## Network Theory MCQ Set 1

1. In a delta connected system, the voltage across the terminals R and Y is 400∠0⁰. Calculate the line voltage V_{RY}. Assume RYB phase sequence.

a) 400∠0⁰

b) 400∠120⁰

c) 400∠-120⁰

d) 400∠240⁰

### View Answer

_{RY}|= |V

_{Ph}|, and it is displaced by 120⁰, therefore the line voltage V

_{RY}is V

_{RY}= 400∠0⁰V.

2. In the question 1 find the line voltage V_{YB}.

a) 400∠120⁰

b) 400∠-120⁰

c) 400∠240⁰

d) 400∠-240⁰

### View Answer

_{YB}|= |V

_{Ph}|, and is displaced by 120⁰, therefore the line voltage V

_{YB}is V

_{YB}= 400∠-120⁰V. A balanced three phase, three wire, delta connected system is referred to as mesh connection because it forms a closed circuit.

3. In the question 1 find the line voltage V_{BR}.

a) 400∠240⁰

b) 400∠120⁰

c) 400∠-240⁰

d) 400∠-120⁰

### View Answer

_{BR}|= |V

_{Ph}|, and is displaced by 120⁰, therefore the line voltage V

_{BR}is V

_{BR}= 400∠-240⁰V. Delta connection is so called because the three branches in the circuit can be arranged in the shape of delta.

4. In delta-connected system, the currents I_{R} , I_{Y} , I_{B} are equal in magnitude and they are displaced by _____ from one another.

a) 0⁰

b) 60⁰

c) 90⁰

d) 120⁰

### View Answer

_{R}, I

_{Y}, I

_{B}are equal in magnitude and they are displaced by 120⁰ from one another. From the manner of interconnection of the three phases in the circuit, it may appear that the three phase are short circuited among themselves.

5. In a delta-connected system, the currents I_{R} = I_{B} = I_{Y} =?

a) I_{Ph}

b) 2I_{Ph}

c) 3I_{Ph}

d) 4I_{Ph}

### View Answer

_{R}= I

_{B}= I

_{Y}= I

_{Ph}. Since the system is balanced, the sum of the three voltages round the closed mesh is zero; consequently no current can flow around the mesh when the terminals are open.

6. The relation between I_{L} and I_{Ph} is in a delta connected system is?

a) I_{L} = I_{Ph}

b) I_{L} =√3 I_{Ph}

c) I_{L} =3 I_{Ph}

d) I_{L} = 3√3I_{Ph}

### View Answer

_{L}and I

_{Ph}is in a delta connected system is I

_{L}=√3 I

_{Ph}. The arrows placed alongside the voltages of the three phases indicate that the terminals are positive during their positive half cycles.

7. The line currents are ___ behind respective phase currents in a delta connected system is?

a) 120⁰

b) 90⁰

c) 60⁰

d) 30⁰

### View Answer

8. In a delta connected system, the expression of power (P) is?

a) V_{L}I_{L}cosφ W

b) √3 V_{L}I_{L}cosφ W

c) 3V_{L}I_{L}cosφ W

d) 3√3V_{L}I_{L}cosφ W

### View Answer

_{L}I

_{L}cosφ W.

9. A balanced delta-connected load of (2+j3) Ω per phase is connected to a balanced three-phase 440V supply. The phase current is 10A. Find the total active power.

a) 7.26W

b) 726W

c) 7260W

d) 72.6W

### View Answer

_{Ph}= √(2

^{2}+3

^{2})=3.6∠56.3⁰Ω. cosφ = R

_{Ph}/Z

_{Ph}= 2/3.6 = 0.55. I

_{L}= √3× I

_{Ph}= 17.32A. Active power = √3 V

_{L}I

_{L}cosφ = √3×440×17.32×0.55= 7259.78W.

10. Find the apparent power in the information provided in the question 9.

a) 10955.67 VAR

b) 10.95567 VAR

c) 109.5567 VAR

d) 1.095567 VAR

### View Answer

_{L}I

_{L}sinφ. V

_{L}= 440V, I

_{L}= 17.32A. On substituting we get reactive power = √3 x 440 x 17.32 = 10955.67 VAR.

