General Sir John Kotelawala Defence University
Bachelor of Science in Engineering Degree Examination – 23 November 2022
EE 1203 THEORY OF ELECTRICITY
Model Answers
Question 1
[LO1]
E = 250V
 =  rad/s
L = 160mH
C = 200F
R = 25
Figure 1abcd
For the circuit shown in figure Q1abcd,
a) Obtain an expression for the parallel impedance of inductor L and capacitor C in terms of .
Impedance of inductor L = j L = j 16010-3 = j 0.16 
Impedance of capacitor C = 1/j C = – j /20010-6 = – j 5000/ 
Parallel impedance = =
−𝑗5000
𝜔
−𝑗5000
0.16𝜔+
𝜔
𝑗 0.16𝜔×
𝑗
=
800𝜔
[4 marks]
𝑗 0.16𝜔2 −𝑗5000
b) Determine an expression for the total impedance seen by the source.
Zcircuit = 25 +
800𝜔
[2 marks]
𝑗 0.16𝜔2 −𝑗5000
c) Determine the value of  for parallel resonance.
𝑗 0.16𝜔2 − 𝑗5000 = 0 → 𝜔2 = 31,250 →  = 176.8 rad/s
[2 marks]
d) Under the condition of resonance, determine the current supplied by the source and the voltage
across the inductor
Is = 0 A
[1 mark]
VL = VS = 250 V
[1 mark]
e) For the circuit shown in figure Q1efg, convert the circuit with mutual coupling to an equivalent
circuit with no mutual coupling and redraw the circuit
jM= j20 
A
Z1 = j15
E1=240  V
 = 250 rad/s
B
Z1 = j30
E4 = 240   /6 V
Z2 = – j25 
Figure Q1efg
j(30 + 20)
j(15 + 20)
A
E1=240   V
 = 250 rad/s
R4= 25 
– j20 
 = 250 rad/s
B
R4= 25 
E4 = 240   /6 V
Z2 = – j25 
 = 250 rad/s
[4 marks]
Page 1 of 6
f)
Mark relevant mesh currents on the non-coupled diagram and write down the Kirchhoff’s
voltage law equation for these meshes.
j50
j35
A
E1=240   V
 = 250 rad/s
B
25 
– j20 
I1
I2
E4 = 240   /6 V
2400 = j35 I1 – j(20 + 25) (I1 – I2)
– 240/6 = – j(20 + 25) (I2 – I1) + (j50 + 25) I2
g) Determine the loop currents
– j10 I1 + j45 I2 = 2400
→
j45 I1 + (25 + j5) I2 = – 240/6 →
[1 mark]
 = 250 rad/s
– j25 
[2 marks]
2 I1 – 9 I2 = j48
9 I1 + (1 – j5) I2 = 482/3 = – 24 + j 41.57
solve and obtain
i1 = 5.0693.6O A
i2 = 4.2188.0O A
Question 2
[3 marks]
[LO2]
Z1 = j15 
Z3= j30 
A
B
R2 = 25 
E1 = 240 V
 = 250 rad/s
Z4 =-j25 
Ig4 = 8 
 = 250 rad/s
E
Figure Q2abc
For the circuit shown in Figure Q2abc,
a) Convert the current source to a voltage source and redraw the circuit for mesh analysis,
including marking the relevant independent loops.
Z1 = j15 
E1 = 240 V
 = 250 rad/s
Im1
Z3= j30 
A
R2 = 25 
B
Im2
Z4 =-j25 
E4 = 200− 
 = 250 rad/s
E
[2 marks]
b) Write down the branch-mesh incidence matrix and the branch voltage vector.
2400
1 0
0
[1 −1] , [
]A
0
0 1
−200 − π/6
0 1
c) Determine the mesh impedance matrix and the mesh voltage vector.
2400
25 + 𝑗25
− 25
]V
[
] , [
− 25
25 + 𝑗5
−200 − π/6
[2x2 marks]
[2x2 marks]
Page 2 of 6
d) For the circuit shown in figure Q2d,
0.5 i1
i. convert the dependent current source, with no direct shunt
admittance, to equivalent dependent voltage sources.
i1
i4 10
ii. write down the Kirchhoff’s circuit equations
iii. determine the currents in all the branches
85 V
i5 60
50
i3
40
i2
Figure Q2d
The current source can be distributed as shown and converted to voltage sources.
