# 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
ππ =
2π√πΏπΆ
1
15π₯103 =
2π√(10π₯10−3 )πΆ
1
√(10π₯10−3 )πΆ =
2π(15π₯103 )
2
(√(10π₯10−3 )πΆ) = (1.0610π10−5 )2
(1.1258π10^ − 10)
(10π₯10−3 )
πΆ = 11.258π10−9 π
πͺ = ππ. πππ ππ­
b. Solving the current through the coil at resonance
Consider the visualization of the circuit on the figure below
πΆ=
ππ  = 120π
Converting the circuit to the domain of frequency
π€π = 2πππ
π€π = 2π(15π₯103 )
π€π = 94.2478 ππ»π§
ππΏ = ππ€πΏ
ππΏ = π(94.2478π₯103 )(100π₯10−3 )
ππΏ = π9424.7780 πβππ  ππ 9425 πβππ
1
ππΆ =
ππ€πΆ
1
ππΆ =
3
π(94.2478π₯10 )(1.1258π₯10−9 )
ππΆ = −π9424.6999 πβππ  ππ − π9425 πβππ
ππ = π + πππΏ − πππΆ
ππ = 20 + π9425 − π9425
ππ = 20 πβππ
ππ
πΌπ =
ππ
120 π
πΌπ =
20 πβππ
π°π» = π π¨
c. Solving Quality Factor
1
π€π πΏ
π=
ππ
π€π πΆπ
π
π=
(94.2478π₯103 )(10π₯10−3 )
1
ππ
(94.2478π₯103 )(11.258π₯10−9 )(20)
20
πΈ = ππ. ππππ ππ ππ. ππππ
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
ππ = πππ1 + πππ2
ππ = 9300π + 5400π
π·π» = πππππ πΎ ππ ππ. πππΎ
√3(πππ1 − πππ2 )
πππ1 + πππ2
√3(9300π − 5400π)
π‘πππ =
9300π + 5400π
13
π = tan−1( √3)
49
π = 24.6799
ππ = πππ π
ππ = πππ 24.6799
ππ = π. ππππ
π‘πππ =
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:
`
πππ =
πππ
√3∠−90&deg;
=
220∠10&deg;
√3∠−90&deg;
; πππ = ππ΅πΆ
π½ππ = πππ∠πππ&deg; π½
Since balance system, we can relate them with the positive sequence
of the network.
ππ΄π΅ = ππ΅πΆ ∠π + 120 = 220∠10 + 120 = 220∠130&deg; π
ππ΄πΆ = ππ΄πΆ ∠π − 120 = 220∠10 − 120 = 220∠ − 110&deg; π
If the value of IbB is 30∠60&deg; A , then by the identity of the balance
system
πΌππ΄ = πΌππ΅ ∠π + 120 = 30∠60 + 120 = 30∠180&deg; π΄
πΌππΆ = ππ΅πΆ ∠π − 120 = 30∠60 − 120 = 30∠ − 60&deg; π΄
Converting phase current to line current.
πΌπ΄π΅ =
πΌππ΄
√3∠−30&deg;
=
30∠180&deg;
√3∠−30&deg;
= 17.3205∠ − 150&deg; π΄ ππ
πΌπ΄π΅ = 17.3205∠210&deg; π΄
πΌπ΅πΆ =
πΌππ΅
√3∠−30&deg;
=
30∠60&deg;
√3∠−30&deg;
= 17.3205∠90&deg; π΄ ππ
πΌπ΅πΆ = 17.3205∠ − 270&deg; π΄
πΌπΆπ΄ =
πΌππΆ
√3∠−30&deg;
=
30∠−60&deg;
√3∠−30&deg;
= 17.3205∠ − 30&deg; π΄ ππ
πΌπΆπ΄ = 17.3205∠330&deg; π΄
πΌπ΄πΆ = −πΌπΆπ΄ = −1(17.3205∠ − 30&deg; π΄)
π°π¨πͺ = (ππ. ππππ∠πππ&deg; π¨) = (ππ. ππππ∠ − πππ&deg; π¨)
Use the gained value to obtain the impedance of any phase since
they are a balance system.
ππ΅πΆ
π=
πΌπ΅πΆ
220∠10&deg; π
π=
17.3205∠90&deg; π΄
π = ππ. ππππ∠ − ππ&deg; π¨
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.