## Network Theory MCQ Set 2

1. Mesh analysis is applicable for non planar networks also.

a) true

b) false

### View Answer

2. A mesh is a loop which contains ____ number of loops within it.

a) 1

b) 2

c) 3

d) no loop

### View Answer

3. Consider the circuit shown below. The number mesh equations that can be formed are?

a) 1

b) 2

c) 3

d) 4

### View Answer

4. In the figure shown in the question 2, the current through loop 1 be I_{1} and through the loop 2 be I_{2}, then the current flowing through the resistor R_{2} will be?

a) I_{1}

b) I_{2}

c) I_{1}-I_{2}

d) I_{1}+I_{2}

### View Answer

_{2}both the currents I

_{1}, I

_{2}are flowing. So the current through R

_{2}will be I

_{1}-I

_{2}.

5. If there are 5 branches and 4 nodes in graph, then the number of mesh equations that can be formed are?

a) 2

b) 4

c) 6

d) 8

### View Answer

6. Consider the circuit shown in the figure. Find voltage V_{x}.

a) 1

b) 1.25

c) 1.5

d) 1.75

### View Answer

_{1}(CW) in the loop 1 and I

_{2}(ACW) in the loop 2. So, the equations will be V

_{x}+I

_{2}-I

_{1}=0. I

_{1}=5/2=2.5A. I

_{2}=4Vx/4= V

_{x}. V

_{x}+V

_{x}-2.5=0. V

_{x}=1.25V.

7. Consider the circuit shown below. Find the current I_{1}.

a) 3.3

b) 4.3

c) 5.3

d) 6.3

### View Answer

_{1}– 3(I

_{2}) – 6(I

_{3}) =10 -3(I

_{1}) + (2+5+3)I

_{2}=4 -6(I

_{1}) + 10(I

_{3}) = -4 +20 On solving the above equations, I

_{1}=4.3A.

8. Find the current I_{2} (A) in the figure shown in the question 7.

a) 1.7

b) 2.6

c) 3.6

d) 4.6

### View Answer

_{1}– 3(I

_{2}) – 6(I

_{3}) = 10. -3(I

_{1}) + (2+5+3)I

_{2}= 4. -6(I

_{1}1) + 10(I

_{3}) = -4 + 20 On solving the above equations, I

_{2}=1.7A.

9. Find the current I_{3} (A) in the figure shown in the question 7.

a) 4

b) 4.7

c) 5

d) 5.7

### View Answer

_{1}– 3(I

_{2}) – 6(I

_{3}) = 10. -3(I

_{1}) + (2+5+3)I

_{2}= 4. -6(I

_{1}) + 10(I

_{3}) = -4 + 20. On solving the above equations, I

_{3}= 4.7A.

10. Find current through R_{2} resistor.

a) 3

b) 3.25

c) 3.5

d) 3.75

### View Answer

_{1}) + 2(I

_{1}-I

_{2}) = 10. 10(I

_{2}) + 2(I

_{2}-I

_{1}) + 40 = 0. On solving, I

_{1}= 0.5A, I

_{2}= -3.25A. So current through R

_{2}resistor is 0.5-(-3.25) = 3.75 A.

## Network Theory MCQ Set 3

1. If there are M branch currents, then we can write ___________ number of independent equations.

a) M-2

b) M-1

c) M

d) M+1

### View Answer

2. If there are M meshes, B branches and N nodes including reference node, the number of mesh currents is given as M=?

a) B + (N+1)

b) B + (N-1)

c) B-(N+1)

d) B-(N-1)

### View Answer

3. Determine the current I_{1} in the circuit shown below using mesh analysis.

a) 0.955∠-69.5⁰

b) 0.855∠-69.5⁰

c) 0.755∠-69.5⁰

d) 0.655∠-69.5⁰

### View Answer

_{1}(j4) + 6(I

_{1}-I

_{2}) = 5∠0⁰. The equation for loop 2 is 6(I

_{1}-I

_{2}) + (j3) I

_{2}+ (2) I

_{2}= 0. Solving the above equations, I

_{1}= 0.855∠-69.5⁰.

4. In the circuit shown in question 3 find the current I_{2}.

a) 0.5∠-90⁰

b) 0.6∠-90⁰

c) 0.7∠-90⁰

d) 0.8∠-90⁰

### View Answer

_{1}(j4) + 6(I

_{1}-I

_{2}) = 5∠0⁰. The equation for loop 2 is 6(I

_{1}-I

_{2}) + (j3) I

_{2}+ (2) I

_{2}= 0. Solving the above equations, I

_{2}= 0.6∠-90⁰.

