UNIVERSITI TUNKU ABDUL RAHMAN
Answer Guideline for Chapter 3 Tutorial
Q1.
(i) Consider voltage source E1,
15 || 10 = 6,
9 || 6 = 3.6,
42
IT =
= 1.94 A
18 + 3.6
Using current divider,
9
9
I1 =
I T = ( 1.94 ) = 1.17 A
9+6
15
15
15
I '=
I1 = ( 1.17 ) = 0.7A
15 +10
25
(ii) Consider voltage source E2,
9 || 18 = 6 ,
24
IT =
= 2A
6+6
1 5 || 10 = 6 ,
U sing current divider,
15
15
I ''=
IT =
( 2 ) = 1 .2 A
15 + 10
25
I 10 = I ' + I ' ' = 0 . 7 + 1 . 2 = 1 . 9 A
Q2.
(i) Consider current source 2A,
2A
5
6
2A
+ v01
12
io 5
+ v01
5
Page 1 of 8
6||3 = 2 , 4||12 = 3
i0 = 5/5 = 1, v01 = 5 i01 = 5 V
(ii) Consider voltage source 12V,
6
12V
6
+
+ v02
12
12V
+ v02
3
v1
3||8 = 24/11, v1 = [(24/11)/(6 + 24/11)]12 = 16/5
v02 = (5/8) v1 = (5/8)(16/5) = 2 V
(iii) Consider current source 2A,
5
6
+ v03
12
3
19V
+ v03
v2
12
+
19V
7||12 = (84/19) , v2 = [(84/19)/(4 + 84/19)]19 = 9.975 V
v03 = (-5/7) v2 = -7.125 V
vo = v01 + v02 + v03 = 5 + 2 7.125 = -125 mV
Q3.
Using source transformations,
2
18 V
12 V
10 V
10
Page 2 of 8
2A
10
3A
5
2A
3A
3.333
3.333
10 V
b
+
Norton Equivalent Circuit
Thevenin Equivalent Circuit
Q4.
To find RTh,
a
6
6
2
2
6
b
(a)
a
18
1.8
a
2
2
18
1.8
18
b
(b)
RT
1.8
(c)
R = 2||18 = 1.8 , RTh = (1.8 + 1.8) || 1.8 = 1.2
Page 3 of 8
To get VTh, apply mesh analysis,
2
a
6
12V
i3
12V
+
+
VTh
6
2
i1
i2
12V
b
(d)
Mesh1:
Mesh2:
Mesh3:
-12 12 + 14i1 6i2 6i3 = 0,
7 i1 3 i2 3i3 = 12
(1)
12 + 12 + 14 i2 6 i1 6 i3 = 0
-3 i1 + 7 i2 3 i3 = -12
(2)
14 i3 6 i1 6 i2 = 0
-3 i1 3 i2 + 7 i3 = 0
(3)
7 3 3 i 1 12
3 7 3 i = 12
3 3 7 i 3 0
7
= 3
3
3
7
3
3
3 = 100
7
7 12
3
2 = 3 12 3 = 120
3
0
7
i2 = /2 = -120/100 = -1.2 A
VTh = 12 + 2i2 = 9.6 V, and IN = VTh/RTh = 8 A
Page 4 of 8
Q5.
To find RTh,
24V
4
V1
RTh
2A
VTh
c
(a)
(b)
RTh = 5||(2 + 3 + 4) = 3.21
To get VTh, at the node V1,
V 1 24
V 1 0
2
=0
234
5
V Th = V 1= 15 V
Q6.
To obtain RN,
6
6
Isc = IN
2A
1
(a)
+
12V
(b)
RN = 6 + 4 = 10
To obtain IN, use mesh analysis:
Mesh1:
i1 = 2 A
Mesh2:
10i2 4i1 + 12 = 0
IN = i2 = -0.4 A
i
IN = 0.4A
RN = 10
5
4A
(c)
i = [10/(10 + 5)] (4 0.4) = 2.4 A
Page 5 of 8
Q7.
4
12V
+
+
RTh
8V
VTh
20V
(a)
(b)
(a)
To obtain RTh and VTh
RTh = 2 + 4 + 6 = 12
i(12)-VTh + 12 + 8 + 20 = 0, or VTh = 40 V(because i = 0)
(b)
iL = VTh/(RTh + R) = 40/(12 + 8) = 2A
(c)
For maximum power transfer,
RL = RTh = 12
(d)
P = VTh2/(4RTh) = (40)2/(4x12) = 33.33 W.
(a)
For maximum power transfer, RL = RTh
To determine RTh,
RTh = 4 || 4 = 2 ohms
RL = RTh = 2 ohms
(b)
To determine VTh, through Superposition,
(i)
Consider voltage source 24V,
Q8.
V Th ' = 24
4 4
= 12V
(ii) Consider current source 5A,
V Th ''= IR T = 5 4 4 = 10V
V Th= V Th '
Th ''=
22 V
V 2Th
22 V 2
P=
=
= 60 .5W
4R Th 4 2
Page 6 of 8
Q9.
(a)
3
32 V
RTh
VTh
RL
12 V
b
Source transformation: v = 8(4) = 32 V
Mesh: 8 i 32 12 = 0
i = 5 .5 A
For VTh (Outer loop from b to a): VTh = 0 + 32 5(5.5) = 4.5 V or
(inner loop from b to a): VTh = 0 12 + 3(5.5) = 4.5 V
For RTh: RTh = (1+4) // 3 = 1.875
4. 5
i = R VR = 1. 875
15 = 0 . 267
Th
Th
(b)
ia
iL1
Turn off V source:
1
8A
4
15
iL2
(8 ) = 4.267 A
4 + [1 + (3 // 15 )]
3
=
(i a ) = 0.711 A
3 + 15
Turn off I source:
ia =
i L1
12
= 1 .778 A
3 + [(1 + 4 ) // 15 ]
5
=
( ib ) = 0 .444 A
5 + 15
ib =
15
12 V
iL2
ib
i = i i
L
(c)
= 0 . 267 A
(i)
R L = RTh = 1. 875
(ii)
P max =
2
V Th
= 2.7 W
4RTh
VTh
RTh
RL
Page 7 of 8
Q10. (i)
For 10 V source:
V 01 =
10 = 2 . 22 V
5 2
4 4
For 2 A source:
V 02 =
2 5 = 2 . 22 V
5 2
4 4
For 5 V source:
2
5
V 03=
= 0. 55 V
5 2
4 4 2
V = V V V = 3. 89 V
0
(ii)
01
03
03
For node V 1 :
V 1 10
V 1 V 2
2
=0
5
2
7V1 5V 2= 40
For node V 2 :
V 2 V 1 V 2 0 V 2 5
=0
2
4
4
2V1 4V 2 = 5
Using Cramers rule:
185
V 1=
= 10 . 28
18
115
V 2=
= 6 . 39
18
V 0 = V 1 V 2= 3. 89 V
Page 8 of 8