ππππ€ = πΌ(π) = 2∠0&deg;π΄(−π1πβππ ) = −π2π
2Ω
8∠ππ&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 + π4)πΌ1 − ππΌ2 = 0
(2 + π4)πΌ1 − ππΌ2 = 8∠30&deg; → πππ’ππ‘πππ 1
For Loop 2
(2 + π4 − π)πΌ2 − ππΌ1 = 0 → πππ’ππ‘πππ 2
Writing equation 2 in terms of I1
(2 + π4 − π)πΌ2 − ππΌ1 = 0
(2 + π4 − π)πΌ2
= πΌ1
π
(3 − π2)πΌ2 − 2 = πΌ1 → πππ’ππ‘πππ 3
Substitute equation 3 to equation 1 to gain I2
(2 + π4)πΌ1 − ππΌ2 = 8∠30&deg;
(2 + π4)[(3 − π2)πΌ2 − 2] − ππΌ2 = 8∠30&deg;
(14 + π8)πΌ2 − 4 − π8 − ππΌ2 = 8∠30&deg;
(14 + π7)πΌ2 − 4 − π8 = 8∠30&deg;
8∠30&deg; + 4 + π8
πΌ2 =
14 + π7
πΌ2 = 1.0369∠21.1113&deg; π΄
Solving Vx
ππ₯ = 2πΌ2
ππ₯ = 2πβππ (1.0369∠21.1113&deg; π΄)
π½π = π. ππππ∠ππ. ππππ&deg; π½
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
ππ΄π΅
πΌπ΄π΅ =
ππ΄π΅
240∠0&deg; π
πΌπ΄π΅ =
25 πβππ
π°π¨π© = π. π∠ − ππ&deg; π¨
Calculating the phase current IBC
ππ΅πΆ
πΌπ΅πΆ =
ππ΅πΆ
240∠ − 120&deg; π
πΌπ΅πΆ =
30∠30&deg; πβππ
π°π©πͺ = π∠ − πππ&deg; π¨
Calculating the phase current ICA
ππΆπ΄
πΌπΆπ΄ =
ππΆπ΄
240∠120&deg; π
πΌπΆπ΄ =
40 πβππ
π°πͺπ¨ = π∠πππ&deg; π¨
Calculating the power used at phase A.
ππ΄ = |πΌπ΄π΅ |2 ππ΄π΅
ππ΄ = |9.6|2 (π25)
ππ΄ = π2304ππ΄
Calculating the power used at phase B.
ππ΅ = |πΌπ΅πΆ |2 ππ΅πΆ
ππ΅ = |8|2 (30∠30&deg; )
ππ΅ = 1662.7688 + π960 ππ΄
Calculating the power used at phase C.
ππΆ = |πΌπΆπ΄ |2 ππΆπ΄
ππΆ = |6|2 (40
ππΆ = 1440ππ΄
Solving total complex power
ππ = ππ΄ + ππ΅ + ππΆ
ππ = π2304ππ΄ + 1662.7688 + π960 ππ΄ + 1440ππ΄
ππ = 3102.7688 + π3264 ππ΄
πΊπ» = ππππ. ππππ∠ππ. ππππ&deg; π½π¨
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
240π
ππ§πππ1 = 240ππ₯
= 240π
240π
480π
ππ§πππ2 = 240ππ₯
= 480π
240π
460π
ππ§πππ2 = 480ππ₯
= 120π
115π
Per unit computation
Generator 1
ππππ£ππ 240π
ππΊ1 =
=
= 1.0 ππ’
ππππ π
240π
Transformer 1
ππππ£ππ 2 ππππ π
240π 2 30πππ΄
π₯π1 = π₯πππ£ππ (
) (
) = π0.1 (
) (
)
ππππ π
ππππ£ππ
240π
30πππ΄
π₯π1 = π0.1 ππ’
Transformer 2
ππππ£ππ 2 ππππ π
240π 2 30πππ΄
π₯π2 = π₯πππ£ππ (
) (
) = π0.1 (
) (
)
ππππ π
ππππ£ππ
240π
20πππ΄
π₯π2 = π0.1378 ππ’
Line
2
ππππ£ππ
ππππ£ππ
ππππ π
π2
π₯πΏπΌππΈ =
; ππππ π =
; π₯ππππ = 2
=
480π 2
ππππ π
ππππ π
ππππ π
30πππ
ππππ π
π₯πΏπΌππΈ = π0.2604 ππ’
2
ππππ£ππ
ππππ£ππ
ππππ π
0.5 + π0.2
π₯πΏππ΄π· =
; ππππ π =
; π₯ππππ = 2
=
120π 2
ππππ π
ππππ π
ππππ π
30πππ
ππππ π
π₯πΏππ΄π· = 1.0417 + π0.4167 ππ’
REACTANCE DIAGRAM
IL
Solving the current of the system through KVL.
1.0 − πΌπΏ (π0.1 + π0.2604 + π0.1377 + 1.0417 + π0.4167 = 0
1.0
πΌπΏ =
1.0417 + π0.9148
πΌπΏ = 0.7213∠ − 41.2890&deg;
Converting per unit value to the actual value.
πΌπ΄ππ‘π’ππ = πΌππ’ ∗ πΌπππ π
ππ π π‘βπ π£πππ‘πππ πππ π π€βπππ π‘βπ ππππ ππ  πππππ‘ππ π€βππβ ππ  π§πππ 3
ππππ π
30ππ£π΄
πΌπππ π =
=
= 250π΄
ππππ π
120π
πΌπ΄ππ‘π’ππ = 0.7213∠ − 41.2890&deg; ∗ 250π΄
π°π¨πππππ = πππ. ππππ∠ − ππ. ππππ&deg;π¨
[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
```