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Circuit Theory - Solved Assignments - Semester Spring 2008

Solution of Assignment 1(Spring 2008)CIRCUIT THEORY (PHY301) MARKS: 35

Question #1: We have usually two sources of electricity Main & Batteries. Here you are given some every day home items categories these in such a way, so that we may understand which one is operated from Main which from Batteries and those operated from both sources. I) Clock II) Kettle III) Light Bulb IV) Radio V) Dishwasher VI) Mobile Phone VII) Fridge.

Solution1. 2. 3. 4. 5. 6. 7. Clock can run on both mains electricity & batteries. The kettle uses mains electricity. Light bulb uses mains electricity. Radio can run on both mains electricity & batteries. Dishwashers use mains electricity. Mobile phone can run on batteries but charge up using mains electricity. Fridge uses mains electricity.

Question #2: Find V1 & V2 through the following circuit given below .Draw the circuit diagram of each step otherwise you will lose your marks. Write each step of the calculation to get maximum marks and also mention the units of each derived value

Question #3: Find total current I passing through circuit shown below. Draw the circuit diagram of each step otherwise you will lose your marks. Write each step of the calculation to get maximum marks and also mention the units of each derived value

WWW.VIRTUALIANS.COMAssignment 2(Spring 2008)(Solution)CIRCUIT THEORY (PHY301) MARKS: 50 Due Date: 05/05/2008

Q.1.

Sol.Part (A):

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VI

RT U

AL

IA N

S. C

From this study, state what you have learned about changing Reference Node in Nodal Analysis. Note: Draw and labeled complete circuit diagram otherwise you will lose your marks. Write each step of the calculation to get maximum marks and also mention the units of each derived value.

OM

You are a) b) c)

given the circuit in Fig (A), Fig (B) & Fig (C). You will find out node voltages V1,V2,V3 for each diagram. You will find out currents IA,IB,IC& ID for each diagram. You will also find out VA,VB,VC for each diagram.

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Write KCL equations at V1.

V1 V2 2

+

V1 V3 4

-3=0

2 V1 - 2 V2 + V1 - V3 -12 = 0 3 V1 - 2 V2 - V3 = 12 Write KCL equations at V2. ________ (1)

Write KCL equations at V3.

+

+ 2 IA = 0

We know

Put (A) in (B)

Solving (1),(2) and (3) simultaneously

V1 = 4.0V

RT U

V3 - V2 +2 V3 -2 V1 + 8 V1 -8 V2 = 0 6 V1 -9 V2 +3 V3 = 0 _______ (3);

ALV2 =2.4V

V1 V2 ______ (A) 2 V3 - V2 +2 V3 -2 V1 + (2 I A )8 = 0 _______ (B)

IA

=

Now For VA

V1 V3

V1 V2 = 1.2 A 2 V V 7.2 IB = 1 3 = = 1.8A 4 4 V V 2.4 + 2.4 IC = 2 3 = = 0.6A 8 8 V I D = 2 = 0.6A 4

VI

IA =

= VA

VA = V1 V3 =4.8-(-2.4)

VA =7.2V

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IA Nand

V3 V2 8

V3 V1 4

V3 = -2.4V

S. C

V2 V1 V2 V3 V2 + + =0 2 8 4 -4 V1 + 7 V2 - V3 = 0 ________ (2)

OM

WWW.VIRTUALIANS.COMFor VB

VB = V2 V3 =2.4-(-2.4) VB =4.8V VC = V1 VC =4.8V

For VC

Part (B)

Write KCL equations at V1.

3 V1 - V3 =12 ----------- (1)

V2 4

VI

RT U

IA =

V2 V1 +3-2( )=0 4 2 V2 +12-4 V1 = 0

Write KCL equations at V3.

V3 V3 V1 + +2 I A = 0 8 4 V IA = 1 2 V3 V3 V1 V1 + +2( )=0 8 4 2 V3 +2 V3 -2 V1 +8 V1 = 0

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AL+3-2 I A = 0

Write KCL equations at V2.

V1 2

V2 -4 V1

IA N= -12 ------------- (2)

V1 V1 V3 + -3 = 0 2 4 2 V1 + V1 - V3 =12

S. C

OM

WWW.VIRTUALIANS.COM6 V1 +3 V3 = 0

V1 =2.4V ; V2 =-2.4V ; V3 =-4.8V V I A = 1 =1.2A 2 V V 2.4 + 4.8 IB = 1 3 = =1.8A 4 4 V 4.8 IC =- 3 = =0.6A 8 8 V 2.4 I D =- 2 = =0.6A 4 4

VA = I B x 4= 1.8 x 4 = 7.2VFor VB

For VC

V1 V3 - V2Part (C)

=0

V3 = V1 V2

= 2.4-(-2.4) =4.8V

Write KCL equations at V1.

VI

RT U

V1 V3 V1 V1 V2 + + =0 4 8 2 2 V1 -2 V2 + V1 - V3 +4 V1 = 07 V1 -2 V2 - V3 = 0 ------------- (1)

Write KCL equations at V2.

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AL

IA N

S. C

VB =8 IC

= 4.8V

OM

For VA

WWW.VIRTUALIANS.COMV2 V1 4Where + 3 -2 I A = 0

IA = -

V2 V1 V1 +3-2()=0 4 2 V2 - V1 +12+4 V1 = 0 -------------- (2)Write KCL equations at V3.