i1
5 i1
30 i1
– +
– +
10
60
40
50
85 V
i3
[4 marks]
i2
85 + 5 i1 = 10 i1 + 50 i2
30 i1 = (60+40) (i1 – i2) – 50 i2
[2 marks]
which may be reduced to
5 i1 + 50 i2 = 85
or
i1 + 10 i2 = 17 and 15 i2 = 7 i1
substitution gives i1 = 3 A, i2 = 1.4 A
and i3 = i1 – i2 = 1.6 A, i4 = i1 – 0.5i1 = 1.5 A, i5 = i3 – 0.5i1 = 0.1 A
[4 marks]
e) Write down the two port parameter matrices for the 2 diagrams shown in figure Q4a
Z
Y
Figure Q2e
[
1
𝑌
0
]
1
[
1 𝑍
]
0 1
[4 marks]
f) Determine the ABDC matrix of the two-port network shown in figure Q2f.
Figure Q2f
𝐴
i.e. [
𝐶
1.3 + j0.4
𝐵
]=[
0.115 + j0.02
𝐷
[3 marks]
6 + j8
1.36017.1°
]=[
1.3 + j0.4
0.1179.9°
1053.1°
]
1.36017.1°
[3 marks]
Page 3 of 6
Question 3
[LO3]
200 
e(t)
i(t)
C = 200 F
1000 
v(t)
Figure Q3abc
a) Laplace transform the circuit shown in figure Q3abc and obtain an expression for the transform of the
current I(s) in terms of the transform of the input voltage E(s).
200
E(s)
I(s)
)
5000
𝑠
1000
V(s)
[2 marks]
𝐸(𝑠)
𝐸(𝑠)
𝐼(𝑠) =
=
5000
5000
200 +
1000 × 𝑠
𝑠+5
200 +
5000
1000 + 𝑠
𝐼(𝑠) =
(𝑠+5)
200𝑠+6000
𝐸(𝑠)
[2 marks]
b) Hence obtain the transfer function relating I(s) with input voltage E(s).
Transfer Function =
c)
(𝑠+5)
200𝑠+6000
[1 mark]
If a 400V step is applied as e(t), determine an expression for the transient current i(t).
(𝑠+5)
(2𝑠+10)
400
1
5
𝐼(𝑠) =
= 𝑠(𝑠+30) = 3𝑠 + 3(𝑠+30)
[3 marks]
200𝑠+6000 𝑠
𝑖(𝑡) =
1
3
(1 + 5𝑒 −30𝑡 ). 𝐻(𝑡)
[2 marks]
Page 4 of 6
Question 4
[LO4]
v(t)
10V
5V
-15
-10
-5
0
5
10
15 t (s)
Figure Q4a
a) For the voltage waveform v(t) shown in figure Q4c, determine three significant terms of the Fourier
series. Determine also the form factor and the peak factor of the voltage waveform.
Waveform v(t) has a mean value of 5V. Therefore Ao/2 in Fourier Series = 5 V.
[2 marks]
Consider new function v(t) – 5. This is an odd waveform. Therefore An = 0 for all n.
[2 marks]
Waveform does not have half wave symmetry.
Period = 10 s, o = 2 /10 =  / 5
There
Bn
[1 mark]
[2 marks]
= 2  2/T  [v(t) – 5 ] sin nt dt from t = 0 to 5 = 4/10  t . sin n t/5 dt
= 0.4 [ t . {cos n t/5}/{ n/5}limits – 1.{cos n t/5}/{ n/5} dt]
= 0.4 [0 – 5  (5/ n)cos n −  + (5/ n) sin n
= (10/ n)cos n
B1 = −10/  B2 = 5/
[3 marks]
Thus, the Fourier Series is
v(t) = 5 − (10/) sin  t/5 + (5/) sin 2 t/5 + ....
[1 mark]
r m s value of waveform = [52 + ½ (10/) + ½ (5/) +    V
[or more accurately  [(1/10)  (5 +t)2 dt] = 5.774V]
For this waveform, average value is identical to the mean value. i.e. average value = 5 V
[1 mark]
 form factor = 5.774/5 = 1.154
[1 mark]
(or 5.60/5 = 1.12)
Peak value = 10 V,  peak factor = 10/5.774 = 1.732
[2 marks]
b) If a current i(t) = 60 sin 200 t – 40 sin 600t + 10 sin 1000 t is passed through a series
combination of R = 10 Ω and L = 50 mH, determine the Fourier series of the voltage across the
combination, and the active power dissipation in the circuit.
