Surname: . . . . . . . . . . . . . . . . . . . . . . . . .
First Name: . . . . . . . . . . . . . . . . . . .
Student ID: . . . . . . . . . . . . . . . . . . .
T E W H A R E W ¯A N A N G A O T E ¯U P O K O O T E I K A A M ¯A U I
VUW VICTORIA
U N I V E R S I T Y O F W E L L I N G T O N
TEST 1 – 2019
TRIMESTER 1
ECEN 203
ANALOGUE CIRCUITS
AND SYSTEMS
Time Allowed:
FIFTY MINUTES
CLOSED BOOK
Permitted materials: You may use a scientific calculator, but not a smartphone or other
similar device.
Instructions:
Attempt ALL THREE Questions
Write your answers on the question paper.
Full marks will be awarded for correct answers.
Partial marks may be awarded to incorrect answers if working is
shown.
The exam will be marked out of a total of 50 marks.
ECEN 203
Page 1 of 8

ECEN 203
Page 2 of 8

1. Mesh current analysis
(20 marks)
Find V by mesh current analysis. Note: you will need to come up with a strategy
to deal with the current source.
10 Ω
V
12 Ω
2 Ω
+
50 V
3 A
−
5 Ω
10 Ω
V
i
i
1
3
12 Ω
2 Ω
+
50 V
3 A
−
i
5
2
Ω
Remember that the key is to make sure that only one current flows through
the current source, hence the need for the loop around the outside, or perhaps
around the bottom half.
(4 marks)
−10i1 − 2i1 − 12i1 + 2i2 + 10i3 =0
i2 =3
10i1 + 5i2 − 5i3 − 10i3 − 50 =0
(6 marks)
−24i1 + 10i3 = − 6
10i1 − 15i3 =35
ECEN 203
Page 3 of 8

12i1 − 5i3 =3
2i1 − 3i3 =7
12 −5 i

1
3
=
2
3
i3
7
(2 marks)

12
−5

=(12)(−3) − (2)(−5) = −36 + 10 = −26

2
−3

3
−5

=(3)(−3) − (7)(−5) = −9 + 35 = 26
7
−3

12
3

=(12)(7) − (2)(3) = 84 − 6 = 78

2
7
(3 marks)
So
26
i1 =
= 1 A
−26
78
i3 =
= −3 A
−26
(2 marks)
So the current through the 10 Ω resistor is i1 − i3 = −1 − −3 = 2 A to the right.
(2 marks)
So V = (10)(2) = +20 V
(1 mark)
ECEN 203
Page 4 of 8

2. Node voltage analysis
(15 marks)
Find V by node voltage analysis. (Note: this is the same circuit as the previous
question).
10 Ω
V
12 Ω
2 Ω
+
50 V
3 A
−
5 Ω
10 Ω
V
v1
v2
12 Ω
2 Ω
+
50 V
3 A
−
5 Ω
(2 marks)
At node 1:
v1 − 50
v
− 3 + 1 − v2 = 0
12
2
(3 marks)
At node 2:
v2 − v1
v
v
+ 2 + 2 − 50 = 0
2
5
10
(3 marks)

7
− 1 v

86
12
2
1
=
12
− 1
8
v
5
2
10
2
(2 marks)
ECEN 203
Page 5 of 8

7
−6 v

1
86
=
−5
8
v2
50
Note that we don’t need to solve this for v1, just v2:

7
86

−

5 50
350 + 430
v2 =
=

7
−6
56 − 30

−

5
8
780
= 26
=30
(3 marks)
So v = 50 − 30 = 20 V
(2 marks)
ECEN 203
Page 6 of 8

3. Thevenin/Norton equivalents
(15 marks)
Find the Thevenin equivalent at terminals A&B:
+ −
A
4 Ω
10 V
2 Ω
6 Ω
38 A
+
130 V
−
B
There are many ways to do this.
One of the easiest is to replace:
+ −
4 Ω
10 V
2 Ω
+
130 V
−
by
6 Ω
+
120 V
−
(5 marks)
and then by its Norton equivalent
20 A
6 Ω
(3 marks)
And then we have
ECEN 203
Page 7 of 8

A
20 A
6 Ω
6 Ω
38 A
B
(2 marks)
and hence
A
58 A
3 Ω
B
(3 marks)
and so the Thevenin equivalent is
A
3 Ω
+
174 V
−
B
(2 marks)
It is actually very easy to get the Thevenin resistance, by zeroing the sources to
get 6k6 = 3 Ω, but finding the open-circuit voltage or the short-circuit current is
possibly a bit harder than the transformations above.
* * * * * * * * * * * * * * *
ECEN 203
Page 8 of 8

Surname: . . . . . . . . . . . . . . . . . . . . . . . . .
First Name: . . . . . . . . . . . . . . . . . . .
Student ID: . . . . . . . . . . . . . . . . . . .
T E W H A R E W ¯A N A N G A O T E ¯U P O K O O T E I K A A M ¯A U I
VUW VICTORIA
U N I V E R S I T Y O F W E L L I N G T O N
TEST 2 – 2019
TRIMESTER 1
ECEN 203
ANALOGUE CIRCUITS
AND SYSTEMS
Time Allowed:
FIFTY MINUTES
CLOSED BOOK
Permitted materials: You may use a scientific calculator, but not a smartphone or other
similar device.
Instructions:
Attempt ALL Questions
Write your answers on the question paper.
Full marks will be awarded for correct answers.
Partial marks may be awarded to incorrect answers if working is
shown.
The exam will be marked out of a total of 50 marks.
ECEN 203
Page 1 of 8

ECEN 203
Page 2 of 8

1. Transient analysis
(20 marks)
i(t)
100 Ω
+
10 V
10 mH
200 Ω
−
The switch closes at t = 0 after being open for a long time. Calculate an expres-
sion for and plot i(t). (It may be easiest to separately calculate the steady state
response and the transient response).
For the steady-state response di = 0 so v
dt
L = 0, so in effect we have
100 Ω
+
−
10 V
200 Ω
So iss(t) = 10 = 0.1 A
100
For the transient response, (i.e., with the source removed) we have:
100 Ω i
10 mH
vL
200 Ω
iL
Using node voltage analysis:
v
i = − L
100
and
vL
v
+ i
L = 0
100
L + 200
3vL + i
200
L = 0
ECEN 203
Page 3 of 8

di
200
v
L
L = L
= 10−2 diL −
i
dt
dt
3 L
diL
20000
= −
i
dt
3
L
−t
i
t
150
L = Ae− 20000
3
= Ae
µs
(We could also have got this from considering the Thevenin equivalent of the
voltage source and resistors.)
v
i
L
tr(t) =iL(t) + 200
1
di
=i
L
L(t) +
L
200
dt
−t
1

−1
−t
=Ae 150 µs +
10 × 10−3
e 150 µs
200
150 µs
−t
=Be 150 µs
i.e., the transient response of i(t) takes the same form as iL(t), and we don’t
actually need to find out what A is.
At the moment the switch closes, iL(0+) = iL(0−) = 0, so i(0) = 10 = 1
300
30
So the total response is
itotal =iss + itr
−t
=0.1 + Be 150 µs
1
1
iL,total(0) =
=
+ B
30
10
so B = − 2
30
so
1
−t

i(t) =
3 − 2e 150 µs
30
i(t)
0.1
8 · 10−2
6 · 10−2
4 · 10−2
2 · 10−2
t(µs)
100
200
300
400
ECEN 203
Page 4 of 8

