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UNIVERSITI TUNKU ABDUL RAHMAN

Answer Guideline for Chapter 3 Tutorial

Q1. (i) Consider voltage source E1,

0.7A1.1725

15

1015

15

1.171.9415

9

69

9

divider,current sing

1.943.618

42

3.669 61015

1

1

=)(=I+

=' I

A=)(=I+

=I

U

A=+

=I

Ω,=||Ω,=||

T

T

(ii) Consider voltage source E2,

A2.1)2(25

15

1015

15''

divider,current sing

A266

42

, 610||51 , 618||9

T

T

==+

=

=+

=

Ω=Ω=

I I

I

U

A9.12.17.0'' ' =+=+=∴ Ω I II 10

Q2. (i) Consider current source 2A,

+ v01 −−−−

5 ΩΩΩΩ

12 ΩΩΩΩ

6ΩΩΩΩ

4 ΩΩΩΩ

3ΩΩΩΩ

2A 2A

io

+ v01 −−−−

5 ΩΩΩΩ

5 ΩΩΩΩ

Page 2 of 8

6||3 = 2 Ω , 4||12 = 3 Ω

i0 = 5/5 = 1, v01 = 5 i01 = 5 V

(ii) Consider voltage source 12V,

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,

7||12 = (84/19) Ω , 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,

6 ΩΩΩΩ

3 ΩΩΩΩ 12V

+

− 12 ΩΩΩΩ

4 ΩΩΩΩ

+ v02 −−−−

5 ΩΩΩΩ

12V

+

−

+

v1

−−−−

6 ΩΩΩΩ

3 ΩΩΩΩ 3 ΩΩΩΩ

+ v02 −−−−

5 ΩΩΩΩ

3 ΩΩΩΩ 12 ΩΩΩΩ

4 ΩΩΩΩ

+ v03 −−−−

5 ΩΩΩΩ

6ΩΩΩΩ 19V

+

− 19V

+

−

+

v2

−−−−

12 ΩΩΩΩ

+ v03 −−−−

5 ΩΩΩΩ

2ΩΩΩΩ

4 ΩΩΩΩ

2

5 ΩΩΩΩ

+ −

10 V

+ −

12 V

+ −

18 V 10 ΩΩΩΩ b a

Page 3 of 8

Q4. To find RTh,

R = 2||18 = 1.8 Ω, RTh = (1.8 + 1.8) || 1.8 = 1.2 Ω

b a

3A

5 ΩΩΩΩ

2A

10 ΩΩΩΩ

2A

Norton Equivalent Circuit

3A

b a 3.333ΩΩΩΩ b a 3.333ΩΩΩΩ

+ −

10 V

Thevenin Equivalent Circuit

6 ΩΩΩΩ

(a)

2 ΩΩΩΩ

b

a

6 ΩΩΩΩ

6 ΩΩΩΩ

2 ΩΩΩΩ

2 ΩΩΩΩ

(b)

2 ΩΩΩΩ

b

a

2 ΩΩΩΩ

2 ΩΩΩΩ

18 ΩΩΩΩ 18 ΩΩΩΩ RT

(c)

1.8 ΩΩΩΩ

b

a

1.8 ΩΩΩΩ

1.8 ΩΩΩΩ 18 ΩΩΩΩ

Page 4 of 8

To get VTh, apply mesh analysis,

Mesh1: -12 – 12 + 14i1 – 6i2 – 6i3 = 0,

7 i1 – 3 i2 – 3i3 = 12 (1)

Mesh2: 12 + 12 + 14 i2 – 6 i1 – 6 i3 = 0

-3 i1 + 7 i2 – 3 i3 = -12 (2)

Mesh3: 14 i3 – 6 i1 – 6 i2 = 0

-3 i1 – 3 i2 + 7 i3 = 0 (3)

−=

−−

−−

−−

0

12

12

i

i

i

733

373

337

3

2

1

100

733

373

337

=

−−

−−

−−

=∆ , ∆2=∣7 12 − 3

− 3 − 12 − 3

− 3 0 7

∣= − 120

i2 = ∆/∆2 = -120/100 = -1.2 A

VTh = 12 + 2i2 = 9.6 V, and IN = VTh/RTh = 8 A

12V

+

−

−

+

VTh

12V

6 ΩΩΩΩ

(d)

