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## Network Theory MCQ Set 1

1.In the circuit shown below, find the transmission parameter A. a) 6/5
b) 5/6
c) 3/4
d) 4/3

Answer: a [Reason:] Open circuiting b-b, V1 = 6 I1, V2 = 5I1. On solving V1/V2 = 6/5. On substituting we get A = V1/V2=6/5.

2. In the circuit shown above, find the transmission parameter C.
a) 4/5
b) 3/5
c) 2/5
d) 1/5

Answer: d [Reason:] C = I1/V2 |I2=0. By open circuiting b-b we get V2 = 5 I1 =>I1/V2 = 1/5. On substituting we get C = I1/V2=1/5 Ω.

3. In the circuit shown above, find the transmission parameter B.
a) 15/5
b) 17/5
c) 19/5
d) 21/5

Answer: b [Reason:] The transmission parameter B is given by B = -V1/I2 |V2=0. Short circuiting b-b, -I2= 5/17 V1 => -V1/I2 = 17/5. On substituting we get B=17/5 Ω.

4. In the circuit shown above, find the transmission parameter D.
a) 1/5
b) 3/5
c) 7/5
d) 9/5

Answer: c [Reason:] D is a transmission parameter and is given by D = -I1/I2 |V2=0. Short circuiting b-b, I1= 7/17 V1 and-I2= 5/17 V1. So we get I1/I2 = 7/5. So D=7/5.

5. The hybrid parameter h11 is called?
a) short circuit input impedance
b) short circuit forward current gain
c) open circuit reverse voltage gain
d) open circuit output admittance

Answer: a [Reason:] h11=V1/I1 |V2=0. So the hybrid parameter h11 is called short circuit input impedance.

6. The hybrid parameter h21 is called?
a) open circuit output admittance
b) open circuit reverse voltage gain
c) short circuit forward current gain
d) short circuit input impedance

Answer: c [Reason:] h21=I2/I1 |V2=0. So the hybrid parameter h21 is called short circuit forward current gain.

7. In the circuit shown below, find the h-parameter h11. a) 1
b) 2
c) 3
d) 4

Answer: b [Reason:] h11=V1/I1 |V2=0. So short circuiting b-b, V1 = I1((2||2)+1) = 2I1 => V1/I1= 2. On substituting we get h11 = V1/I1= 2Ω.

8. In the circuit shown in question 7, find the h-parameter h21.
a) 1
b) -1
c) 1/2
d) -1/2

Answer: d [Reason:] Short circuiting b-b, h21 = I2/I1 when V2=0 and -I2= I1/2 => I2/I1 = -1/2. So h21 = -1/2.

9. In the circuit shown in question 7, find the h-parameter h12.
a) 1/2
b) -1/2
c) 1
d) -1

Answer: a [Reason:] Open circuiting a-a we get V1=Iy×2 and Iy=I2/2 and V2=Ix×4 and Ix=I2/2. On solving and substituting, we get h12 =V1/V2=1/2.

10. In the circuit shown in question 7, find the h-parameter h22.
a) 1
b) 2
c) 1/2
d) 3/2

Answer: c [Reason:] Open circuiting a-a we get V1=Iy×2 and Iy=I2/2 and V2=Ix×4 and Ix=I2/2. On solving and substituting, we get h22 =I2/V2=1/2 Ω.

## Network Theory MCQ Set 2

1. A graph is said to be a directed graph if ________ of the graph has direction.
a) 1 branch
b) 2 branches
c) 3 branches
d) every branch

Answer: d [Reason:] If every branch of the graph has direction, then the graph is said to be a directed graph. If the graph does not have any direction then that graph is called undirected graph.

2. The number of branches incident at the node of a graph is called?
a) degree of the node
b) order of the node
c) status of the node
d) number of the node

Answer: a [Reason:] Nodes can be incident to one or more elements. The number of branches incident at the node of a graph is called degree of the node.

3. If no two branches of the graph cross each other, then the graph is called?
a) directed graph
b) undirected graph
c) planar graph
d) non-planar graph

Answer: c [Reason:] If a graph can be drawn on a plane surface such that no two branches of the graph cross each other, then the graph is called planar graph .