2. LCR circuits and frequency response
(30 marks)
iin(t)
R
L
C
vout(t)
(a) (10 marks) (Phasors) For the circuit above, write a general expression for
the output voltage phasor Vout in terms of the input current phasor Iin.
The output voltage is the input current times the combined impedance of the
three parallel components:

!
1
Vout = Iin
sC + 1 + 1
R
sL
Vout(s)
sRL
=
Iin(s)
s2RLC + sL + R
Since we are next told that the input is a sinusoid of constant magnitude,
s = jω, and the above is equivalent to
Vout(ω)
jωRL
=
Iin(ω)
−ω2RLC + jωL + R
ECEN 203
Page 5 of 8

(b) (10 marks) (Impedance) If R = 50 Ω, L = 0.1 H C = 10 µF, and iin is a
sinusoid of rms value 10 A and frequency 63.662 Hz, calculate the amplitude
and phase of the output voltage vout and its phase relative to iin. Hence write
an expression for vout.
ω = 2π f = 2π63.662 = 400.0001 ≈ 400 rad s−1
Also the peak amplitude is given by
√
10 2 = 14.142 A
So we have
j(50)(400)(0.1)
√
Vout =
10 2
−4002(50)(0.1)(10−5) + 50 + j(400)(0.1)
j2000
√
=
10 2
50 − 8 + j40
j2000
√
=
10 2
42 + j40
j2000(21 − 20j)
√
=
10 2
2 × 292
1000(20 + 21j)
√
=
10 2
292
=(23.781 + 24.970j)14.142
=336.3 + 353.1j
=487.76 0.8097
=487.76 46.4◦
So vout(t) = 487.7 cos(400t + 0.8097) V
ECEN 203
Page 6 of 8

(c) (10 marks) (Frequency response) If the frequency of the source retains the
same amplitude, but varies in frequency between 1 Hz and 10 kHz, plot the
output voltage Vout amplitude and phase as a function of frequency on the
graphs below.
The plots do not need to be particularly accurate, but should include certain
features that we would expect.
103
102
|V|
101
100100
101
102
103
104
f(Hz)
150
100
50
0
angle(V)
50
100
150
100
101
102
103
104
f(Hz)
At very low frequencies, the inductor is a short circuit, while the capacitor is
an open circuit. So we expect the circuit to behave like a high pass filter with
a 20 dB/decade slope, and phase approaching +90◦.
At very high frequencies, the inductor is an open circuit, while the capacitor
is a short circuit. So we expect the circuit to behave like a low pass filter with
a 20 dB/decade slope, and phase approaching −90◦
At f =
1
√
=
1
√
≈ 159 Hz, the real part of the denominator is zero,
2π LC
2π 106
ECEN 203
Page 7 of 8

so the phase is zero, and the output reaches its maximum value of 707.1 V.
103
102
|V|
101
100100
101
102
103
104
f(Hz)
150
100
50
0
angle(V)
50
100
150
100
101
102
103
104
f(Hz)
* * * * * * * * * * * * * * *
ECEN 203
Page 8 of 8

Surname: . . . . . . . . . . . . . . . . . . . . . . . . .
First Name: . . . . . . . . . . . . . . . . . . .
Student ID: . . . . . . . . . . . . . . . . . . .
T E W H A R E W ¯A N A N G A O T E ¯U P O K O O T E I K A A M ¯A U I
VUW VICTORIA
U N I V E R S I T Y O F W E L L I N G T O N
TEST 3 – 2019
TRIMESTER 1
ECEN 203
ANALOGUE CIRCUITS
AND SYSTEMS
Time Allowed:
FIFTY MINUTES
CLOSED BOOK
Permitted materials: Only silent non-programmable calculators or silent programmable
calculators with their memories cleared are permitted.
Instructions:
Attempt ALL Questions
Write your answers on the question paper.
Full marks will be awarded for correct answers.
Partial marks may be awarded to incorrect answers if working is
shown.
The exam will be marked out of a total of 50 marks.
ECEN 203
Page 1 of 8

1. Frequency Response
(10 marks)
47K
+
v
47 nF
i
(a) (5 marks) vi(t) is a 80 Hz cosinusoid having amplitude of 5 V. What steady
state current will flow through the resistor? Write your answer as a function
of time.
V
I(
i(ω)
ω) = R + 1jω
5
= 47 × 103 +
1
j2π80×47×10−9
5
= 47000 − 42328.44j
5
= 63.251 × 1036 − 42.0◦
=(5.874 + 5.290) × 10−5
=7.905 × 10−56 42.0◦
so
i(t) =79.05 cos(160π + 42.0◦) µA
(b) (5 marks) Does the current lead or lag the input voltage of this circuit?
Using a phasor diagram, explain why your answer is correct.
The current leads the voltage. Since both voltage and current can be con-
sidered to be rotating counter-clockwise, the phasor diagram shows that for
any given angle, the current arrow “arrives first.”
Im
I
φ
V
Re
ECEN 203
Page 2 of 8

2. Frequency response
(10 marks)
A circuit has transfer function G(jω) = Vo =
500
. If an input voltage of
Vi
jω+200
vi(t) = 10 cos (400t − 70◦) is applied to the circuit, what will the steady state
output be? Express your answer as a function of time.
ω = 400
So
500
G(jω) = 400j + 200
500
= 447.2146 1.107rad
500
= 447.2146 63.435◦
=1.1186 − 63.435◦
Vi = 106 − 70◦
So
Vo =(106 − 70◦)(1.1186 − 63.435◦)
=11.186 − 133.435◦
So
vo(t) =11.18 cos(400t − 133.435◦) V
ECEN 203
Page 3 of 8

3. Circuit transfer function
(10 marks)
R
+
vi
L
C
vo
(a) (5 marks) What is the transfer function Vo/Vi of the circuit shown above?
Write your answer so that the coefficient of the highest power of ω in the
denominator is 1.
Vo(s)
Z
=
LkZC
Vi(s)
R + ZLkZC
sL
sC
sL+ 1
=
sC
sL
R +
sC
sL+ 1
sC
sL
=
s2LC+1
R +
sL
s2LC+1
sL
= s2LCR + sL + R
s 1
=
RC
s2 + s 1 + 1
RC
LC
V
1
o(jω)
jω
=
RC
V
1
i(jω)
−ω2 + jω
+ 1
RC
LC
−j
1
ω
=
RC
1
ω2 − jω
− 1
RC
LC
ECEN 203
Page 4 of 8

(b) (5 marks) By considering the response at various frequencies (for R = L =
C = 1 for example) sketch the form of the magnitude and phase responses
of this circuit
0
−5
(dB)
−10
Amplitude
−15
−20
10−1
100
101
Angular frequency ω (rad s−1)
200
100
rees)
(deg
0
Phase
−100
−200
10−1
100
101
Angular frequency ω (rad s−1)
ECEN 203
Page 5 of 8

