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## 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 VRY. Assume RYB phase sequence.
a) 400∠0⁰
b) 400∠120⁰
c) 400∠-120⁰
d) 400∠240⁰

Answer: a [Reason:] In a balanced delta-connected system we know |VRY |= |VPh|, and it is displaced by 120⁰, therefore the line voltage VRY is VRY = 400∠0⁰V.

2. In the question 1 find the line voltage VYB.
a) 400∠120⁰
b) 400∠-120⁰
c) 400∠240⁰
d) 400∠-240⁰

Answer: b [Reason:] As |VYB |= |VPh|, and is displaced by 120⁰, therefore the line voltage VYB is VYB = 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 VBR.
a) 400∠240⁰
b) 400∠120⁰
c) 400∠-240⁰
d) 400∠-120⁰

Answer: c [Reason:] We know, |VBR|= |VPh |, and is displaced by 120⁰, therefore the line voltage VBR is VBR = 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 IR , IY , IB are equal in magnitude and they are displaced by _____ from one another.
a) 0⁰
b) 60⁰
c) 90⁰
d) 120⁰

Answer: d [Reason:] In delta-connected system, the currents IR , IY , IB 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 IR = IB = IY =?
a) IPh
b) 2IPh
c) 3IPh
d) 4IPh

Answer: a [Reason:] In a delta-connected system, the currents IR = IB = IY = IPh. 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 IL and IPh is in a delta connected system is?
a) IL = IPh
b) IL =√3 IPh
c) IL =3 IPh
d) IL = 3√3IPh

Answer: b [Reason:] The relation between IL and IPh is in a delta connected system is IL =√3 IPh. 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

Answer: d [Reason:] In a delta connected system, all the line currents are equal in magnitude but displaced by 120⁰ from one another and the line currents are 30⁰ behind the respective phase currents.

8. In a delta connected system, the expression of power (P) is?
a) VLILcosφ W
b) √3 VLILcosφ W
c) 3VLILcosφ W
d) 3√3VLILcosφ W

Answer: b [Reason:] The total power in the delta circuit is the sum of the powers in the three phases. In a delta connected system, the expression of power (P) is P = √3VLILcosφ 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

Answer: c [Reason:] ZPh = √(22+32 )=3.6∠56.3⁰Ω. cosφ = RPh /ZPh = 2/3.6 = 0.55. IL = √3× IPh = 17.32A. Active power = √3 VLILcosφ = √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

Answer: a [Reason:] Sinφ = 0.83. Reactive power = √3 VLILsinφ. VL = 440V, IL = 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

Answer: b [Reason:] Mesh analysis is applicable only for planar networks. A circuit is said to be planar if it can be drawn on a plane surface without crossovers.

2. A mesh is a loop which contains ____ number of loops within it.
a) 1
b) 2
c) 3
d) no loop

Answer: d [Reason:] A loop is a closed path. A mesh is defined as a loop which does not contain any other loops within it.

3. Consider the circuit shown below. The number mesh equations that can be formed are? a) 1
b) 2
c) 3
d) 4

Answer: b [Reason:] We know if there are n loops in the circuit, n mesh equations can be formed. So as there are 2 loops in the circuit. So 2 mesh equations can be formed.

4. In the figure shown in the question 2, the current through loop 1 be I1 and through the loop 2 be I2, then the current flowing through the resistor R2 will be?
a) I1
b) I2
c) I1-I2
d) I1+I2

Answer: c [Reason:] Through the resistor R2 both the currents I1, I2 are flowing. So the current through R2 will be I1-I2.

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

Answer: a [Reason:] Number of mesh equations= B-(N-1). Given number of branches = 5 and number of nodes = 4. So Number of mesh equations = 5-(4-1) =2.