4. Consider the graph given below. Which of the following is a not a tree to the graph?  Answer: d [Reason:] Tree is sub graph which consists of all node of original graph but no closed paths. So, ‘d’ is not a tree to the graph.

5. Number of twigs in a tree are? n- number of nodes
a) n
b) n+1
c) n-1
d) n-2

Answer: c [Reason:] Twig is a branch in a tree. Number of twigs in a tree are n-1. If there are 4 nodes in a tree then number of possible twigs are 3.

6. Loops which contain only one link are independent are called?
a) open loops
b) closed loops
c) basic loops
d) none of the above

Answer: c [Reason:] The addition of subsequent link forms one or more addition al loops. Loops which contain only one link are independent are called basic loops.

7. If the incident matrix of a graph is given below. The corresponding graph is?

```     a    b   c  d   e    f
1   +1   0  +1   0   0   +1
2   -1  -1   0  +1   0    0
3    0  +1   0   0   +1  -1
4    0   0  -1  -1   -1   0```  Answer: b [Reason:] For the given incidence matrix,

```     a    b   c  d   e    f
1   +1   0  +1   0   0   +1
2   -1  -1   0  +1   0    0
3    0  +1   0   0   +1  -1
4    0   0  -1  -1   -1   0```

the corresponding graph is ‘b’ considering the directions specified in the graph.

8. If A represents incidence matrix, I represents branch current vectors, then?
a) AI = 1
b) AI = 0
c) AI = 2
d) AI= 3

Answer: b [Reason:] If A represents incidence matrix, I represents branch current vectors, then the relation is AI= 0 that is its characteristic equation must be equated to zero

9. If a graph consists of 5 nodes, then the number of twigs in the tree are?
a) 1
b) 2
c) 3
d) 4

Answer: d [Reason:] Number of twigs= n-1. As given number of nodes are 5 then n = 5. On substituting in the equation, number of twigs =5 -1 = 4.

10. If there are 4 branches, 3 nodes then number of links in a co-tree are?
a) 2
b) 4
c) 6
d) 8

Answer: a [Reason:] Number of links = b-n+1. Given number of branches = 4 and number of nodes = 3. On substituting in the equation, number of links in a co-tree = 4 – 3 + 1 = 2.

## Network Theory MCQ Set 3

1. In star connected system, VRY is equal to?
a) VYR
b) -VYR
c) 2VYR
d) 3VYR

Answer: b [Reason:] The voltage available between any pair of terminals is called the line voltage. The double script notation is purposefully used to represent voltages and currents in poly phase circuits. In star connected system, VRY = – VYR.

2. In three phase system, the line voltage VRY is equal to?
a) phasor sum of VRN and VNY
b) phasor difference of VRN and VNY
c) phasor sum of VRN and VNY
d) algebraic sum of VRN and VNY

Answer: a [Reason:] In three phase system, the line voltage VRY is equal to the phasor sum of VRN and VNY which is also equal to the phasor difference of VRN and VYN.

3. The relation between the lengths of the phasors VRN and – VYN is?
a) |VRN| > – |VYN|
b) |VRN| < – |VYN|
c) |VRN| = – |VYN|
d) |VRN| >= – |VYN|

Answer: c [Reason:] The voltage VRY is found by compounding VRN and VYN reversed. The relation between the lengths of the phasors VRN and – VYN is |VRN |= – |VYN |.

4. In a star connected system, the phasors VRN , VYN are ____ apart.
a) 15⁰
b) 30⁰
c) 45⁰
d) 60⁰

Answer: d [Reason:] In a star connected system, the phasors VRN , VYN are separated by θ= 60⁰. To subtract VYN from VRN, we reverse the phase VYN and find its phasor sum with VRN.

5. The relation between VRY ,Vph in a star connected system is?
a) VRY =Vph
b) VRY =√3Vph
c) VRY =3√3Vph
d) VRY =3Vph

Answer: b [Reason:] The two phasors VYN and VBN are equal in length and are 60⁰apart. The relation between VRY ,Vph in a star connected system is VRY = √3Vph.