4. Frequency response
(11 marks)
5
0
(dB)
−5
Amplitude
−10
100
101
Angular frequency ω (106 rad s−1)
Consider the magnitude response of a resonant system shown above.
(a) (2 marks) What are the half power frequencies of the system?
The peak is 6 dB, so half power will be at 3 dB. These are at 2.7016 × 106 and
3.7016 × 106 rad s−1
(b) (2 marks) What is the bandwidth of the system?
So the bandwidth is 1 × 106 rad s−1
(c) (2 marks) What is the Q of the system?
√
The centre frequency is
10 × 106 = 3.1623 × 106 rad s−1, so the Q is 3.1623/1
= 3.1623
ECEN 203
Page 6 of 8

(d) (5 marks)
You are required to design a parallel resonant circuit having
this magnitude response using a 10 µH inductor. What value resistor and
capacitor would you select? (You are not restricted to use E12 values.)
1
ω0 = √LC
so
1
1
C =
=
= 10−8
2
ω L
10 × 1012 ∗ 10 × 10−6
0
so use a 10 nF capacitor.
q
For a parallel circuit, Q = R
C so
L
r
r
L
√
10−5
R = Q
=
10
= 3162.3
C
10−8
so use a 3.2 kΩ resistor.
ECEN 203
Page 7 of 8

5. Differential equation modes
(9 marks)
The solutions to the constant coefficient linear differential equations that we en-
counter when dealing with circuit analysis have solutions that are made up of
complex exponential terms e(σ+jω)t. For each of the following, sketch the REAL
PARTS of the modes (the waveforms) that correspond to the particular choice
of σ + jω. (You don’t need to be precise about this; the important thing is to
distinguish between the different plots, not to plot them accurately).
(a) (3 marks) σ = −1, σ = −3 and σ = 1, each with ω = 0.
1
1
20
0.8
0.8
15
0.6
0.6
10
0.4
0.4
5
0.2
0.2
t
t
t
0.5
1
1.5
2
2.5
3
0.5
1
1.5
2
2.5
3
0.5
1
1.5
2
2.5
3
(b) (2 marks) ω = 2, ω = 3, both with σ = −1.
1
1
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2
0.5
1
1.5
2
2.5
3
0.5
1
1.5
2
2.5
3
t
t
−0.2
−0.2
(c) (2 marks) ω = 1, ω = 2, both with σ = 0.
1
1
0.5
0.5
0.5
1
1.5
2
2.5
3
0.5
1
1.5
2
2.5
3
t
−0.5
t
−0.5
−1
(d) (2 marks) σ = −4 and σ = 3, both with ω = 3.
1
0.8
0.5
1
1.5
2
2.5
3
0.6
−2,000
0.4
t
−4,000
0.2
−6,000
t
0.5
1
1.5
2
2.5
3
* * * * * * * * * * * * * * *
ECEN 203
Page 8 of 8

Surname: . . . . . . . . . . . . . . . . . . . . . . . . .
First Name: . . . . . . . . . . . . . . . . . . .
Student ID: . . . . . . . . . . . . . . . . . . .
T E W H A R E W ¯A N A N G A O T E ¯U P O K O O T E I K A A M ¯A U I
VUW VICTORIA
U N I V E R S I T Y O F W E L L I N G T O N
TEST 4 – 2019
TRIMESTER 1
ECEN 203
ANALOGUE CIRCUITS
AND SYSTEMS
Time Allowed:
FIFTY MINUTES
CLOSED BOOK
Permitted materials: Only silent non-programmable calculators or silent programmable
calculators with their memories cleared are permitted.
Instructions:
Attempt ALL Questions
Write your answers on the question paper.
Full marks will be awarded for correct answers.
Partial marks may be awarded to incorrect answers if working is
shown.
The exam will be marked out of a total of 50 marks.
ECEN 203
Page 1 of 10

1. Solving differential equations
(12 marks)
(a) (5 marks)
A system is governed by the differential equation
y00(t) + 5y0(t) + 4y(t) = 6x00(t) + 16x0(t) + 4x(t)
Assuming zero initial conditions, show that the transfer function of the system
can be written as
Y(s)
6s2 + 16s + 4
G(s) :=
=
X(s)
s2 + 5s + 4
Laplace transform both sides of the equations, remembering that all initial
conditions are zero.
s2Y(s) + 5sY(s) + 4Y(s) = 6s2X(s) + 16sX(s) + 4X(s)
h
i
h
i
s2 + 5s + 4 Y(s) = 6s2 + 16s + 4 X(s)
Y(s)
6s2 + 16s + 4
=
X(s)
s2 + 5s + 4
1 mark for Laplace transforming the expression
2 marks for getting the transform right
2 marks for arranging into a transfer function
ECEN 203
Page 2 of 10

(b) (7 marks) A unit step input is applied to the system (i.e., x(t) = u(t)). What
is the output of the system in the time domain?
6s2 + 16s + 4
1
Y(s) =
×
s2 + 5s + 4
s
6s2 + 16s + 4
= s(s + 1)(s + 4)
A
B
C
=
+
+
s
s + 1
s + 4
Use coverup technique (or alternative) to find A, B and C.
6s2 + 16s + 4
4
A =

=
= 1
(s + 1)(s + 4) s=0
4
6s2 + 16s + 4
−6
B =

=
= 2
s(s + 4)

−
s→−1
3
6s2 + 16s + 4
36
C =

=
= 3
s(s + 1)
s→−4
12
1
2
3
=⇒ Y(s) =
+
+
s
s + 1
s + 4
h
=⇒ y(t) = 1 + 2e−t + 3e−4ti u(t)
2 marks for finding L {u(t)} = 1 , and multiplying with G(s)
s
4 marks for doing the partial fractions expansion correctly
1 mark for inverting to find y(t)
ECEN 203
Page 3 of 10

2. Transfer functions and partial fractions
(12 marks)
An input of x(t) = 10 sin(3t) is applied to a filter having transfer function
1
G(s) = s + 1
(a) (3 marks) Show that the output Y(s) = G(s)X(s) can be described by
30
Y(s) = (s + 1)(s2 + 9)
X(s) = L {10 sin(3t)}
3
= 10 s2 + 32
30
= s2 + 9
1
30
Y(s) = s + 1 s2 + 9
30
= (s + 1)(s2 + 9)
2 marks for Laplace transforming x(t) correctly.
1 mark for multiplying by G(s)
ECEN 203
Page 4 of 10

(b) (9 marks) Find y(t)
30
Y(s) = (s + 1)(s2 + 9)
A
B
=
1s + A0 +
s2 + 9
s + 1
30
30
B =

=
= 3
s2 + 9 s=−1
10
30
So, Y(s) = (s + 1)(s2 + 9)
A
3
=
1s + A0 +
s2 + 9
s + 1
(A
=
1s + A0)(s + 1) + 3s2 + 27
(s + 1)(s2 + 9)
A
=
1s2 + A1s + A0s + A0 + 3s2 + 27
(s + 1)(s2 + 9)
s2 : A

1 + 3 = 0 =⇒ A1 = −3

=⇒
s1
: A1 + A0 = 0 =⇒ A0 = 3

s0
: A0 + 27 = 30
−3s + 3
3
Y(s) =
+
s2 + 9
s + 1
3s − 3
1
= −
+
s2 + 9
s + 1
3s
3
1
= −
+
+
s2 + 9
s2 + 9
s + 1
=⇒ y(t) = −3 cos(3t) + sin(3t) + 3e−t u(t)
2 marks for correctly identifying the form of the partial fractions expansion.
6 marks for finding the correct coefficients (half marks if algebraic mistakes
made along the way.
1 mark for correctly converting to the time domain (even if the answer is
wrong because of mistakes made previously).
ECEN 203
Page 5 of 10

