load as shown below. Find the phase currents and the
total complex power. (10 points)
SOLUTION:
The system is unbalanced, so computation of the phase current
will be done individually. Assuming VAB is the reference
voltage.
Calculating the phase current IAB
𝑉𝐴𝐵
𝐼𝐴𝐵 =
𝑍𝐴𝐵
240∠0&deg; 𝑉
𝐼𝐴𝐵 =
25 𝑂ℎ𝑚𝑠
𝑰𝑨𝑩 = 𝟗. 𝟔∠ − 𝟗𝟎&deg; 𝑨
Calculating the phase current IBC
𝑉𝐵𝐶
𝐼𝐵𝐶 =
𝑍𝐵𝐶
240∠ − 120&deg; 𝑉
𝐼𝐵𝐶 =
30∠30&deg; 𝑂ℎ𝑚𝑠
𝑰𝑩𝑪 = 𝟖∠ − 𝟏𝟓𝟎&deg; 𝑨
Calculating the phase current ICA
𝑉𝐶𝐴
𝐼𝐶𝐴 =
𝑍𝐶𝐴
240∠120&deg; 𝑉
𝐼𝐶𝐴 =
40 𝑂ℎ𝑚𝑠
𝑰𝑪𝑨 = 𝟔∠𝟏𝟐𝟎&deg; 𝑨
Calculating the power used at phase A.
𝑆𝐴 = |𝐼𝐴𝐵 |2 𝑍𝐴𝐵
𝑆𝐴 = |9.6|2 (𝑗25)
𝑆𝐴 = 𝑗2304𝑉𝐴
Calculating the power used at phase B.
𝑆𝐵 = |𝐼𝐵𝐶 |2 𝑍𝐵𝐶
𝑆𝐵 = |8|2 (30∠30&deg; )
𝑆𝐵 = 1662.7688 + 𝑗960 𝑉𝐴
Calculating the power used at phase C.
𝑆𝐶 = |𝐼𝐶𝐴 |2 𝑍𝐶𝐴
𝑆𝐶 = |6|2 (40
𝑆𝐶 = 1440𝑉𝐴
Solving total complex power
𝑆𝑇 = 𝑆𝐴 + 𝑆𝐵 + 𝑆𝐶
𝑆𝑇 = 𝑗2304𝑉𝐴 + 1662.7688 + 𝑗960 𝑉𝐴 + 1440𝑉𝐴
𝑆𝑇 = 3102.7688 + 𝑗3264 𝑉𝐴
𝑺𝑻 = 𝟒𝟓𝟎𝟑. 𝟒𝟐𝟖𝟕∠𝟒𝟔. 𝟒𝟓𝟎𝟔&deg; 𝑽𝑨
27. Three zones of a single phase circuit are identified in the figure. The zones are
connected by the treansformers T1 and T2, whose rating shown below.
Using base values of 30 kVA and 240 volts in zone 1, draw the pu
circuit and determine the pu impedance and pu source voltage.
