2
SEEE 1013 / SKEE 1013
Q.1
(a)
For the circuit shown in Figure Q.1(a), by using Basic Laws of electrical circuits,
determine:(i)
the equivalent resistance between the terminals a-b, Rab.
(5 marks)
(ii)
the voltage drop across the resistor 6 , 𝑉𝑥 .
(3 marks)
(iii)
the power absorbed by the 6 Ω resistor.
(2 marks)
8
a
8
+
6
9V
80
b
Vx
40
12
Figure Q.1(a)
(b)
A linear network is supplied by a sinusoidal voltage, 𝑣(𝑡) with its input current, 𝑖(𝑡). Figure
Q.1(b) shows the voltage and current waveforms of the aforementioned network.
(i)
Express the voltage, 𝑣(𝑡) and the current, 𝑖(𝑡) in time domain.
(6 marks)
(ii)
State whether the current, 𝑖(𝑡) is ‘lagging’ or ‘leading’ the voltage, 𝑣(𝑡).
(1 mark)
3
SEEE 1013 / SKEE 1013
15 V/i
10
v(t)
5
i(t)
0
-0.01
-5
0
0.01
0.02
0.03
0.04
0.05
second (s)
-10
-15
Figure Q.1(b)
(c)
In Figure Q.1(c), the voltage, 𝑉𝑠 is given by 15 ∠5° 𝑉. By using the Basic Laws, determine:(i)
The current, 𝐼𝑠 .
(6.5 marks)
(ii)
The voltage across the j30  inductive reactance, 𝑉𝑧 .
(1.5 marks)
j30
-
Vz
- Vs +
-
+
20
+
Is
-j10
Figure Q.1(c)
5
4
SEEE 1013 / SKEE 1013
Q.2
(a)
Considering the circuit in Figure Q.2(a), use the mesh analysis OR nodal analysis to find
:(i)
The voltage drop across the 10 Ω resistor, 𝑉𝑥 .
(11 marks)
(ii)
The power absorbed by the 14 Ω resistor.
(2 marks)
5A
ia
Va
5V
5
10
Vb
ib
-
Vx
4Vx
ic
Vc
+
14
6
Figure Q.2(a)
(b)
For the circuit shown in Figure Q.2(b),
(i)
Find the Norton or Thevenin equivalent circuit with respect to the terminals a-b.
(9 marks)
(ii)
Determine the value of RL that yields the maximum power transfer at terminals a-b.
(1 mark)
(iii)
Find the maximum power dissipated across the resistor, RL.
(2 marks)
5
SEEE 1013 / SKEE 1013
5V
6
18
9
3
a
20 V
RL
b
Figure Q.2(b)
Q.3
(a) Figure Q.3(a) shows a circuit in which the switch has been at position ‘a’ for a long time.
At t = 0 s, the switch moves to position ‘b’, and the resistor, R is 10 .
(i)
Find the initial and final value of capacitor voltage, vC (0) and vC (∞) respectively
(4 marks)
(ii)
Find the time constant,
 and the expression of vC (t) for circuit at t  0 s
(5 marks)
(iii)
Find the new value of the resistor R such that the response of vC(t) is twice as fast as the
previous value.
(4 marks)
6Ω
t=0s
a
+
10 V
vC (t)
R
b
10 mF
5A
12 Ω
_
Figure Q.3(a)
2Ω
6
SEEE 1013 / SKEE 1013
(b) The switch in Figure Q.3(b) has been closed for a long time and at t = 0 s, the switch is
open. Then, at t ≥ 0 s, the circuit is required to operate in ‘critically damped’ response.
(i)
Find the value of the resistor R.
(3 marks)
(ii)
Find the values of vC (0−), vC (0+),iL (0−), iL (0+),
dvC (0 + )
and vC (∞)
dt
(7 marks)
(iii)
Find the expression of vC (t) in terms of A1 and A2 for t  0 s. [DO NOT SOLVE for A1 and
A2].
(2 marks)
R
t=0s
1A
200 Ω
1 mH
iL (t)
25 nF
+
vC (t)
−
Figure Q.3(b)
80 V
7
SEEE 1013 / SKEE 1013
Q.4
(a) Use Nodal Analysis to find voltage Vx and current Iy in Figure Q.4(a)
(13 marks)
V1
Iy
60 Ω
−j70 Ω
+ Vx −
V2
600 V
V3
80 Ω
130 A
j170 Ω
50 Ω
Figure Q.4(a)
(b) Find v1(t) in Figure Q.4(b) by using appropriate method of analysis and theorem. The value of the
sources are as follows
vac (t) = 100 cos (300t) V
iac (t) = 2 cos (500t) A
vdc = 50 V
(12 marks)
vdc
+
v1 (t)
33 Ω
15 Ω
−
100 F
vac (t)
50 mH
Figure Q.4(b)
iac (t)
8
SEEE 1013 / SKEE 1013
APPENDIX 1
Trigonometric Identities
Integrals
1) sin  = cos (90-) = - cos (90+)
1)
2) sin  = sin(180-)
2)
3) sin (-) = -sin 
3)

4) cos (-) = cos 
4)
 sin t dt = −cos t
5) tan  = -tan(180-)
5)
 sin t dt =
6) sin (   ) = sincos  cossin
6)
 sin t dt
7) cos(   ) = coscos  sinsin
7)
 cos t dt = sin t
8)
 cos t dt =
9)
 cos t dt = sin t
8) tan(   ) =
9) sin 2  =
tan   tan 
1  tan  tan 
(1 − cos2 )
10) cos 2 =
2
(1 + cos2 )
2
11) sin 2 = 2sin  cos 
12) cos 2 = cos2  - sin2 
13) sin2  + cos2  = 1
14) sin (t + /2) = cos t
15) cos(t − /2) = sin t
Note:
x(t ) = A1e s1t + A2 e s2t + x()
x(t ) = ( A1 + A2t )e −t + x()
x(t ) = ( A1 cos d t + A2 sin d t )e −t + x()
t
 dt = t
0
t
0
t
 10dt = 10t
0
t
0
tdt =
t2
2
t
0
t
0
t
0
t
0
t
- cos t
0

t
0
= −cos t 0 t
t
0
t
0
t
sin t
0

t
0
t
0
t
0