3. Partial fractions
(12 marks)
2
(a) (6 marks) Find the inverse Laplace transform of G(s) =
.
(s + 1)2(s + 2)
2
G(s) = (s + 1)2(s + 2)
A
B
C
=
+
+
(s + 1)2
s + 1
s + 2
2
A =

= 2
s + 2 s=−1
2

C =

= 2
(s + 1)2 s=−2
There are multiple methods to find B.
d
2
B = lim
s→−1 ds s + 2
d
= lim
2(s + 2)−1
s→−1 ds
−2
= lim
s→−1 (s + 2)2
= −2
Or, substitute s = 0 (or some other value, s /
∈ {−1, −2})
2
2
B
2
G(0) =
=
+
+
2
12
1
2
1 = 3 + B
=⇒ B = −2
2
2
2
Either way we have, G(s) =
−
+
(s + 1)2
s + 1
s + 2
h
=⇒ g(t) = 2te−t − 2e−t + 2e−2ti u(t)
2 marks for correctly identifying the partial fractions form
3 mark for finding the residuals
1 mark for converting to the time domain
ECEN 203
Page 6 of 10

s + 6
(b) (6 marks) A signal has an s-domain representation of H(s) =
.
(s + 2)3
What is h(t)?
A
B
C
H(s) =
+
+
(s + 2)3
(s + 2)2
s + 2
A = lim s + 6 = 4
s→−2
d
B = lim
s + 6 = lim 1 = 1
s→−2 ds
s→−2
1 d
C = lim
1 = 0
s→−2 2! ds
4
1
=⇒ H(s) =
+
(s + 2)3
(s + 2)2
4

So, h(t) =
t2e−2t + te−2t u(t)
2

= 2t2 + t e−2tu(t)
2 marks for correctly identifying the partial fractions form.
3 marks for finding the residuals.
1 marks for converting to the time domain.
ECEN 203
Page 7 of 10

4. Unit steps
(14 marks)
(a) (3 marks)
Sketch the waveform of x(t) = tu(t − 1) − 2u(t − 4)
x(t)
6
4
2
t
2
4
6
8
(b) (3 marks)
What is the Laplace transform of x(t)?
x(t) = (t − 1)u(t − 1) + u(t − 1) − 2u(t − 4)
e−s
e−s
2e−4s
=⇒ X(s) =
+
−
s2
s
s
1 mark for rewriting the ramp part of the function in a form for which the time
shift formula can be used (even if not written explicitly)
1 mark for correctly using the time shift formula to get the expected coeffi-
cients of s in the exponents.
1 mark for the correct denominators in X(s).
ECEN 203
Page 8 of 10

(c) (8 marks)
2
If x(t) is applied to a system having transfer function G(s) =
, what will
s + 2
the output be?
Warning: This question will take longer than reflected in the number of
marks. Don’t spend too long on this.
2
Y(s) =
X(s)
s + 2
2e−s
2e−s
4e−4s
=
+
−
s2(s + 2)
s(s + 2)
s(s + 2)
1
A
B
Now,
=
+
s(s + 2)
s
s + 2
1
1
A =

=
s + 2 s=0
2
1
1
B =

= −
s s=−2
2
2e−s
e−s
e−s
2e−4s
2e−4s
Y(s) =
+
−
−
+
s2(s + 2)
s
s + 2
s
s + 2
1
E
F
G
Similarly,
=
+
+
s2(s + 2)
s2
s
s + 2
1
1
E =

=
s + 2 s=0
2
1
1
G =

=
s2 s=−2
4
1
1 1
F
1
1
=
+
+
s2(s + 2)
2 s2
s
4 s + 2
Substitute s = 1 (or something else)
1
1
1
=
+ F +
3
2
12
4 − 6 − 1 = F
12
1
F = − 4
1
1 1
1 1
1
1
i.e.,
=
−
+
s2(s + 2)
2 s2
4 s
4 s + 2
So finally, we get
e−s
1 e−s
1 e−s
e−s
e−s
2e−4s
2e−4s
Y(s) =
−
+
+
−
−
+
s2
2 s
2 s + 2
s
s + 2
s
s + 2
1
1
and, y(t) = (t − 1)u(t − 1) − u(t − 1) + e−2(t−1)u(t − 1)
2
2
+ u(t − 1) − e−2(t−1)u(t − 1) − 2u(t − 4) + 2e−2(t−4)
ECEN 203
Page 9 of 10

1 mark for correctly identifying the partial fractions form
3 marks for finding the residuals
1 marks for converting to the time domain
* * * * * * * * * * * * * * *
ECEN 203
Page 10 of 10

Surname: . . . . . . . . . . . . . . . . . . . . . . . . .
First Name: . . . . . . . . . . . . . . . . . . .
Student ID: . . . . . . . . . . . . . . . . . . .
T E W H A R E W ¯A N A N G A O T E ¯U P O K O O T E I K A A M ¯A U I
VUW VICTORIA
U N I V E R S I T Y O F W E L L I N G T O N
ASSIGNMENT 1 – 2020
TRIMESTER 1
ECEN 203
ANALOGUE CIRCUITS
AND SYSTEMS
Time Allowed:
FIFTY MINUTES
CLOSED BOOK
Permitted materials: You may use a scientific calculator, but not a smartphone or other
similar device.
Instructions:
Attempt ALL THREE Questions
Write your answers on the question paper.
Full marks will be awarded for correct answers.
Partial marks may be awarded to incorrect answers if working is
shown.
The exam will be marked out of a total of 50 marks.
ECEN 203
Page 1 of 9

ECEN 203
Page 2 of 9

7
−6 v

1
86
=
−5
8
v2
50
Note that we don’t need to solve this for v1, just v2:

7
86

−

5 50
350 + 430
v2 =
=

7
−6
56 − 30

−

5
8
780
= 26
=30
(3 marks)
So v = 50 − 30 = 20 V
(2 marks)
ECEN 203
Page 6 of 9

* * * * * * * * * * * * * * *
ECEN 203
Page 9 of 9

Surname: . . . . . . . . . . . . . . . . . . . . . . . . .
First Name: . . . . . . . . . . . . . . . . . . .
Student ID: . . . . . . . . . . . . . . . . . . .
CLASS TEST 2 – 2020
TRIMESTER 1
ECEN 203
ANALOGUE CIRCUITS
AND SYSTEMS
Time Allowed:
THIRTY TWO HOURS
OPEN BOOK
Permitted materials: Any.
Instructions:
Attempt ALL Questions
Full marks will be awarded for correct answers.
Partial marks may be awarded to incorrect answers if working is
shown.
Along with your worked answers, please submit a text file with the
name answers.txt, which has one numerical answer per line.
The test will be marked out of a total of 50 marks.
ECEN 203
Page 1 of 9

