set2

Set-1RAJEEV GANDHI MEMORIAL COLLEGE OF ENGINEERING AND TECHNOLOGY

( AUTONOMUS)

BRANCH: ECEMID TEST: 1

CLASS: I-B.TECHMAX MARKS: 25

SUBJECT: NETWORK ANALYSISDURATION:2 HOURS

DATE:13/10/2015NOTE:Attempt any five questions but first question is compulsory

1. (a) Calculate the value of i in the circuit given below ? [2M]

1. A solid copper sphere, 10 cm in diameter is deprived

of 1020 electrons by a charging scheme. The charge on

the sphere is

(A) 160.2 C (B) 160.2 C

(C) 16.02 C (D) 16.02 C

2. A lightning bolt carrying 15,000 A lasts for 100 s. If

the lightning strikes an airplane flying at 2 km, the

charge deposited on the plane is

(A) 13.33 C (B) 75 C

(C) 1500 C (D) 1.5 C

3. If 120 C of charge passes through an electric

conductor in 60 sec, the current in the conductor is

(A) 0.5 A (B) 2 A

(C) 3.33 mA (D) 0.3 mA

4. The energy required to move 120 coulomb through

3 V is

(A) 25 mJ (B) 360 J

(C) 40 J (D) 2.78 mJ

5. i ?

(A) 1 A (B) 2 A

(C) 3 A (D) 4 A

6. In the circuit of fig P1.1.6 a charge of 600 C is

delivered to the 100 V source in a 1 minute. The value

of v1 must be

(A) 240 V (B) 120 V

(C) 60 V (D) 30 V

7. In the circuit of the fig P1.1.7, the value of the

voltage source E is

(A) 16 V (b) 4 V

(C) 6 V (D) 16 V

CHAPTER

1.1

BASIC CONCEPTS

Page

3

100 V

60 V

v1

20

Fig. P.1.1.6

1 A

2 A 5 A

4 A3 A

i

Fig. P.1.1.5

2 V 1 V

5 V

10 V

0 V

4 V

E

+ +

+

+

+

Fig. P.1.1.7

GATE EC BY RK Kanodia

www.gatehelp.com(b) i. What is the relation between tie-set and cut-set matrix ? [0.5M]

ii. What is the relation between tie-set and incident matrix ? [0.5M]iii. What is the rank of complete incident matrix ? [0.5M]iv. What is the size of tie-set matrix ? [0.5M]

(c) Convert the following polar form into rectangular form

i. 1060 [1M]ii.

6545 [1M](d) i. Given the sinusoidal function 5 sin(4pit 60), calculate its phase angle,angular frequency, time period, and fre-

quency ? [1M]ii. Calculate the phase angle between v1 = 10 cos(t+ 50) and v2 = sin(t 10). State which sinusoidal is

leading. [1M]

(e) Calculate the value of va in the circuit given below ? [2M]

1. v1 ?

(A) 0.4vs (B) 1.5vs

(C) 0.67vs (D) 2.5vs

2. va ?

(A) 11 V (B) 11 V

(C) 3 V (D) 3 V

3. v1 ?

(A) 120 V (B) 120 V

(C) 90 V (D) 90 V

4. va ?

(A) 4.33 V (B) 4.09 V

(C) 8.67 V (D) 8.18 V

5. v2 ?

(A) 0.5 V (B) 1.0 V

(C) 1.5 V (D) 2.0 V

6. ib ?

(A) 0.6 A (B) 0.5 A

(C) 0.4 A (D) 0.3 A

CHAPTER

1.3

Page

23

METHODS OF ANALYSIS

6R 3R

4vs vsv16R

+

Fig. P1.3.1

3 A

2 va

3 1 A

Fig. P1.3.2

12 V

10

4

4 A1 2

10 V

va

Fig. P1.3.4

v2

60

30 0.5 A 10 V 30

20

+

Fig. P1.3.5

10

30 V 3 A

20

30

v1

+

6 A 60 9 A60

Fig. P1.3.3

10 V

36 69

0.5 A

37 64

ib

Fig. P1.3.6


www.gatehelp.com

2. (a) Find how many trees will exist for the graph given below. [2M]

(b) Draw all the possible tress. [3M]

114 | Network Theory

2.4 Matrix representation of a graph

For a given oriented graph, there are several representative matrices. They are extremely importantin the analytical studies of a graph, particularly in the computer aided analysis and synthesis oflarge scale networks.