6. Consider the circuit shown in the figure. Find voltage Vx. a) 1
b) 1.25
c) 1.5
d) 1.75

Answer: b [Reason:] Consider current I1 (CW) in the loop 1 and I2 (ACW) in the loop 2. So, the equations will be Vx+I2-I1=0. I1=5/2=2.5A. I2=4Vx/4= Vx. Vx+Vx-2.5=0. Vx =1.25V.

7. Consider the circuit shown below. Find the current I1. a) 3.3
b) 4.3
c) 5.3
d) 6.3

Answer: b [Reason:] According to mesh analysis, (1+3+6)I1 – 3(I2) – 6(I3) =10 -3(I1) + (2+5+3)I2 =4 -6(I1) + 10(I3) = -4 +20 On solving the above equations, I1=4.3A.

8. Find the current I2 (A) in the figure shown in the question 7.
a) 1.7
b) 2.6
c) 3.6
d) 4.6

Answer: a [Reason:] According to mesh analysis, (1+3+6)I1 – 3(I2) – 6(I3) = 10. -3(I1) + (2+5+3)I2 = 4. -6(I11) + 10(I3) = -4 + 20 On solving the above equations, I2 =1.7A.

9. Find the current I3 (A) in the figure shown in the question 7.
a) 4
b) 4.7
c) 5
d) 5.7

Answer: b [Reason:] According to mesh analysis, (1+3+6)I1 – 3(I2) – 6(I3) = 10. -3(I1) + (2+5+3)I2 = 4. -6(I1) + 10(I3) = -4 + 20. On solving the above equations, I3 = 4.7A.

10. Find current through R2 resistor. a) 3
b) 3.25
c) 3.5
d) 3.75

Answer: d [Reason:] Applying mesh analysis, 5(I1) + 2(I1-I2) = 10. 10(I2) + 2(I2-I1) + 40 = 0. On solving, I1 = 0.5A, I2 = -3.25A. So current through R2 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

Answer: c [Reason:] If there are M branch currents, then we can write M number of independent equations. Number of independent equations = M.

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)

Answer: d [Reason:] If there are M meshes, B branches and N nodes including reference node, the number of mesh currents is given as M= B-(N-1).

3. Determine the current I1 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

Answer: b [Reason:] The equation for loop 1 is I1(j4) + 6(I1-I2) = 5∠0⁰. The equation for loop 2 is 6(I1-I2) + (j3) I2 + (2) I2 = 0. Solving the above equations, I1 = 0.855∠-69.5⁰.

4. In the circuit shown in question 3 find the current I2.
a) 0.5∠-90⁰
b) 0.6∠-90⁰
c) 0.7∠-90⁰
d) 0.8∠-90⁰

Answer: b [Reason:] The equation for loop 1 is I1(j4) + 6(I1-I2) = 5∠0⁰. The equation for loop 2 is 6(I1-I2) + (j3) I2 + (2) I2 = 0. Solving the above equations, I2 = 0.6∠-90⁰.

5. Find Z11, Z12, Z13 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

Answer: d [Reason:] Z11= self impedance of loop 1 = (5 + 3 – j4) Ω. Z12 = Impedance common to both loop 1 and loop2 = -5Ω. Z13 = No common impedance between loop1 and loop 3 = 0Ω.

6. Determine Z21, Z22, Z23 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

Answer: b [Reason:] Z21 = common impedance to loop 1 and loop 2 = -5 Ω. Z22 = self impedance of loop2 = (5+j5-j6) Ω. Z23 = common impedance between loop2 and loop 3 = – (-j6) Ω.

7. Find Z31, Z32, Z33 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

Answer: a [Reason:] Z31 = common impedance to loop 3 and loop 1 = 0 Ω. Z32 = common impedance between loop3 and loop 2 = – (-j6) Ω. Z33 = self impedance of loop 3 = (4-j6) Ω.

8. Find the values of Z11, Z22, Z33 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

Answer: b [Reason:] Z11= self impedance of loop 1 = (4 + j3) Ω. Z22 = self impedance of loop2 = (j3+3+j4-j5) Ω. Z33 = self impedance of loop 3 = (-j5+5) Ω.