Then calculate the load current in pu ans in amperes. (15 points)
SOLUTION:
Sbase=30 kVA
Vbase per region
240𝑉
𝑉𝑧𝑜𝑛𝑒1 = 240𝑉𝑥
= 240𝑉
240𝑉
480𝑉
𝑉𝑧𝑜𝑛𝑒2 = 240𝑉𝑥
= 480𝑉
240𝑉
460𝑉
𝑉𝑧𝑜𝑛𝑒2 = 480𝑉𝑥
= 120𝑉
115𝑉
Per unit computation
Generator 1
𝑉𝑔𝑖𝑣𝑒𝑛 240𝑉
𝑉𝐺1 =
=
= 1.0 𝑝𝑢
𝑉𝑏𝑎𝑠𝑒
240𝑉
Transformer 1
𝑉𝑔𝑖𝑣𝑒𝑛 2 𝑆𝑏𝑎𝑠𝑒
240𝑉 2 30𝑘𝑉𝐴
𝑥𝑇1 = 𝑥𝑔𝑖𝑣𝑒𝑛 (
) (
) = 𝑗0.1 (
) (
)
𝑉𝑏𝑎𝑠𝑒
𝑆𝑔𝑖𝑣𝑒𝑛
240𝑉
30𝑘𝑉𝐴
𝑥𝑇1 = 𝑗0.1 𝑝𝑢
Transformer 2
𝑉𝑔𝑖𝑣𝑒𝑛 2 𝑆𝑏𝑎𝑠𝑒
240𝑉 2 30𝑘𝑉𝐴
𝑥𝑇2 = 𝑥𝑔𝑖𝑣𝑒𝑛 (
) (
) = 𝑗0.1 (
) (
)
𝑉𝑏𝑎𝑠𝑒
𝑆𝑔𝑖𝑣𝑒𝑛
240𝑉
20𝑘𝑉𝐴
𝑥𝑇2 = 𝑗0.1378 𝑝𝑢
Line
2
𝑍𝑔𝑖𝑣𝑒𝑛
𝑍𝑔𝑖𝑣𝑒𝑛
𝑉𝑏𝑎𝑠𝑒
𝑗2
𝑥𝐿𝐼𝑁𝐸 =
; 𝑍𝑏𝑎𝑠𝑒 =
; 𝑥𝑙𝑖𝑛𝑒 = 2
=
480𝑉 2
𝑍𝑏𝑎𝑠𝑒
𝑆𝑏𝑎𝑠𝑒
𝑉𝑏𝑎𝑠𝑒
30𝑘𝑉𝑎
𝑆𝑏𝑎𝑠𝑒
𝑥𝐿𝐼𝑁𝐸 = 𝑗0.2604 𝑝𝑢
2
𝑍𝑔𝑖𝑣𝑒𝑛
𝑍𝑔𝑖𝑣𝑒𝑛
𝑉𝑏𝑎𝑠𝑒
0.5 + 𝑗0.2
𝑥𝐿𝑂𝐴𝐷 =
; 𝑍𝑏𝑎𝑠𝑒 =
; 𝑥𝑙𝑜𝑎𝑑 = 2
=
120𝑉 2
𝑍𝑏𝑎𝑠𝑒
𝑆𝑏𝑎𝑠𝑒
𝑉𝑏𝑎𝑠𝑒
30𝑘𝑉𝑎
𝑆𝑏𝑎𝑠𝑒
𝑥𝐿𝑂𝐴𝐷 = 1.0417 + 𝑗0.4167 𝑝𝑢
REACTANCE DIAGRAM
IL
Solving the current of the system through KVL.
1.0 − 𝐼𝐿 (𝑗0.1 + 𝑗0.2604 + 𝑗0.1377 + 1.0417 + 𝑗0.4167 = 0
1.0
𝐼𝐿 =
1.0417 + 𝑗0.9148
𝐼𝐿 = 0.7213∠ − 41.2890&deg;
Converting per unit value to the actual value.
𝐼𝐴𝑐𝑡𝑢𝑎𝑙 = 𝐼𝑝𝑢 ∗ 𝐼𝑏𝑎𝑠𝑒
𝑈𝑠𝑒 𝑡ℎ𝑒 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑏𝑎𝑠𝑒 𝑤ℎ𝑒𝑟𝑒 𝑡ℎ𝑒 𝑙𝑜𝑎𝑑 𝑖𝑠 𝑙𝑜𝑐𝑎𝑡𝑒𝑑 𝑤ℎ𝑖𝑐ℎ 𝑖𝑠 𝑧𝑜𝑛𝑒 3
𝑆𝑏𝑎𝑠𝑒
30𝑘𝑣𝐴
𝐼𝑏𝑎𝑠𝑒 =
=
= 250𝐴
𝑉𝑏𝑎𝑠𝑒
120𝑉
𝐼𝐴𝑐𝑡𝑢𝑎𝑙 = 0.7213∠ − 41.2890&deg; ∗ 250𝐴
𝑰𝑨𝒄𝒕𝒖𝒂𝒍 = 𝟏𝟖𝟎. 𝟑𝟐𝟖𝟐∠ − 𝟒𝟏. 𝟐𝟖𝟗𝟎&deg;𝑨
[Total Points = 60]
Prepared by:
Engr. Jefrey Jay S. Claus
Instructor, EE Dept.
Date: May 20, 2022
Engr. Analyn Cueto
Instructor, EE Dept.
Date: May 25,2022
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