2.4.1 Incidence Matrix An

It is also known as augmented incidence matrix. The element node incidence matrix A indicatesin a connected graph, the incidence of elements to nodes. It is an N B matrix with elements ofAn = (akj)

akj = 1, when the branch bj is incident to and oriented away from the kth node.

= 1, when the branch bj is incident to and oriented towards the kth node.= 0, when the branch bj is not incident to the kth node.

As each branch of the graph is incident to exactly two nodes,

nk=0

akj = 0 for j = 1, 2, 3, B.

That is, each column of An has exactly two non zero elements, one being +1 and the other1. Sum of elements of any column is zero. The columns ofAn are lineraly dependent. The rankof the matrix is less than N .

Significance of the incidence matrix lies in the fact that it translates all the geometrical featuresin the graph into an algebraic expression.

Using the incidence matrix, we can writeKCL as

An iB = 0, where iB = branch current vector.

But these equations are not linearly independent. The rank of the matrix A is N 1. Thisproperty ofAn is used to define another matrix called reduced incidence matrix or bus incidencematrix.

For the oriented graph shown in Fig. 2.3(a), the incidence matrix is as follows:

Nodes branches1 2 3 4 5

An =

abcd

1 1 1 0 01 0 0 1 00 1 0 1 10 0 1 0 1

Note that sum of all elements in each column is zero.Figure 2.3(a)

1

3. One cycle of a periodic voltage waveform is depicted in the below figure.

(a) Find the average value. [2M]

(b) Find the RMS value. [3M]

11.29 Calculate the effective value of the current waveformin Fig. 11.60 and the average power delivered to a

resistor when the current runs through theresistor.12-

11.33 Determine the rms value for the waveform in Fig. 11.64.

494 Chapter 11 AC Power Analysis

Figure 11.60For Prob. 11.29.

11.30 Compute the rms value of the waveform depicted inFig. 11.61.


11.31 Find the rms value of the signal shown in Fig. 11.62.


11.32 Obtain the rms value of the current waveform shownin Fig. 11.63.


50

15 25 t

10

10

i(t)

10 20 30

210

4 6 8 10 t

2

v (t)

v (t)2

0

4

1 2 3 4 5 t

10 2 3 4 5 t

10

i(t)

10t2


11.34 Find the effective value of f(t) defined in Fig. 11.65.


11.35 One cycle of a periodic voltage waveform is depictedin Fig. 11.66. Find the effective value of the voltage.Note that the cycle starts at and ends at t 6 s.t 0


11.36 Calculate the rms value for each of the followingfunctions:

(a) (b) (c) (d)

11.37 Design a problem to help other students betterunderstand how to determine the rms value of thesum of multiple currents.

Section 11.5 Apparent Power and Power Factor

11.38 For the power system in Fig. 11.67, find: (a) theaverage power, (b) the reactive power, (c) the powerfactor. Note that 220 V is an rms value.

v(t) 5 sin t 4 cos t Vi(t) 8 6 sin 2t Av(t) 4 3 cos 5t Vi(t) 10 A

0 1 2 3 4 5 6 7 8 9 10

i(t)10

t

f (t)10

1 0 1 2 3 4 5 t

10 2 3 4 65 t

10

20

30

v (t)

ale29559_ch11.qxd 07/08/2008 12:02 PM Page 494

4. (a) Using nodal analysis find the value of v0 in the fig given below. [2M]

(b) calculate the power dissipated by 2k , 5k , 4k , Resistors. [3M]