9. Find the common impedances Z12, Z13, Z21, Z23, Z31, Z32 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

Answer: a [Reason:] The common impedances Z12 and Z21 are Z12= Z21 = -j3Ω. The common impedances Z13 and Z31 are Z13 = Z31 =0Ω. The common impedances Z23 and Z32 are Z23 = Z32 = j5Ω.

10. Find the value V2 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

Answer: b [Reason:] The three loop equations are (4+j3)I1 –(j3)I2 = 20∠0⁰. (-j3)I1 + (3+j2)I2 + (j5)I3 = 0. (j5)I2 + (5-j5)I3 = -V2. The current through (3+j4) Ω is zero, I2= ∆2/∆ =0

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

On solving, V2 = 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’

Answer: b [Reason:] Millman’s Theorem states that if there are voltage sources V1, V2,…… Vn with internal resistances R1, R2,…..Rn, 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=(V1G1+V2G2+⋯.+VnGn)
b) V=((V1G1+V2G2+⋯.+VnGn))/((1/G1+1/G2+⋯1/Gn))
c) V=((V1G1+V2G2+⋯.+VnGn))/(G1+G2+⋯Gn)
d) V=((V1/G1+V2/G2+⋯.+Vn/Gn))/( G1+G2+⋯Gn)

Answer: c [Reason:] The value of equivalent voltage source is V= ((V1G1+V2G2+⋯.+VnGn))/(G1+G2+⋯Gn).

3. In the question above the value of equivalent resistance is?
a) R’= G1+G2+⋯Gn
b) R’=1/G1+1/G2+⋯1/Gn
c) R’=1/((G1+G2+⋯Gn) )
d) R’=1/(1/G1+1/G2+⋯1/Gn)

Answer: c [Reason:] Let the equivalent resistance is R’. The value of equivalent resistance is R’=1/((G1+G2+⋯Gn) ).

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’

Answer: c [Reason:] Millman’s Theorem states that if there are current sources I1,I2,…… In with internal conductances G1,G2,…..Gn, 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=((I1R1+I2R2+⋯.+InRn))/(R1+R2+⋯Rn)
b) I’= I1R1+I2R2+⋯.+InRn
c) I’=((I1/R1+I2/R2+⋯.+In/Rn))/( R1+R2+⋯Rn)
d) I’=I1/R1+I2/R2+⋯.+In/Rn

Answer: a [Reason:] The value of equivalent current source is I=((I1R1+I2R2+⋯.+InRn))/(R1+R2+⋯Rn).

6. In the question above, the value of equivalent conductance is?
a) G’= R1+R2+⋯Rn
b) G’=1/(1/R1+1/R2+⋯1/Rn)
c) G’=1/((R1+R2+⋯Rn) )
d) G’=1/R1+1/R2+⋯1/Rn

Answer: c [Reason:] Let the equivalent conductance is G’. The value of equivalent conductance is G’=1/((R1+R2+⋯Rn) ).

7. Calculate the current through 3Ω resistor in the circuit shown below. a) 1
b) 2
c) 3
d) 4

Answer: c [Reason:] Applying Nodal analysis the voltage V is given by (10-V)/2+(20-V)/5=V/3. V=8.7V. Now the current through 3Ω resistor in the circuit is I = V/3 = 8.7/3 = 2.9A ≅ 3A.

8. Find the current through 3Ω resistor in the circuit shown above using Millman’s Theorem.
a) 4
b) 3
c) 2
d) 1

Answer: b [Reason:] V=((V1G1+V2G2))/(G1+G2)=(10(1/2)+20(1/5))/(1/2+1/5)=12.86V. R’=1/((G1+G2) )=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

Answer: b [Reason:] Applying Nodal analysis the voltage V is given by (5-V)/1+(10-V)/3=V/4. V=6V. The current through 4Ω resistor I = V/4 = 6/4 = 1.5A.