6. In a star connected system, the relation between VYB ,Vph is?
a) VYB =Vph
b) VYB =3√3Vph
c) VYB =3Vph
d) VYB =√3Vph

Answer: d [Reason:] In a star connected system, the relation between VYB ,Vph is VYB = √3Vph. The line voltage VYB is equal to the phasor difference of VYN and VBN and is equal to √3Vph.

7. The voltages, VBR ,Vph are related in star connected system is?
a) VBR =3Vph
b) VBR =3√3Vph
c) VBR =√3Vph
d) VBR =Vph

Answer: c [Reason:] The voltages, VBR ,Vph in star connected system are related as VBR =√3Vph. The line voltage VYB is equal to the phasor difference of VBN and VRN and is equal to √3Vph.

8. A symmetrical star connected system has VRN =230∠0⁰. The phase sequence is RYB. Find VRY. a) 398.37∠30⁰
b) 398.37∠-30⁰
c) 398.37∠90⁰
d) 398.37∠-90⁰

Answer: a [Reason:] Since the system is a balanced system, all the phase voltages are equal in magnitude but displaced by 120⁰. VRN = 230∠0⁰V. VRY = √3×230∠(0o+30o)V=398.37∠30oV.

9. Find VYB in the figure shown in question 8.
a) 398.37∠-30⁰
b) 398.37∠210⁰
c) 398.37∠90⁰
d) 398.37∠-90⁰

Answer: d [Reason:] Corresponding line voltages are equal to √3 times the phase voltages and are 30⁰ ahead of the respective phase voltages. VYN = 230∠-120⁰V. VYB = √3×230∠(-120o+30o)V=398.37∠-90⁰V.

10. Find VBR in the figure shown in question 8.
a) 398.37∠210⁰
b) 398.37∠-210⁰
c) 398.37∠120⁰
d) 398.37∠-120⁰

Answer: b [Reason:] All the line voltages are equal in magnitude and are displaced by 120⁰. VBN = 230∠-240⁰V. VBR = √3×230∠(-240o+30o)V=398.37∠-210oV.

## Network Theory MCQ Set 4

1. Two ports containing no sources in their branches are called?
a) active ports
b) passive ports
c) one port
d) three port

Answer: b [Reason:] Two ports containing no sources in their branches are called passive ports; among them are power transmission lines and transformers.

2. Two ports containing sources in their branches are called?
a) three port
b) one port
c) passive ports
d) active ports

Answer: d [Reason:] Two ports containing sources in their branches are called active ports. A voltage and current is assigned to each of the two ports.

3. In determining open circuit impedance parameters, among V1, V2, I1, I2, which of the following are dependent variables?
a) V1 and V2
b) I1 and I2
c) V1 and I2
d) I1 and V2

Answer: a [Reason:] In determining open circuit impedance parameters, among V1, V2, I1, I2; V1 and V2 are dependent variables and I1, I2 are independent variables i.e., dependent variables depend on independent variables.

4. In determining open circuit impedance parameters, among V1, V2, I1, I2, which of the following are independent variables?
a) I1 and V2
b) V1 and I2
c) I1 and I2
d) V1 and V2

Answer: c [Reason:] In determining open circuit impedance parameters, among V1, V2, I1, I2; I1 and I2 are independent variables and V1, V2 are dependent variables. Independent variables are the variables that do not depend on any other variable.

5. Which of the following expression is true in case of open circuit parameters?
a) V1 = Z11 V1 + Z12 I2
b) V1 = Z11I1 + Z12 V2
c) V1 = Z11I1 + Z12 I2
d) V2 = Z11I1 + Z12 I2

Answer: c [Reason:] The expression relating the open circuit parameters Z11, Z12 and currents I1, I2 and voltage V1 is V1 = Z11I1 + Z12 I2.