Problems 115



3.6 Use nodal analysis to obtain in the circuit of Fig. 3.55.

vo

30 V +

2 k

20 V +

5 k4 k vo

+

6 2 12 V

10 V

+

+ 4

I3I2

vo

I1

10 2 A 0.2Vx20 Vx

+


3.5 Obtain in the circuit of Fig. 3.54.vo

3.7 Apply nodal analysis to solve for in the circuit ofFig. 3.56.

Vx

3.8 Using nodal analysis, find in the circuit of Fig. 3.57.vo


3 V4vo2 vo

+

1

3 5

+

+

3.9 Determine in the circuit of Fig. 3.58 using nodalanalysis.

Ib


12 V +

50 150

60Ib250 +

Ib

3.10 Find in the circuit of Fig. 3.59.Io


3.11 Find and the power dissipated in all the resistorsin the circuit of Fig. 3.60.

Vo


2 4 8

1

4 A 2 Io

Io

36 V +

+2 12 V

1 Vo 4

3.12 Using nodal analysis, determine in the circuit inFig. 3.61.

Vo


2 5

10 1

30 V + 4 Ix

Ix

Vo

+

ale29559_ch03.qxd 07/08/2008 10:47 AM Page 115

5. (a) The current flowing through the 2 F capacitor is shown in below figure. Draw the corresponding voltage wave form,v(0)=0 [4M]

Time(ms)

ic(A)

0

1

2

3

5A

(b) Convert the below star to delta. [1M]

56 Chapter 2 Basic Laws

Transform the wye network in Fig. 2.51 to a delta network.

Answer: Ra 140 , Rb 70 , Rc 35 .

Practice Problem 2.14

Figure 2.51For Practice Prob. 2.14.

20

R2ba

c

10

R1

R3 40

Obtain the equivalent resistance for the circuit in Fig. 2.52 and useit to find current i.

Solution:

1. Define. The problem is clearly defined. Please note, this partnormally will deservedly take much more time.

2. Present. Clearly, when we remove the voltage source, we endup with a purely resistive circuit. Since it is composed of deltasand wyes, we have a more complex process of combining theelements together. We can use wye-delta transformations as oneapproach to find a solution. It is useful to locate the wyes (thereare two of them, one at n and the other at c) and the deltas(there are three: can, abn, cnb).

3. Alternative. There are different approaches that can be used tosolve this problem. Since the focus of Sec. 2.7 is the wye-deltatransformation, this should be the technique to use. Anotherapproach would be to solve for the equivalent resistance byinjecting one amp into the circuit and finding the voltagebetween a and b; we will learn about this approach in Chap. 4.

The approach we can apply here as a check would be to usea wye-delta transformation as the first solution to the problem.Later we can check the solution by starting with a delta-wyetransformation.

4. Attempt. In this circuit, there are two Y networks and three networks. Transforming just one of these will simplify the circuit.If we convert the Y network comprising the 5- 10- and20- resistors, we may select

Thus from Eqs. (2.53) to (2.55) we have

Rc R1 R2 R2 R3 R3 R1

R3

3505 70

Rb R1 R2 R2 R3 R3 R1

R2

35020

17.5

35010

35

Ra R1 R2 R2 R3 R3 R1

R1

10 20 20 5 5 1010

R1 10 , R2 20 , R3 5

,,

RabExample 2.15

a ai

bb

c n120 V5

30

12.5

15

10

20

+

Figure 2.52For Example 2.15.

ale29559_ch02.qxd 07/09/2008 11:19 AM Page 56

2

SCHEME OF EVALUATION

1.(a)

SOLUTIONS

1. (C) n 1020, Q ne e 10 16 0220 . C

Charge on sphere will be positive.

2. (D) Q i t 15000 100 15. C

3. (B) idQ

dt

120

602 A

4. (B) W Qv 360 J

6. (A)

6. (A) In order for 600 C charge to be delivered to the

100 V source, the current must be anticlockwise.

idQ

dt

600

6010 A

Applying KVL we get

v1 60 100 10 20 or v1 240 V

7. (A) Going from 10 V to 0 V

10 5 1 E 0 or E 16 V

8. (D) 100 65 352 2 v v V

v v v3 2 330 65 V

105 65 04 3 v v v4 25 V

v v4 115 55 0 v1 15 V

9. (B) Voltage is constant because of 15 V source.

10. (C) Voltage across 60 resistor 30 V

Current 30

600 5. A

Voltage across R1 is 70 20 50 V

R150

0 5100

.

11. (C) The current i will be distributed in the cube

branches symmetrically

vi i i

iab 6

3

6

6

6

35 ,

Rv

ieq

ab 5

12. (C) If we go from +side of 1 k through 7 V, 6 V and

5 V, we get v1 7 6 5 8 V

13. (D) It is not possible to determine the voltage across

1 A source.

14. (D) RR

Req

eq

eq

510 5

10 5

( )

R R R Req eq eq eq2 15 5 75 10 50

Req 125 1118.