Student ID: . . . . . . . . . . . . . . . . . . . .
ECEN 203
Page 2 of 9

Student ID: . . . . . . . . . . . . . . . . . . . .
1. DC Circuits
(5 marks)
40 Ω
+
28 V
−
7 A
46 Ω
v
Use the principle of superposition to find the voltage v (in volts).
We first include just the voltage source
40 Ω
+
28 V
−
46 Ω
v1
For which we obtain v1 = (28)(46) = 14.9767 Volts
40+46
Then we include just the current source
40 Ω
7 A
46 Ω
v
For this we have v2 = −740×46 = −149.7674
40+46
So the required voltage is v1 + v2 = 14.9767 + −149.7674 = −134.7907
2. DC circuits
(2 marks)
20 KΩ
+
22 V
−
50 KΩ
What is the voltage (in volts) between the terminals of the switch before it is
closed?
There is no current flowing in the resistors, so zero voltage drop across
them, so the voltage across the switch is simply the source voltage. i.e.,
Vswitch = 22.
ECEN 203
Page 3 of 9

Student ID: . . . . . . . . . . . . . . . . . . . .
3. Transient analysis
(20 marks)
t=0
R2=10 Ω
R1=60 Ω
+
+
V1=17 V
R3=90 Ω
L=19 mH
V
−
2=24 V
−
iL
The switch has been open for a long time. Then at time t = 0 the switch is
closed.
(a) (2 marks) What is the value (in amperes) of the inductor current iL at time
t = 0+
While the switch is open, the only relevant part of the circuit is the right
part:
R2=10 Ω
+
L=19 mH
V2=24 V −
iL
At steady state, the inductor behaves as a short circuit, and so the
current is V2 = 24 = 2.4000 A
R2
10
(b) (10 marks) What will be the final value (in amperes) of the inductor current?
(Hint: Start by using Norton’s and Thevenin’s theorems).
Once the switch is closed, we can analyse the behaviour by finding the
Thevenin equivalent of all parts of the circuit except for the inductor.
The left part has VT = V1R3 = 17×90 = 10.2000 V
R1+R3
60+90
and RT = R1kR3 = R1R3 = 60×90 = 36.0000 Ω
R1+R3
60+90
The Norton equivalent of this is a current source of I1 = VT/RT =
10.2000/36.0000 = 0.2833 A in parallel with RT = 36.0000 Ω
The Norton equivalent of the right part of the circuit is I2 = V2 = 24 =
R2
10
2.4000 A in parallel with R2 = 10 Ω
So the combined Norton equivalent is I3 = I1 + I2 = 0.2833 + 2.4000 =
2.6833 A in parallel with RN3 = R2kRT = R2RT = 7.8261 Ω, and the
R2+RT
Thevenin equivalent is the same resistance in series with V3 = 2.6833 ×
7.8261 = 21.0000 V
Once the inductor is again in it’s steady state (behaving as a short cir-
cuit), the current it carries will be I3 = 2.6833 A
ECEN 203
Page 4 of 9

Student ID: . . . . . . . . . . . . . . . . . . . .
(c) (3 marks) What is the time constant (in milliseconds) of the change in the
inductor current?
The time constant is τ = L/RN3 = 2.4278 ms
(d) (5 marks) What will be the inductor current (in amperes) at time t = 1 ms?
The formula for the current will be
iL(t) = Ifinal + (Iinitial − Ifinal)e− tτ
If we plug in the values we get
iL(1 ms) =2.6833 + (2.4000 − 2.6833)e− 1
2.4278
=2.4957
ECEN 203
Page 5 of 9

Student ID: . . . . . . . . . . . . . . . . . . . .
4. Capacitor Circuits
(5 marks)
C1=7µF
A
F
1µ
=
C4=6µF
C 5
C2=5µF
B
C3=9µF
What is the capacitance between terminals A and B (in microfarads)?
It’s probably easier to first express this if we first think of these as impedances:
Z = ((Z2 + Z3)kZ5 + Z1)kZ4
But because capacitors add in parallel and the reciprocals add when they
are in series, we have
C = (((C2kC3) + C5)kC1) + C4
Plugging in the numbers we get C = 8.6306 µF
5. AC Analysis
(7 marks)
3:1
I
120 Ω
+
Vi
74µF
61 mH
Vi is a sinusoidal AC voltage source having frequency 100 Hz, amplitude 5 V and
phase 0.
(a) (5 marks) What is the amplitude (in milliamperes) of the current I?
First note that the capacitor is symmetrical, so the position of the dot
makes no difference for this question.
The angular frequency is ω = 2 ∗ π ∗ f = 628.3185 rad s−1
The impedance of the capacitor is ZC = 1 = −21.5074j Ω
jωC
But this is transformed by the square the turns ratio to ZC2 = 32 ×
(−21.5074j) = −193.5668j
The impedance of the inductor is ZL = jωL = 38.3274j Ω
So the total impedance of the resistor, inductor and (transformed) ca-
pacitor is Z = ZL + ZC2 + R = 120.0000 + −155.2394j Ω
The current is Vi/Z = 5/(120.0000 + −155.2394j) = 0.0156 + 0.0202j A
This has magnitude 25.4826 mA
ECEN 203
Page 6 of 9

Student ID: . . . . . . . . . . . . . . . . . . . .
(b) (2 marks)
What is the phase (in degrees) of the current I in part (a)?
And phase 52.2960◦
ECEN 203
Page 7 of 9

Student ID: . . . . . . . . . . . . . . . . . . . .
6. Controlled sources
(6 marks)
R1=2 KΩ
R2=2 KΩ
+
+
1 V
−
R3=3 KΩ
−
800.0i3
i3
What is the voltage (in volts) of the controlled source?
Probably easiest to use the node voltage approach. Putting node v be-
tween R1 and R2, we apply KCL to get:
v − V1
v
v − ki
+
+
3 = 0
R1
R3
R2
we also can see that i3 = v . We combine these to get
R3
v − V1
v
v − kv/R
+
+
3 = 0
R1
R3
R2
1
1
1
k

V
v
+
+
−
=
R1
R3
R2
R2R3
R1
Plugging in the numbers we get v = 0.4167 V
We need to divide this by R3 to get i3 and then multiply to k to get the
voltage across the controlled source: 0.1111 V
7. Op-amps
(5 marks)
R2=52 K
R1=18 K
−
+
+
v1=7 mV −
R3=10 K
+
RL=84 K
v2=14 mV −
vo
Assuming the op-amp is ideal, what is the value of vo (in millivolts)?
ECEN 203
Page 8 of 9

Student ID: . . . . . . . . . . . . . . . . . . . .
First note that the value of R3 is irrelevant, because (almost) no current
flows through it. So v− = v2. Also, we have negative feedback, and hence
a virtual short, so v+ = v− = v2

We then have vo−v2 = v2−v1 which we can rearrange to give v
1 + R2 v
R
o =
2 −
2
R1
R1
R2 v
R
1
1
Substituting in the numbers, we get vo = 34.2222 mV
* * * * * * * * * * * * * * *
ECEN 203
Page 9 of 9