5. Find Z_{11}, Z_{12}, Z_{13} obtained from the mesh equations in the circuit shown below.

a) (8+j4) Ω, 5 Ω, 0Ω

b) (8-j4) Ω, 5 Ω, 0Ω

c) (8+j4) Ω, – 5 Ω, 0Ω

d) (8-j4) Ω, -5 Ω, 0Ω

### View Answer

_{11}= self impedance of loop 1 = (5 + 3 – j4) Ω. Z

_{12}= Impedance common to both loop 1 and loop2 = -5Ω. Z

_{13}= No common impedance between loop1 and loop 3 = 0Ω.

6. Determine Z_{21}, Z_{22}, Z_{23} in the circuit shown in question 5.

a) 5Ω, (5-j1) Ω, j6 Ω

b) -5Ω, (5-j1) Ω, j6 Ω

c) -5Ω, (5+j1) Ω, j6 Ω

d) -5Ω, (5-j1) Ω, – j6 Ω

### View Answer

_{21}= common impedance to loop 1 and loop 2 = -5 Ω. Z

_{22}= self impedance of loop2 = (5+j5-j6) Ω. Z

_{23}= common impedance between loop2 and loop 3 = – (-j6) Ω.

7. Find Z_{31}, Z_{32}, Z_{33} in the circuit shown in question 5.

a) 0Ω, j6Ω, (4-j6) Ω

b) 0Ω, -j6Ω, (4+j6) Ω

c) 0Ω, -j6Ω, (4-j6) Ω

d) 0Ω, j6Ω, (4+j6) Ω

### View Answer

_{31}= common impedance to loop 3 and loop 1 = 0 Ω. Z

_{32}= common impedance between loop3 and loop 2 = – (-j6) Ω. Z

_{33}= self impedance of loop 3 = (4-j6) Ω.

8. Find the values of Z_{11}, Z_{22}, Z_{33} in the circuit shown below.

a) (4+j3) Ω, (3-j2) Ω, (5-j5) Ω

b) (4+j3) Ω, (3+j2) Ω, (5-j5) Ω

c) (4-j3) Ω, (3-j2) Ω, (5-j5) Ω

d) (4+j3) Ω, (3-j2) Ω, (5+j5 ) Ω

### View Answer

_{11}= self impedance of loop 1 = (4 + j3) Ω. Z

_{22}= self impedance of loop2 = (j3+3+j4-j5) Ω. Z

_{33}= self impedance of loop 3 = (-j5+5) Ω.

9. Find the common impedances Z_{12}, Z_{13}, Z_{21}, Z_{23}, Z_{31}, Z_{32} respectively in the circuit shown in question 8.

a) -j3Ω, 0Ω, -j3Ω, j5Ω, 0Ω, j5Ω

b) j3Ω, 0Ω, -j3Ω, j5Ω, 0Ω, j5Ω

c) j3Ω, 0Ω, -j3Ω, j5Ω, 0Ω,- j5Ω

d) j3Ω, 0Ω, -j3Ω, -j5Ω, 0Ω, j5Ω

### View Answer

_{12}and Z

_{21}are Z

_{12}= Z

_{21}= -j3Ω. The common impedances Z

_{13}and Z

_{31}are Z

_{13}= Z

_{31}=0Ω. The common impedances Z

_{23}and Z

_{32}are Z

_{23}= Z

_{32}= j5Ω.

10. Find the value V_{2} in the circuit shown in question 9 if the current through (3+j4) Ω is zero.

a) 16∠-262⁰

b) 17∠-262⁰

c) 18∠-262⁰

d) 19∠-262⁰

### View Answer

_{1}–(j3)I

_{2}= 20∠0⁰. (-j3)I

_{1}+ (3+j2)I

_{2}+ (j5)I

_{3}= 0. (j5)I

_{2}+ (5-j5)I

_{3}= -V

_{2}. The current through (3+j4) Ω is zero, I

_{2}= ∆

_{2}/∆ =0

4+j3 20∠0⁰ 0 ∆2 = | -j3 0 j5 | 0 -V2 5-j5

On solving, V_{2} = 17∠-262⁰.

## Network Theory MCQ Set 4

1. According to Millman’s Theorem, if there are n voltage sources with n internal resistances respectively, are in parallel, then these sources are replaced by?

a) single current source I’ in series with R’

b) single voltage source V’ in series with R’

c) single current source I’ in parallel to R’

d) single voltage source V’ in parallel to R’

### View Answer

_{2},…… Vn with internal resistances R

_{1}, R

_{2},…..R

_{n}, respectively, are in parallel, then these sources are replaced by single voltage source V’ in series with R’.