V1 2

9V1 + 3V3 = 0 ------------ (3)

For VA For VB For VC

VB = I C x8=0.6x8=4.8V

Overall summary

VI

VC = - V2 = -(-4.8)= 4.8VPart (B) 2.4V -2.4V -4.0V 1.2A 1.8A 0.6A 0.6A Part (C) -2.4V -4.8V -7.2V 1.2A 1.8A 0.6A 0.6A

V1V2

Part (A) 4.0V 2.4V -2.4V 1.2A 1.8A 0.6A 0.6A

V3 IAIB

IC ID

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RT U

V1 = -2.4V , V2 =-4.8V and V3 =-7.2V V 2.4 I A =- 1 = =1.2A 2 2 V (7.2) I B =- 3 ==1.8A 4 4 V V 2.4 (7.2) IC = 1 3 = =0.6A 8 8 V V 2.4 + 4.8 ID = 1 2 = =0.6A 4 4 VA = I B x4=1.8x4=7.2V

AL

IA N

Solving eq (1), (2) and (3) simultaneously

S. C

V3 +2 I A = 0 4 V IA = - 1 Where 2 V3 V1 V3 V1 + +2()=0 8 4 2 V3 V1 +2 V3 8V1 = 0+

OM

V3 V1 8

WWW.VIRTUALIANS.COMVA7.2V 4.8V 4.8V 7.2V 4.8V 4.8V 7.2V 4.8V 4.8V

VB VC

Conclusion: Changing Reference nodes, changes the reference node voltages. However, all currents and voltages across elements remain the same.

Q.2.

Sol.

Write KCL equations at Vx.

Vx 24V Vx Vx 60 I b + + =0 250 150 50 3 Vx -72+15 Vx -300 I b =0 ---------- (I) 24 Vx Ib = ------------(A) 250Put A in (I) So,

VI

23 Vx -300(

24 Vx ) = 72 250 23 Vx - 28.8 +1.2 Vx = 7224.2 Vx = 100.8

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RT U

ALVx = 4.165V

IA N

S. C

Use nodal analysis to find Current Ib in the network given below. Write each step of the calculation to get maximum marks and also mention the units of each derived value.

OM

WWW.VIRTUALIANS.COMIb ==

24 Vx 250 24 4.165 250

=79.3mA

Q.3.First Identify and label each node in the network. Use nodal analysis to find voltage at each node in the network given below. Draw and labeled complete circuit diagram otherwise you will lose your marks. Write each step of the calculation to get maximum marks and also mention the units of each derived value.

Sol.

Write KCL equations at V1.

VI

V1 V1 12 V1 V2 + +3+ =0 8 4 2 4 V1 -48+ V1 - V2 +24+2 V1 = 04

Write KCL equations at V2.

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RT U

AL

First we will identify nodes and labeled them,

V1 - V2 = 24 ----------- (A)

IA N

S. C

OM

WWW.VIRTUALIANS.COMV2 + 5V0 V2 V1 + -3 = 0 1 8 V2 V1 +8 V2 +40 V0 -24 = 0 -------------- (1)But

V1 + V0 -12= 0 V0 = 12- V1---------- (2)

Put (2) in (1) -41 Solving (A) and (B)

V1 + 9 V2 +40(12- V1 )=24

V1 +9 V2 = -450 ------------- (B),

------ Good Luck -----

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VI

RT U

AL

IA N

S. C

OM

V1 = -10.91V

V2 =-100.36V

WWW.VIRTUALIANS.COMAssignment 3(Spring 2008)(Solution)CIRCUIT THEORY (PHY301) MARKS: 50 Due Date: 16/05/2008

Q.1.

Sol.

Write KCL equations at the super node 1+ 2 V0 = But

VI

RT U

V1 V2 V1V3 + + 4 1 1 V0 = V1 V3

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AL

First we will identify nodes and labeled them,

IA N----------------- (1)

S. C

OM

First Identify and label each node in the network. Use nodal analysis to find voltage at each node in the network given below. Draw and labeled complete circuit diagram otherwise you will lose your marks. Write each step of the calculation to get maximum marks and also mention the units of each derived value.

WWW.VIRTUALIANS.COMHence eq. (1) becomes 4=-8 V1 + 4 Write KCL equations at node V3.

V2 -4 V3 +8 V3 + V1 +4 V1 4=-3 V1 + 4 V2 +4 V3 -------------------- (2) 2V0 +V3 10 V3 = V1 V3 + 4 2 V3 V3 10 V3 V1 + + - 2V0 =0 4 1 2 4 V1 - V3 =20 --------------- (3)

At the super node constraint equation will be

But

V2 = V1 + V3Solving eq.(4) and (2) we get

-------------- (4)

V1 +8 V3 =4 -------------- (5)Solving eq. (5) and eq. (3)

V3 = -0.12vPut this value in eq.(3) we get

V1 =4.97vAs

Putting values we get

V2 =4.85v

Q.2.

Use Mesh analysis to determine Currents in each Mesh in the network given below. Identify and label each mesh otherwise you will lose your marks. Draw and labeled complete circuit diagram. Write each step of the calculation to get maximum marks and also mention the units of each derived value.

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VI

RT U

AL

V2 = V1 + V3

IA N

S. C

Hence from constraint eq. we have

OM

V2 V1 = 4 I 0 V I0 = 3 4

------------- (A)

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Sol.First we will identify mesh and labeled them,

For mesh I -10-2 I x +10 I1 -6 I 2 =0 But

Hence,

10=-2 I1 +2 I 2 +10 I1 -6 I 2 5=4 I1 -2 For mesh II

Solving (1) and (2)

RT U

I1 =0.8A

Q.3.

.

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VI

Use Mesh analysis to find Voltage Vx and Ix in the

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