v(t) = L d i(t)/dt +R i(t)
= 0.05200[60 cos 200 t – 403cos 600t + 105cos 1000 t] + 10[60 sin 200 t – 40 sin 600t +10 sin 1000 t]
= 10[60 cos 200 t – 120 cos 600t + 50 cos 1000 t] + 10[60 sin 200 t – 40 sin 600t + 10 sin 1000 t]
= (600 cos 200 t + 600 sin 200 t) – (1200 cos 600t .+ 400 sin 600t) + (500 cos 1000 t + 100 sin 1000 t)
v(t) = 848.5 sin (200t +450) + 1264.9 sin (600t +108.430) + 509.9 sin(1000t+78.690) [3 marks]
Active power dissipation = R  |I|2 = 10 [602/2 + 402/2 +102/2] = 26.5 kW
[2 marks]
0
(OR P = sum of harmonic active power = 0.5x[848.5x60 cos 45 +1264.9x40 cos 71.570 +
509.9x10 cos 78.690 ] = 17,999 + 7,998 + 500 = 26,497 = 26.5 kW)
Page 5 of 6
Question 5
[LO5]
A 415 V, 50 Hz, balanced 3 phase source supplies two 3-phase loads (i) a balanced star-connected load with
each arm consisting of a series connection of a resistance R = 20  and an inductance of L = 40mH, and (ii) a
three-phase ac motor of 5 kW at a power factor of 0.65 lag.
(a) Sketch the equivalent three-phase diagram for the system.
[1 mark]
(b) Determine the line current supplied, the active and reactive power consumed by load 1.
Line current 𝐼𝐿1
=
415⁄
239.6<00
√3
=
= 10.14– 32.140 A
20+𝑗2500.040 23.62<32.14 𝑜
Active power P = 341510.14cos(32.140) = 6172 W
[1 mark]
Reactive power Q = 341510.14sin(32.140) = 3877 var (inductive)
(c) Determine the line current supplied and the reactive power consumed by load 2.
Line current 𝐼𝐿2
=
5000
√3×4150.65
=
[2 marks]
10.70– 49.460 A
[1 mark]
[1 mark]
0
Reactive power Q = 5000tan(49.46 ) = 5846 var (inductive)
[1 mark]
(d) Determine the total line current, the overall load power factor, the total active power and the total
reactive power consumed.
Total line current IL = 10.14– 32.140 +10.70– 49.460
= 8.59 – j 5.39 + 6.95 – j 8.13 = 15.54 – j 13.52 = 20.60 −41.020 A
[1 mark]
Overall load power factor = cos(41.020) = 0.754 lag
[1 mark]
Total active power = 341520.600.754 = 11.16 kW
[1 mark]
Total Reactive power = 341520.60sin(41.020) = 9.72 kvar (inductive)
[1 mark]
(e) Determine the rating of each of the 3 capacitors required to improve the load
power factor to 0.98 by the use of a 3-phase, delta connected capacitor bank.
Q
new
with pf = 0.98 lag → 11.48o , new reactive power = 11.16×tan 11.48o = 2.27 kvar (inductive)
Rating of each capacitor QC = (9.72 – 2.27)/3 = 2.48 kvar
Q
[or value of each capacitor C = 2485/(4152×100) = 45.9 F not required]
[3 marks]
(f) An unbalanced system with phase sequence A-B-C has the phase components of the current given as IA
= 3.0300A, IB = 4.800A and IC = 2.0–300A. Determine the sequence components of the currents.
IA0
1
1
[IA1] = 3 [1
IA2
1
1
𝛼
𝛼2
1
3.0300
𝛼 2 ] [ 4.800 ]
𝛼 2.0 − 300
[1 mark]
IA0 = (3.030o+4.80o+2.0−30o)/3 = (2.598+j1.5 +4.8 + 1.732 -j1.0)/3 = 3.043+j0.167 = 3.0483.1o A
IA1 =(3.030o+4.8120o+2.0210o)/3=(2.598+j1.5 –2.4+j4.156 –1.732–j1.0)/3= –0.511+j1.552=1.634108.3o A
IA2 = (3.030o+4.8240o+2.090o)/3 = (2.598+j1.5 –2.4 –j4.156 +j2.0)/3 = 0.066 – j0.219 = 0.229 – 73.90 A
[3 marks]
(g) Determine the total active power in the unbalanced system if the sequence components of the voltage
are given as VA0 = 1000V, VA1 = 2301200V and VA2 = 50–600V.
P = 3(VA0  IA0  cos A0 + VA1  IA1  cos A1 + VA2  IA2  cos A2)
= 3 (103.048cos 3.1o + 2301.634cos 11.7o + 500.229cos 13.9o) = 3 (30 + 368 + 11) = 1227W
[2 marks]
End of answer script
Page 6 of 6
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