2. In the question above, the value of equivalent voltage source is?

a) V^{‘}=(V_{1}G_{1}+V_{2}G_{2}+⋯.+V_{n}G_{n})

b) V^{‘}=((V_{1}G_{1}+V_{2}G_{2}+⋯.+V_{n}G_{n}))/((1/G_{1}+1/G_{2}+⋯1/G_{n}))

c) V^{‘}=((V_{1}G_{1}+V_{2}G_{2}+⋯.+V_{n}G_{n}))/(G_{1}+G_{2}+⋯G_{n})

d) V^{‘}=((V_{1}/G_{1}+V_{2}/G_{2}+⋯.+V_{n}/G_{n}))/( G_{1}+G_{2}+⋯G_{n})

### View Answer

^{‘}= ((V

_{1}G

_{1}+V

_{2}G

_{2}+⋯.+V

_{n}G

_{n}))/(G

_{1}+G

_{2}+⋯G

_{n}).

3. In the question above the value of equivalent resistance is?

a) R’= G_{1}+G_{2}+⋯G_{n}

b) R’=1/G_{1}+1/G_{2}+⋯1/G_{n}

c) R’=1/((G_{1}+G_{2}+⋯G_{n}) )

d) R’=1/(1/G_{1}+1/G_{2}+⋯1/G_{n})

### View Answer

_{1}+G

_{2}+⋯G

_{n}) ).

4. According to Millman’s Theorem, if there are n current sources with n internal conductances respectively, are in series, then these sources are replaced by?

a) single voltage source V’ in parallel with G’

b) single current source I’ in series with G’

c) single current source I’ in parallel with G’

d) single voltage source V’ in series with G’

### View Answer

_{1},I

_{2},…… In with internal conductances G

_{1},G

_{2},…..G

_{n}, respectively, are in series, then these sources are replaced by single current source I’ in parallel with G’.

5. In the question above, the value of equivalent current source is?

a) I^{‘}=((I_{1}R_{1}+I_{2}R_{2}+⋯.+I_{n}R_{n}))/(R_{1}+R_{2}+⋯R_{n})

b) I’= I_{1}R_{1}+I_{2}R_{2}+⋯.+I_{n}R_{n}

c) I’=((I_{1}/R_{1}+I_{2}/R_{2}+⋯.+I_{n}/R_{n}))/( R_{1}+R_{2}+⋯R_{n})

d) I’=I_{1}/R_{1}+I_{2}/R_{2}+⋯.+I_{n}/R_{n}

### View Answer

^{‘}=((I

_{1}R

_{1}+I

_{2}R

_{2}+⋯.+I

_{n}R

_{n}))/(R

_{1}+R

_{2}+⋯R

_{n}).

6. In the question above, the value of equivalent conductance is?

a) G’= R_{1}+R_{2}+⋯R_{n}

b) G’=1/(1/R_{1}+1/R_{2}+⋯1/R_{n})

c) G’=1/((R_{1}+R_{2}+⋯R_{n}) )

d) G’=1/R_{1}+1/R_{2}+⋯1/R_{n}

### View Answer

_{1}+R

_{2}+⋯R

_{n}) ).

7. Calculate the current through 3Ω resistor in the circuit shown below.

a) 1

b) 2

c) 3

d) 4

### View Answer

8. Find the current through 3Ω resistor in the circuit shown above using Millman’s Theorem.

a) 4

b) 3

c) 2

d) 1

### View Answer

^{‘}=((V

_{1}G

_{1}+V

_{2}G

_{2}))/(G

_{1}+G

_{2})=(10(1/2)+20(1/5))/(1/2+1/5)=12.86V. R’=1/((G

_{1}+G

_{2}) )=1/(1/2+1/5)=1.43Ω. Current through 3Ω resistor=I=12.86/(3+1.43)=2.9A≅3A.

9. Consider the circuit shown below. Find the current through 4Ω resistor.

a) 2

b) 1.5

c) 1

d) 0.5

### View Answer

10. In the circuit shown in the question 9 find the current through 4Ω resistor using Millman’s Theorem.

a) 0.5

b) 1

c) 1.5

d) 2

### View Answer

^{‘}=((V

_{1}G

_{1}+V

_{2}G

_{2}))/(G

_{1}+G

_{2})=(5(1/1)+10(1/3))/(1/1+1/3)=6.25V. R’=1/((G

_{1}+G

_{2}) )=1/(1/1+1/3)=0.75Ω. I=6.25/(4+0.75)=1.5A.