Surname: . . . . . . . . . . . . . . . . . . . . . . . . .
First Name: . . . . . . . . . . . . . . . . . . .
Student ID: . . . . . . . . . . . . . . . . . . .
ASSIGNMENT 1 – 2021
TRIMESTER 1
EEEN 203
ANALOGUE CIRCUITS
AND SYSTEMS
Time Allowed:
FIFTY MINUTES
CLOSED BOOK
Permitted materials: You may use a scientific calculator, but not a smartphone or other
similar device.
Instructions:
Attempt Questions 1–3.
Attempt Question 4 ONLY if you have time and are confident about
your answers to Questions 1–3.
Write your answers on the question paper.
Full marks will be awarded for correct answers.
Partial marks may be awarded to incorrect answers if working is
shown.
EEEN 203
Page 1 of 10

EEEN 203
Page 2 of 10

1. Mesh current analysis
(20 marks)
Find V by mesh current analysis. Note: you will need to come up with a strategy
to deal with the current source.
10 Ω
V
12 Ω
2 Ω
+
50 V
3 A
−
5 Ω
EEEN 203
Page 3 of 10

EEEN 203
Page 4 of 10

2. Node voltage analysis
(15 marks)
Find V by node voltage analysis. (Note: this is the same circuit as the previous
question).
10 Ω
V
12 Ω
2 Ω
+
50 V
3 A
−
5 Ω
EEEN 203
Page 5 of 10

EEEN 203
Page 6 of 10

3. Thevenin/Norton equivalents
(15 marks)
Find the Thevenin equivalent at terminals A&B:
+ −
A
4 Ω
10 V
2 Ω
6 Ω
38 A
+
130 V
−
B
EEEN 203
Page 7 of 10

EEEN 203
Page 8 of 10

4. Superposition
(15 marks)
Find V by using the principle of superposition. (Note: this is the same circuit as
questions 1 and 2).
10 Ω
V
12 Ω
2 Ω
+
50 V
3 A
−
5 Ω
EEEN 203
Page 9 of 10

* * * * * * * * * * * * * * *
EEEN 203
Page 10 of 10

Surname: . . . . . . . . . . . . . . . . . . . . . . . . .
First Name: . . . . . . . . . . . . . . . . . . .
Student ID: . . . . . . . . . . . . . . . . . . .
CLASS TEST 2 – 2021
TRIMESTER 1
ECEN 203
ANALOGUE CIRCUITS
AND SYSTEMS
Time Allowed:
FIFTY MINUTES
CLOSED BOOK
Permitted materials: Calculators.
Instructions:
Attempt ALL Questions
Full marks will be awarded for correct answers.
Partial marks may be awarded to incorrect answers if working is
shown.
The test will be marked out of a total of 50 marks.
ECEN 203
Page 1 of 7

Student ID: . . . . . . . . . . . . . . . . . . . .
ECEN 203
Page 2 of 7

Student ID: . . . . . . . . . . . . . . . . . . . .
1. Transient analysis
(24 marks)
t=0
R2=2 Ω
R1=3 Ω
+
+
V1=45 V
R3=6 Ω
L=7 mH
V
−
2=26 V
−
iL
The switch has been open for a long time. Then at time t = 0 the switch is
closed.
(a) (4 marks) What is the value (in amperes) of the inductor current iL at time
t = 0+ (Hint: Only part of the circuit is relevant for finding this value).
(b) (10 marks) What will be the final value (in amperes) of the inductor current?
(Hint: Start by using both Norton’s and Thevenin’s theorems).
ECEN 203
Page 3 of 7

Student ID: . . . . . . . . . . . . . . . . . . . .
(c) (4 marks) What is the time constant (in milliseconds) of the change in the
inductor current?
(d) (6 marks) What will be the inductor current (in amperes) at time t = 1 ms?
ECEN 203
Page 4 of 7

Student ID: . . . . . . . . . . . . . . . . . . . .
2. Capacitor Circuits
(6 marks)
C1=8µF
A
F
2µ
=
C4=5µF
C 5
C2=2µF
B
C3=1µF
What is the capacitance between terminals A and B (in microfarads)?
ECEN 203
Page 5 of 7

Student ID: . . . . . . . . . . . . . . . . . . . .
3. AC Analysis
(10 marks)
1:1
I
586 Ω
+
Vi
19µF
36 mH
Vi is a sinusoidal AC voltage source having frequency 110 Hz, amplitude 90 V
and phase 0.
(a) (5 marks) What is the amplitude (in milliamperes) of the current I?
(b) (5 marks)
What is the phase (in degrees) of the current I in part (a)?
ECEN 203
Page 6 of 7

Student ID: . . . . . . . . . . . . . . . . . . . .
4. Op-amps
(10 marks)
R2=30 K
R1=62 K
−
+
+
v1=31 mV −
R3=6 K
+
RL=38 K
v2=93 mV −
vo
Assuming the op-amp is ideal, what is the value of vo (in millivolts)?
* * * * * * * * * * * * * * *
ECEN 203
Page 7 of 7

2022 EEEN 203 Circuit Analysis

Test 4
Total marks: 85 (Worth 16%)

Below are the problems for this assignment. Do your calculation as needed and then put
your final answers as well any discussion or plots in the spaces required. Submit this
document with a filename:
EEEN203_Test4_2022_”your surname”-“your initial” on the Wiki submission system no later
than Saturday 13 June by 8.00 pm.

Name: ……………………………………………………………………………………………..
Student Number:……………………………………………………………………………….

Q1. An ideal transformer is rated at 2400/220 V, 9.6 kVA, and has 50 turns on the
secondary side. Calculate: (a) the turns ratio, (b) the number of turns on the primary
side, and (c) the current ratings for the primary and secondary windings.

Answer

(10)

Q2. A distribution transformer is used to supply a household as in Figure. The load
consists of eight 100-W bulbs, a 350-W TV, and a 15-kW kitchen range. If the
secondary side of the transformer has 72 turns, calculate: (a) the number of turns of
the primary winding, and (b) the current Ip in the primary winding.

Answer

(10)

Q3. The voltage
is applied to a 0.1-H inductor. Find the steady-
state current through the inductor.
Answer

(6)

Q4. Find Real, Reactive and Apparent power for following load configuration.

Answer

(15)

Q5. When connected to a 240-V (rms), 50-Hz power line, a load absorbs 4 kW at a
lagging power factor of 0.8. Find the value of capacitance necessary to raise the pf
to 0.95.
Answer

(20)

Q6. Determine the total average, reactive, and complex powers at the source and at
the load for the following figure.

Answer

(24)

2022/23 Test 1-3 Sample

2023 EEEN 203 Circuit Analysis
Test 4
Total marks: 90 (Worth 17%)
Below are seven problems for this test. Do your calculations as needed and then put your
answers (both working and final) as well any discussion or plots in a new document.
Submit your answers with a filename:
EEEN203_Test4_2023_” your surname”-“your initial” on the Wiki submission system no
later than Monday 12 June by 8.00 pm
Note: You are allowed to work in a team and share knowledge but let me know who you worked
with so that I don’t penalise if the two solutions are similar.
Name: ……………………………………………………………………………………………..
Student Number:…………………………………………………………

Q1. An ideal transformer is rated at 2400/220 V, 9.6 kVA, and has 50 turns on the
secondary side. Calculate: (a) the turns ratio, (b) the number of turns on the primary
side, and (c) the current ratings for the primary and secondary windings.
[10 Marks]
Q2. A distribution transformer is used to supply a household as in Figure. The load
consists of twenty-five 100-W bulbs, a 350-W TV, and a 3.5-kW kitchen range. If the
secondary side of the transformer has 72 turns, calculate: (a) the number of turns of
the primary winding, and (b) the current Ip in the primary winding.
[10 Marks]
25 Lamps
Q3. Find Real, Reactive and Apparent Power for the following load configuration.
[15 Marks]

Q4. When connected to a 240-V (rms), 50-Hz power line, a load absorbs 4 kW at a
lagging power factor of 0.6. Find the value of capacitance necessary to raise the
power factor to 0.95.