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

Answer: c [Reason:] V=((V1G1+V2G2))/(G1+G2)=(5(1/1)+10(1/3))/(1/1+1/3)=6.25V. R’=1/((G1+G2) )=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

Answer: a [Reason:] Voltage transfer ratio is the ratio of voltage transform at first port to the voltage transform at the second port and is denoted by G(s). G21 = V2(s)/V1(s) G12 = V1(s)/V2(s).

2. The ratio of the current transform at one port to current transform at other port is called?
b) Transfer impedance
c) Current transfer ratio
d) Voltage transfer ratio

Answer: c [Reason:] Current transfer ratio is the ratio of the current transform at one port to current transform at other port and is denoted by α(s). α12(s) = I1(s)/I2(s) α21(s) = I2(s)/I1(s).

3. The ratio of voltage transform at first port to the current transform at the second port is called?
a) Voltage transfer ratio
c) Current transfer ratio
d) Transfer impedance

Answer: d [Reason:] Transfer impedance is the ratio of voltage transform at first port to the current transform at the second port and is denoted by Z(s). Z21(s) = V2(s)/I1(s) Z12(s) = V1(s)/I2(s).

4. For the network shown in the figure, find the driving point impedance. a) (s2-2s+1)/s
b) (s2+2s+1)/s
c) (s2-2s-1)/s
d) (s2+2s-1)/s

Answer: b [Reason:] Applying Kirchoff’s law at port 1, Z(S)=V(S)/I(S), where V(s) is applied at port 1 and I(s) is current flowinmg through the network. Then Z(S)=V(S)/I(S) = 2+S+1/S = (s2+2s+1)/s.

5. Obtain the transfer function G21 (S) in the circuit shown below. a) (s+1)/s
b) s+1
c) s
d) s/(s+1)

Answer: d [Reason:] Applying Kirchhoff’s law V1 (S) = 2 I1 (S) + 2 sI1 (S) V2 (S) = I1 (S) X 2s Hence G21 (S) = V2(s)/V1(s) =2 s/(2+2 s)=s/(s+1).

6. Determine the transfer function Z21 (S) in the circuit shown in question 5.
a) s
b) 2 s
c) 3 s
d) 4 s

Answer: b [Reason:] The transfer function Z21 (S) is Z21 (S) = V2(S)/I1(S). V2 (S) = I1 (S) X 2s. V2(S)/I1(S)=2s. On substituting Z21 (S) = 2s.

7. Find the driving point impedance Z11 (S) in the circuit shown in question 5.
a) 2(s+2)
b) (s+2)
c) 2(s+1)
d) (s+1)

Answer: c [Reason:] The driving point impedance Z11 (S) is Z11 (S)=V1(S)/I1(S). V1 (S) = 2 I1 (S) + 2 sI1 (S) => V1(S) = (2+2s)I1(S) => V1(S)/I1(S) = 2(s+1). On substituting Z11 (S) = 2(S+1).

8. Obtain the transfer function G21 (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)

Answer: d [Reason:] From the circuit, the parallel combination of resistance and capacitance can be combined into equivalent in impedance. Zeq(S) = 1/(2 S+1/2)=2/(4 S+1). Applying Kirchhoff’s laws, we have V2 (S) = 2 I1(S) => V1 (S) = I1 (S)[2/(4 S+1)+2] = I1 (S)[(8 S+4)/(4 S+1)] The transfer function G21 (s) = V2(s)/V1(s) =2 I1(S)/((8 S+4)/(4 S+1))I1(S) =(8 S+2)/(8 S+4).

9. Obtain the transfer function Z21(s) in the circuit shown in question 8.
a) 1
b) 2
c) 3
d) 4

Answer: b [Reason:] The transfer function Z21(s) is Z21 (S) = V2(S)/I1(S). V2 (S) = 2 I1(S) => V2 (S)/I1 =2. On substituting Z21(s) = 2.

10. Determine the driving point impedance Z11(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)