6. Which of the following expression is true in case of open circuit parameters?
a) V2 = Z21I2 + Z22 I2
b) V2 = Z21I1 + Z22 I2
c) V1 = Z21I2 + Z22 I2
d) V1 = Z21I1 + Z22 I2

Answer: b [Reason:] The expression relating the currents I1, I2 and voltage V1 and open circuit parameters Z21, Z22 is V2 = Z21I1 + Z22 I2.

7. Find the Z – parameter Z11 in the circuit shown below. a) 1
b) 1.5
c) 2
d) 2.5

Answer: d [Reason:] For determining Z11, the current I2 is equal to zero. Now we obtain Zeq as 1+ (6×2)/(6+2)=2.5Ω. So, Z11 = 2.5Ω.

8. The value of Z21 in the circuit shown in the question 7 is?
a) 0
b) 1
c) 2
d) 3

Answer: b [Reason:] V2 is the voltage across the 4Ω impedance. The current through 4Ω impedance is I1/4. And V2 = (I1/4) x 4 = I1. So, Z21 = 1Ω.

9. Find the value of Z12 in the circuit shown in the question 7.
a) 3
b) 2
c) 1
d) 0

Answer: c [Reason:] The current through vertical 2Ω resistor is = I2/2. So, V1 = 2 x (I2/2). On solving and substituting we get Z12 = 1Ω.

10. Determine the value of Z22 in the circuit shown in the question 7.
a) 0
b) 1
c) 2
d) 3

Answer: c [Reason:] Open circuiting port 1, we get V2 = I2((2+2)||4) => V2 = I2 x 2 =>V2/I2 = 2. Therefore the value of Z22 is 2Ω.

## Network Theory MCQ Set 5

1. Consider the impedance function Z(s)=3(s+2)(s+4)/(s+1)(s+3) . Find the value of R1 after realizing by first Foster method.
a) 9/2
b) 2/9
c) 9
d) 1/9

Answer: a [Reason:] The first Foster form can be realized by taking the partial fraction of Z (s). On solving, we get Z (s)=3+(9/2/(s+1))+(3/2/(s+3)). So the value of R1 is 9/2Ω.

2. Find the value of C1 in question 1.
a) 1/9
b) 9
c) 2/9
d) 9/2

Answer: c [Reason:] After taking the partial fractions, as P1 = 1/C1, we get the 1/C1 value as 9/2. So the value of C1 is 2/9 F. C1 = 2/9 F.

3. Find the value of C2 in question 1.
a) 1/3
b) 3
c) 3/2
d) 2/3

Answer: d [Reason:] We know P2 = 1/C2. The value of P2 is 3/2. So the value of C2 is 2/3 F. C2 = 2/3 F.

4. Find the value of R2 in question 1.
a) 1
b) 1/2
c) 1/4
d) 1/8

Answer: b [Reason:] We got 1/R2C2 = 3. And the value of C2 is 2/3 F. So the value of R2 is ½Ω. R2 = ½Ω.

5. Find the value of R in question 1.
a) 1
b) 2
c) 3
d) 4

Answer: c [Reason:] On taking partial fraction of Z(s), we get the value of R is 3Ω. R = 3Ω.

6. Consider the impedance function Y(s)=(s2+4s+3)/(3s2+18s+24). Find the value of R0 after realizing by second Foster method.
a) 4
b) 8
c) 12
d) 16

Answer: b [Reason:] The second Foster Form can be realized by taking the reciprocal of the impedance function and by taking partial fractions. So we get 1/R0 as 1/8. So the value of R0 is 8Ω. R0 = 8Ω.

7. Find the value of R1 in question 6.
a) 16
b) 12
c) 8
d) 4

Answer: b [Reason:] On taking the partial fractions we get P1 as 1/12. And we know that R1=1/P1. So the value of R1 is 12Ω. R1 = 12Ω.

8. Find the value of C1 in question 6.
a) 1/24
b) 1/12
c) 1/6
d) 1/3

Answer: a [Reason:] The value of C1 is 1/((2)(R1)). As R1 is 12Ω, the value of C1 is 1/24 F. C1 = 1/24 F.

9. Find the value of R2 in question 6.
a) 5
b) 6
c) 7
d) 8