[15 Marks]
Q5. A load Z draws 15 kVA at a power factor of 0.7 lagging from a 240-V rms
sinusoidal source. Calculate: (a) the average and reactive powers delivered to the
load, (b) the peak current, and (c) the load impedance.

[10 Marks]
Q6. A 1200-kW load supplied at 13 kV (rms) operates 410 hours a month at 76 percent
power factor. Calculate the average cost per month based on this simplified tariff:
•
Energy charge: 8 cents per kWh
•
Power-factor penalty: 0.2 percent of energy charge for every 0.01 that
pf falls below 0.85.
•
Power-factor credit: 0.24 percent of energy charge for every 0.01 that
pf exceeds 0.85.
[10 Marks]
Q7. Determine the total average, reactive, and complex powers at the source and at
the load for the following figure.

[20 Marks]
----------------------------------------------------------------------------------

TEST 1
TRIMESTER 1 2024
EEEN203 CIRCUIT ANALYSIS
CRN 33055
Time allowed
: One Hour
Instructions
: Write your answers in the answer booklet provided.
  To get a full score, you must provide detail steps.
  Not all questions have equal marks.
  Dictionary and non-programmable calculator are permitted.
  There are FIVE questions. Answer ALL questions.
Question 1:
Nodal Analysis and Thevenin Equivalent.
20 Marks
Question 2:
Inductor Circuit.
8 Marks
Question 3:
AC Circuit Parameters.
8 Marks
Question 4:
Filter Circuit.
6 Marks
Question 5:
Op Amp Amplifiers.
8 Marks

THIS PAGE IS INTENTIONALLY LEFT BLANK.

Section A. DC Circuit Analysis
[20 Marks]
1. Use the suggested circuit analysis methods to calculate the stated unknown
parameters of the circuit.
a. Applying node-voltage method, find 𝑣𝑜 in the circuit given in the figure below.
[6 marks]
b. Find the Thevenin equivalent of the circuit given in the figure below.
[14 Marks]
Section B. AC Circuit Analysis
[22 Marks]
2. The voltage at the terminals of the 750 μH inductor circuit in figure (a) is shown in
the figure (b). The inductor current 𝑖(𝑡) is known to be zero for 𝑡 < 0. Derive the
expressions for 𝑖(𝑡) when 𝑡 > 0.

[8 marks]
EEEN203
3

3. Find the steady-state expression for 𝑖𝑜(𝑡) in the circuit in the figure below if 𝑣𝑠 =
80 cos 2000𝑡  V.
[8 marks]
4. Using the filter circuit shown in the figure below, find the cut-off frequency (in Hertz)
of the filter. Describe briefly the characteristic of the filter.

[6 Marks]
Section C. Operational Amplifiers
[8 Marks]
5. Considering ideal op amp used in the circuits below, the voltage at the inverting
pin is equal to the voltage at the noninverting pin (𝑣𝑛 = 𝑣𝑝). Find the current through
5 kΩ resistor 𝑖𝑜. If a practical op amp is used instead in the circuit, describe briefly
its gain  and  input/output  impedance  characteristics and how these  influence the
circuit.



[8 marks]
EEEN203
4

USEFUL FORMULAE
A. Basic Electricals
For Capacitor:
1
𝜀
𝐸 = 𝑄𝑉
0𝜀𝑟𝐴

𝑄 = 𝐶𝑉
𝐶 =
𝐶
2
𝑑
𝑇 = 𝐶1 + 𝐶2 + ⋯
𝜏 = 𝑅𝐶
1
1
1
=
+
+ ⋯
𝐶𝑇
𝐶1
𝐶2
For Inductor:
Δ𝐼
Δ𝜙
1
𝐿
𝜙 = 𝐵𝐴
ℇ = −𝐿 (
)
ℇ = −
𝐸 = 𝐿𝐼2
𝜏 =
𝑑𝑡
𝑑𝑡
2
𝑅
𝑁𝑝
𝑉
= 𝑝
𝑁𝑠
𝑉𝑠
Electrical Parameters:
1
𝐼 = 𝐼𝑚𝑎𝑥 sin 𝜔𝑡
𝑉 = 𝑉𝑚𝑎𝑥 sin 𝜔𝑡
𝐼𝑚𝑎𝑥 = √2𝐼𝑟𝑚𝑠
𝑉𝑚𝑎𝑥 = √2𝑉𝑟𝑚𝑠
𝑋𝑐 = 𝜔𝐶
1
𝑋𝐿 = 𝜔𝐿
𝑉 = 𝐼𝑍
𝜔 = 2𝜋𝑓
𝑓0 = 2𝜋√𝐿𝐶
B. Circuit Theorems
Kirchhoff Voltage Law (KVL)
𝑛
∑ 𝑉𝑖 = 𝑉1 + 𝑉2 + 𝑉3 + ⋯ = 0
𝑖=1
Kirchhoff Current Law (KCL)
𝑛
∑ 𝐼𝑖 = 𝐼1 + 𝐼2 + 𝐼3 + ⋯ = 0
𝑖=1
C. Devices
Capacitor
𝑑𝑣
𝑖(𝑡) = 𝐶 (
)
𝑑𝑡
Inductor
𝑑𝑖
𝑣(𝑡) = 𝐿 (
)
𝑑𝑡
D. Modelling of Electrical Systems
Voltage
Current
Resistors
𝑣(𝑡) = 𝑖(𝑡)𝑅
𝑣(𝑡)
𝑖(𝑡) = 𝑅
Capacitors
1
𝑡
𝑑𝑣(𝑡)
𝑣(𝑡) =
∫ 𝑖(𝜏) 𝑑𝜏
𝑖(𝑡) = 𝐶 [
]
𝐶
𝑑𝑡
0
5