2 ΩΩΩΩ

b

a

6 ΩΩΩΩ

6 ΩΩΩΩ

2 ΩΩΩΩ i1

i2

i3

2 ΩΩΩΩ

12V

+

−

+

Page 5 of 8

Q5. To find RTh,

RTh = 5||(2 + 3 + 4) = 3.21 Ω

To get VTh, at the node V1, V

1− 24

234− 2

V1− 0

5= 0

VTh

= V 1= 15 V

Q6. To obtain RN,

RN = 6 + 4 = 10 Ω

To obtain IN, use mesh analysis:

Mesh1: i1 = 2 A

Mesh2: 10i2 – 4i1 + 12 = 0

IN = i2 = -0.4 A

i = [10/(10 + 5)] (4 – 0.4) = 2.4 A

b

c

(a)

RTh

4 ΩΩΩΩ

1ΩΩΩΩ 3ΩΩΩΩ

5 ΩΩΩΩ

2 ΩΩΩΩ V1

(b)

VTh

+

4 ΩΩΩΩ

1ΩΩΩΩ 3ΩΩΩΩ

5 ΩΩΩΩ

b

c

2A

2 ΩΩΩΩ

24V

+

−

6 ΩΩΩΩ

4 ΩΩΩΩ

(a)

12V

+

−

Isc = IN

6 ΩΩΩΩ

4 ΩΩΩΩ

(b)

2A

i

RN = 10

Ω

(c)

IN = 0.4A 5 Ω

4A

1 2

Page 6 of 8

Q7.

(a) To obtain RTh and VTh

RTh = 2 + 4 + 6 = 12 Ω

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 Ω

(d) P = VTh2/(4RTh) = (40)

2/(4x12) = 33.33 W.

Q8. (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,

VTh

'= 244

44= 12V

(ii) Consider current source 5A,

VTh

''= IRT= 54∣∣4= 10V

V

Th= V

Th' V

Th''= 22V

P=

VTh

2

4R Th

=

22V2

42= 60 .5W

6 ΩΩΩΩ

2 ΩΩΩΩ

(a)

RTh

4 ΩΩΩΩ

20V

+

VTh

−−−−

8V

+

−

− +

+ −

2 ΩΩΩΩ 6 ΩΩΩΩ

(b)

4 ΩΩΩΩ 12V

i

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A 5.5

01232 8

=⇒

=−−

i

i

( )[ ]( )

( ) A 711.0153

3

A 267.4815//314

4

1 =+

=

=++

=

aL

a

ii

i

( )[ ]

( ) A 444.0155

5

A 778.115//413

12

2 −=−+

=

=++

=

bL

b

ii

i

Q9. (a)

Source transformation: v = 8(4) = 32 V

Mesh:

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

∴iL=

VTh

RThRL

=

4. 5

1.87515= 0 .267 A

(b)

Turn off V source:

Turn off I source:

∴iL= i

L1i

L2= 0 . 267 A

(c) (i) RL= R

Th= 1. 875

(ii) Pmax

=

V Th

2

4RTh

= 2 .7 W

a 4 Ω 1Ω

3 32 V

12 V

i

b

RTh

VTh

a

b

RL

4Ω

1Ω

3Ω

8 A

1Li

15Ω

ai

4Ω

1Ω

3Ω

2Li

15Ω

12 V

bi

RTh VTh

RL

Page 8 of 8

Q10. (i) For 10 V source:

V01

=

2

524∣∣4×10= 2 .22 V

For 2 A source:

V02

=

2

524∣∣4×2× 5= 2 .22 V

For 5 V source:

V03

=

2

524∣∣4×− 5

2= − 0.55 V

∴V0= V

01V

03V

03= 3.89 V

(ii) For node V1 :

V1− 10

5− 2V

1− V

2

2= 0

7V1− 5V

2= 40

For node V2 :

V2− V

1

2V

2− 0

4V

2− 5

4= 0

− 2V14V

2= 5

Using Cramer’s rule:

V 1=185

18= 10 .28

V 2=115

18= 6 . 39∴V 0= V 1− V 2= 3.89 V