## Network Theory MCQ Set 5

1. The ratio of voltage transform at first port to the voltage transform at the second port is called?

a) Voltage transfer ratio

b) Current transfer ratio

c) Transfer impedance

d) Transfer admittance

### View Answer

_{21}= V

_{2}(s)/V

_{1}(s) G

_{12}= V

_{1}(s)/V

_{2}(s).

2. The ratio of the current transform at one port to current transform at other port is called?

a) Transfer admittance

b) Transfer impedance

c) Current transfer ratio

d) Voltage transfer ratio

### View Answer

_{12}(s) = I

_{1}(s)/I

_{2}(s) α

_{21}(s) = I

_{2}(s)/I

_{1}(s).

3. The ratio of voltage transform at first port to the current transform at the second port is called?

a) Voltage transfer ratio

b) Transfer admittance

c) Current transfer ratio

d) Transfer impedance

### View Answer

_{21}(s) = V

_{2}(s)/I

_{1}(s) Z

_{12}(s) = V

_{1}(s)/I

_{2}(s).

4. For the network shown in the figure, find the driving point impedance.

a) (s^{2}-2s+1)/s

b) (s^{2}+2s+1)/s

c) (s^{2}-2s-1)/s

d) (s^{2}+2s-1)/s

### View Answer

^{2}+2s+1)/s.

5. Obtain the transfer function G_{21} (S) in the circuit shown below.

a) (s+1)/s

b) s+1

c) s

d) s/(s+1)

### View Answer

_{1}(S) = 2 I

_{1}(S) + 2 sI

_{1}(S) V

_{2}(S) = I

_{1}(S) X 2s Hence G

_{21}(S) = V

_{2}(s)/V

_{1}(s) =2 s/(2+2 s)=s/(s+1).

6. Determine the transfer function Z_{21} (S) in the circuit shown in question 5.

a) s

b) 2 s

c) 3 s

d) 4 s

### View Answer

_{21}(S) is Z

_{21}(S) = V

_{2}(S)/I

_{1}(S). V

_{2}(S) = I

_{1}(S) X 2s. V

_{2}(S)/I

_{1}(S)=2s. On substituting Z

_{21}(S) = 2s.

7. Find the driving point impedance Z_{11} (S) in the circuit shown in question 5.

a) 2(s+2)

b) (s+2)

c) 2(s+1)

d) (s+1)

### View Answer

_{11}(S) is Z

_{11}(S)=V

_{1}(S)/I

_{1}(S). V

_{1}(S) = 2 I

_{1}(S) + 2 sI

_{1}(S) => V

_{1}(S) = (2+2s)I

_{1}(S) => V

_{1}(S)/I

_{1}(S) = 2(s+1). On substituting Z

_{11}(S) = 2(S+1).

8. Obtain the transfer function G_{21} (s) in the circuit shown below.

a) (8 S+2)/(8 S+1)

b) (8 S+2)/(8 S+2)

c) (8 S+2)/(8 S+3)

d) (8 S+2)/(8 S+4)

### View Answer

_{eq}(S) = 1/(2 S+1/2)=2/(4 S+1). Applying Kirchhoff’s laws, we have V

_{2}(S) = 2 I

_{1}(S) => V

_{1}(S) = I

_{1}(S)[2/(4 S+1)+2] = I

_{1}(S)[(8 S+4)/(4 S+1)] The transfer function G

_{21}(s) = V

_{2}(s)/V

_{1}(s) =2 I

_{1}(S)/((8 S+4)/(4 S+1))I

_{1}(S) =(8 S+2)/(8 S+4).

9. Obtain the transfer function Z_{21}(s) in the circuit shown in question 8.

a) 1

b) 2

c) 3

d) 4

### View Answer

_{21}(s) is Z

_{21}(S) = V

_{2}(S)/I

_{1}(S). V

_{2}(S) = 2 I

_{1}(S) => V

_{2}(S)/I

_{1}=2. On substituting Z

_{21}(s) = 2.

10. Determine the driving point impedance Z_{11}(S) in the circuit shown in question 8.

a) (8 S+4)/(4 S+4)

b) (8 S+4)/(4 S+3)

c) (8 S+4)/(4 S+2)

d) (8 S+4)/(4 S+1)

### View Answer

_{11}(S) is Z

_{11}(S) = V

_{1}(s)/I

_{1}(s). V

_{1}(s) = I

_{1}(s)((2/(4s+1))+2) = I

_{1}(s)((8s+4)/(4s+1)) => V

_{1}(s)/I

_{1}(s) = ((8s+4)/(4s+1)). On substituting we get Z

_{11}(S) = (8S+4)/(4S+1).