Inductors
𝑑𝑖(𝑡)
1 𝑡
𝑣(𝑡) = 𝐿 [
]
𝑖(𝑡) = ∫ 𝑣(𝜏) 𝑑𝜏
𝑑𝑡
𝐿 0
E. Trigonometric Identities
𝜋
𝜋
sin 𝜃 = ± cos (𝜃 ∓ )
cos 𝜃 = ± sin (𝜃 ± )
2
2
sin 𝜃 = − sin(𝜃 ± 𝜋)
cos 𝜃 = − cos(𝜃 ± 𝜋)
sin(−𝜃) = − sin 𝜃
cos(−𝜃) = cos 𝜃
sin(𝜃 ± 𝜙) = sin 𝜃 cos 𝜙 ± cos 𝜃 sin 𝜙
sin(2𝜃) = 2 sin 𝜃 cos 𝜃
cos(𝜃 ± 𝜙) = cos 𝜃 cos 𝜙 ∓ sin 𝜃 sin 𝜙
2 sin 𝜃 sin 𝜙 = cos(𝜃 − 𝜙) − cos(𝜃 + 𝜙)
cos(2𝜃) = cos2 𝜃 − sin2 𝜃
  = 2 cos2(𝜃) − 1
  = 1 − 2 sin(𝜃)
F. Time and Phasor Transformation
Time [𝒙(𝒕)]
Phasor [𝐗]
𝐴 cos 𝜔𝑡
𝐴
𝐴 cos(𝜔𝑡 + 𝜙)
𝐴𝑒𝑗𝜙
−𝐴 cos(𝜔𝑡 + 𝜙)
𝐴𝑒𝑗(𝜙±𝜋)
𝐴 sin 𝜔𝑡
𝐴𝑒−𝑗𝜋/2 = −𝑗𝐴
𝐴 sin(𝜔𝑡 + 𝜋)
𝐴𝑒𝑗(𝜙−𝜋/2)
−𝐴 sin(𝜔𝑡 + 𝜋)
𝐴𝑒𝑗(𝜙+𝜋/2)
𝑑
𝑗𝜔𝐗
[𝑥(𝑡)]
𝑑𝑡
𝑑
𝑗𝜔𝐴𝑒𝑗𝜙
[𝐴 cos(𝜔𝑡 + 𝜙)]
𝑑𝑡
1
∫ 𝑥(𝑡) 𝑑𝑡
𝐗
𝑗𝜔
1
∫ 𝐴 cos(𝜔𝑡 + 𝜙) 𝑑𝑡
𝐴𝑒𝑗𝜙
𝑗𝜔
G. Complex Number Relationship
Euler Identity:
𝑒𝑗𝜃 = cos 𝜃 + 𝑗 sin 𝜃
𝑒𝑗𝜃−𝑒−𝑗𝜃
𝑒𝑗𝜃+𝑒−𝑗𝜃
Where: sin 𝜃 =
and    cos 𝜃 =
2𝑗
2
Vector to Complex Number:
𝐳 = 𝑥 + 𝑗𝑦 = |𝐳|𝑒𝑗𝜃
𝐳∗ = 𝑥 − 𝑗𝑦 = |𝐳|𝑒−𝑗𝜃
𝑥 = ℜe(𝑧) = |𝐳| cos 𝜃
|𝐳| = √𝐳𝐳∗ = √𝑥2 + 𝑦2
𝑦 = ℑm(𝑧) = |𝐳| sin 𝜃
𝜃 = arctan(𝑦/𝑥)
𝐳𝑛 = |𝐳|𝑛𝑒𝑗𝑛𝜃
𝐳1/2 = ±|𝐳|1/2𝒆𝑗𝜃/2
Operations of Vector:
For 𝐳𝟏 = 𝑥1 + 𝑗𝑦1    and    𝐳𝟐 = 𝑥2 + 𝑗𝑦2
𝐳𝟏𝐳𝟐 = |𝐳𝟏||𝐳𝟐|𝑒𝑗(𝜃1+𝜃2)
6

𝐳
|
𝟏 = 𝐳𝟐    iff   𝑥1 = 𝑥2      and     𝑦1 = 𝑦2
𝐳𝟏
𝐳
=
𝟏| 𝑒𝑗(𝜃1−𝜃2)
𝐳𝟏 + 𝐳𝟐 = (𝑥1 + 𝑥2) + 𝑗(𝑦1 + 𝑦2)
𝐳𝟐
|𝐳𝟐|
Arithmetic of Complex Number:
−1 = 𝑒𝑗𝜋 = 𝑒−𝑗𝜋 = 1∠ ± 180°
(1 + 𝑗)
√𝑗 = ±𝑒𝑗𝜋/4 = ±
𝑗 = 𝑒𝑗𝜋/2 = 1∠90°
√2
𝑗𝜋
(1 − 𝑗)
−𝑗 = 𝑒− 2 = 1∠ − 90°
√−𝑗 = ±𝑒𝑗𝜋/4 = ±
√2
H. Op Amp
Inverting Amplifier
𝑅𝑓
𝐴𝑣 =   − 𝑅𝑖𝑛
Non-inverting Amplifier
𝑅
𝐴
2
𝑣 =  1 + 𝑅1
Summing Amplifier
𝑅𝑓
𝑉𝑜𝑢𝑡 = −
(𝑉
𝑅
1 + 𝑉2 + 𝑉3 + ⋯ )
𝑖𝑛
Differentiator
𝑑𝑉
𝑉
𝑖𝑛
𝑜𝑢𝑡 = −𝑅𝐶 (
)
𝑑𝑡
Integrator
1
𝑉𝑜𝑢𝑡 = −
∫ 𝑉
𝑅𝐶
𝑖𝑛 𝑑𝑡
Difference Amplifier
𝑅
𝑅
𝑉
4(𝑅1 + 𝑅2)
4
𝑜𝑢𝑡 =
𝑉
) 𝑉
𝑅
2 − (
1
1(𝑅3 + 𝑅4)
𝑅3
7

2024 EEEN 203 Circuit Analysis
Test 2
Total marks: 100 (Worth 24%)
Below are seven problems for this test. Do your calculations as needed and then put your
answers (both working and final) as well any discussion or plots in a new document.
Submit your answers with a filename:
EEEN203_Test2_2024_” your surname”-“your initial” on the Wiki submission system no
later than Friday 14 June.
Note: You are allowed to work in a team and share knowledge but let me know who you worked
with so that I don’t penalise if the two solutions are similar.
Name: ……………………………………………………………………………………………..
Student Number:…………………………………………………………

Q1. An ideal transformer is rated at 2400/220 V, 9.6 kVA, and has 50 turns on the
secondary side. Calculate: (a) the turns ratio, (b) the number of turns on the primary
side, and (c) the current ratings for the primary and secondary windings.
[10 Marks]
Q2. A distribution transformer is used to supply a household as in Figure. The load
consists of twenty-five 100-W bulbs, a 250-W Television, and a 3.5-kW kitchen
range. If the secondary side of the transformer has 72 turns, calculate: (a) the
number of turns of the primary winding, and (b) the current Ip in the primary winding.
[10 Marks]
25 Lamps
Q3. Find Real, Reactive and Apparent Power for the fol owing load configuration.
[20 Marks]

Q4. When connected to a 240-V (rms), 50-Hz power line, a load absorbs 4 kW at a
lagging power factor of 0.5. Find the value of capacitance necessary to raise the
power factor to 0.95.

[15 Marks]
Q5. A load Z draws 15 kVA at a power factor of 0.6 lagging from a 240-V rms
sinusoidal source. Calculate: (a) the average and reactive powers delivered to the
load, (b) the peak current, and (c) the load impedance.

[15 Marks]
Q6. A 1200-kW load supplied at 13 kV (rms) operates 410 hours a month at 76 percent
power factor. Calculate the average cost per month based on this simplified tariff:
•
Energy charge: 8 cents per kWh
•
Power-factor penalty: 0.2 percent of energy charge for every 0.01 that
pf fal s below 0.85.
•
Power-factor credit: 0.24 percent of energy charge for every 0.01 that
pf exceeds 0.85.
[10 Marks]
Q7. Determine the total average, reactive, and complex powers at the source and at
the load for the fol owing figure.

[20 Marks]
----------------------------------------------------------------------------------

Document Outline