Chap 1.1Basic Concepts

Page

9

ia

bi

3i

3i

6i6i

3i

3i

Fig. S. 1.1.11

1 A

2 A 5 A

4 A3 A

i = 1 A

2 A

1 A6 A

Fig. S 1.1.5

+

v2

+

v4

+ v3

+ 105 V 15 V + 10 V +

+

55 V

+

65 V

v1+ +

30V

+

30V

100 V

Fig. S 1.1.8

2 V 1 V

5 V

10 V

0 V

4 V

E

+ +

+

+

+

Fig. S 1.1.7

5

10

5

ReqReq

Fig. S 1.1.14


www.gatehelp.com

1.(b).(i) QBT = 0

1.(b).(ii) ABT = 0

1.(b).(iii) n-1

1.(b).(iv) (b-n+1) b1.(c).(i) 5 53i1.(c).(ii) 7.448 + 3.085i

1.(d).(i) phase angle= -60 , angular frequency= 4pi, time period=0.5 sec, frequency= 2 Hz.

1.(d).(ii) given v1 = 10 cos(t+ 50) v2 = sin(t 10)

v1 = 10 cos(t+ 50)

= 10 cos(t 130)

= 10 sin(t 40)

v2 = sin(t 10)

v2 leads v1 by 30

1.(e)va 10

4+va2

= 4 va = 8.67

2. A.At = 8 total 8 trees will exist.

3

116 | Network Theory

Fig. 2.3(b) shows all possible trees corresponding to the matrix A.

Figure 2.3(b)

To verify the property that the determinant of sub matrix At of A = At ; Ai is +1 or 1.For tree [2, 3, 4]

From


A =abc

+1 1 0 1 00 0 1 1 01 0 1 0 1

= At;Ai

Det Ai =

1 +1 00 0 1

1 0 1

= 1For another tree [2, 4, 5]


A =abc

1 0 0 1 10 1 0 1 01 1 1 0 0

= At;Ai

Det Ai =

1 0 00 1 0

1 1 1

= 1

2.5 Loop equations and fundamental loop matrix (Tie-set Matrix)

From the knowledge of the basic loops (tie-sets), we can obtain loop matrix. In this matrix, theloop orientation is to be the same as the corresponding link direction. In order to construct thismatrix, the following procedure is to be followed.

3. Average value=20 and RMS value=21.6 and Form factor=1.08 and Peak factor=1.389

4. Apply KCL to the top node.30 v0

2k+

20 v05k

=v04k v0 = 20

5.

Case 1:- 0 t 1 msec

i(t) = 5000t

v(t) =1

c

t

i(t)dt

=1

2

0

i(t)dt+1

2

t0

i(t)dt

= v(0) +1

2

t0

5000tdt

= 0 +5000

2

[t2

2

] v(t) = 12501 t2 v(1m) = 1250.

Case 2:- 1 t 2 msec

i(t) = 5

v(t) = v(1) +1

2

t1m

5tdt

v(t) = 1250 + 2.51 [t 1m] v(2m) = 1250 + 2500 = 3750

4

Case 3:- 2 t 3 msec

i(t) = 5000(t 3m)

v(t) = v(2) 50002

t2m

(t 3m)dt

= 3750 50002

[(t2

2

)t2m

3m(t 2m)]

v(t) = 3750 50002

[t2

2 2 3mt+ 6]

v(3) = 3750 50002

[4.5 2 9+ 6]

= 3750 + 1250

= 5000

The voltage wave form is shown in the below figure.

0 0.5 1 1.5 2 2.5 3

x 10-3

0

2

4

6

X: 0.001Y: 5

time

curre

nt(a

mps

)

DONE BY S.SREEKANTHA REDDYM.TECH(I.I.T.R)

X: 0.002Y: 5

X: 0.003Y: 0

0 0.5 1 1.5 2 2.5 3

x 10-3

0

2000

4000

6000

X: 0.001Y: 1250

time

volta

ge

X: 0.002Y: 3750

X: 0.003Y: 5000

0 0.5 1 1.5 2 2.5 3

x 10-3

0

1

2x 104

X: 0.001Y: 6250

time

pow

er

X: 0.002Y: 1.875e+004

X: 0.003Y: 0

5

set2

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