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Network Theory MCQ Set 1

1. If there are 8 nodes in network, we can get ____ number of equations in the nodal analysis.
a) 9
b) 8
c) 7
d) 6

Answer: c [Reason:] Number of equations=N-1= 7. So as there are 8 nodes in network, we can get 7 number of equations in the nodal analysis.

2. Nodal analysis can be applied for non planar networks also.
a) true
b) false

Answer: a [Reason:] Nodal analysis is applicable for both planar and non planar networks. Each node in a circuit can be assigned a number or a letter.

3. In nodal analysis how many nodes are taken as reference nodes?
a) 1
b) 2
c) 3
d) 4

Answer: a [Reason:] In nodal analysis only one node is taken as reference node. And the node voltage is the voltage of a given node with respect to one particular node called the reference node.

4. Find the voltage at node P in the figure shown.

a) 8V
b) 9V
c) 10V
d) 11V

Answer: b [Reason:] I1= (4-V)/2, I2= (V+6)/3. The nodal equation at node P will be I1+3=I2. On solving, V=9V.

5. Find the resistor value R1(Ω) in the figure shown below.

a) 10
b) 11
c) 12
d) 13

Answer: c [Reason:] 10=(V1-V2)/14+(V1-V3)/R1. From the circuit, V1=100V, V2=15×2=30V, V3=40V. On solving, R1=12Ω.

6. Find the value of the resistor R2 (Ω) in the circuit shown in the question 5.
a) 5
b) 6
c) 7
d) 8

Answer: b [Reason:] As V1=100V, V2=15×2=30V, V3=40V. (V1-V2)/14+(V1-V3)/R2=15. On solving we get R2= 6Ω.

7. Find the voltage (V) at node 1 in the circuit shown.

a) 5.32
b) 6.32
c) 7.32
d) 8.32

Answer: b [Reason:] At node 1, (1/1+1/2+1/3)V1-(1/3)V2= 10/1. At node 2, -(1/3)V1+(1/3+1/6+1/5)V2= 2/5+5/6. On solving above equations, we get V1=6.32V.

8. Find the voltage (V) at node 2 in the circuit shown in the question 7.
a) 2.7
b) 3.7
c) 4.7
d) 5.7

Answer: c [Reason:] At node 1, (1/1+1/2+1/3)V1-(1/3)V2= 10/1. At node 2, -(1/3)V1+(1/3+1/6+1/5)V2= 2/5+5/6. On solving above equations, we get V2=4.7V.

9. Find the voltage at node 1 of the circuit shown below.

a) 32.7
b) 33.7
c) 34.7
d) 35.7

Answer: b [Reason:] Applying Kirchhoff’s current law at node 1, 10= V1/10+(V1-V2)/3. At node 2, (V2-V1)/3+V2/5+(V2-10)/1=0. On solving the above equations, we get V1=33.7V.

10. Find the voltage at node 2 of the circuit shown in the question 9.
a) 13
b) 14
c) 15
d) 16

Answer: b [Reason:] Applying Kirchhoff’s current law at node 1, 10= V1/10+(V1-V2)/3. At node 2, (V2-V1)/3+V2/5+(V2-10)/1=0. On solving the above equations, we get V2=14V.

Network Theory MCQ Set 2

1. If there are N nodes in a circuit, then the number of nodal equations that can be formed are?
a) N+1
b) N
c) N-1
d) N-2

Answer: c [Reason:] If there are N nodes in a circuit, then the number of nodal equations that can be formed are N-1. Number of nodal equations = N-1.

2. In the network shown below, find the voltage at node ‘a’.

a) 5.22∠104.5⁰
b) 5.22∠-104.5⁰
c) 6.22∠104.5⁰
d) 6.22∠-104.5⁰

Answer: b [Reason:] Applying nodal analysis at node ‘a’, (Va-10∠0o)/j6+Va/(-j6)+(Va-Vb)/3=0. Applying nodal analysis at node ‘b’, (Vb-Va)/3+Vb/j4+Vb/j1=0. Solving the above equations we get, Va = 5.22∠-104.5⁰V.

3. Determine the voltage at node ‘b’ in the circuit shown in the question 2.
a) -1.34∠-180⁰
b) 1.34∠-180⁰
c) -0.34∠-180⁰
d) 0.34∠-180⁰

Answer: a [Reason:] Applying nodal analysis at node ‘a’, (Va-10∠0o)/j6+Va/(-j6)+(Va-Vb)/3=0. Applying nodal analysis at node ‘b’, (Vb-Va)/3+Vb/j4+Vb/j1=0. Solving the above equations we get, Vb = -1.34∠-180⁰V.

4. In the circuit shown below we get a nodal equation as (1/3+1/j4-1/j6)Va—(-1/j6)Vb=x. Find the value of ‘x‘.

a) (5∠0o)/3
b) – (5∠0o)/3
c) (10∠0o)/3
d) – (10∠0o)/3

Answer: c [Reason:] The general equations are YaaVa+YabVb = I1, YbaVa+YbbVb = I2. We get Yaa=1/3+1/j4+1/(-j6) and the self admittance at node a is the sum of admittances connected to node a. Yab=-(1/(-j6)). I1= (10∠0o)/3=x.

5. Find the value of ‘y’ in the equation –(-1/j6)Va+(1/5+1/j5-1/j6)Vb=y obtained from the circuit shown in the question 4.
a) (10∠30o)/5
b) -(10∠30o)/5
c) (5∠30o)/5
d) (-5∠30o)/5

Answer: b [Reason:] We got Ybb=1/5+1/j5-1/j6 and Yab=–(-1/j6). The mutual admittance between node b and a is the sum of the admittances between nodes b and a. I2=-(10∠30o)/5=y.

6. In the circuit shown below find the power dissipated by 2Ω resistor.

a) 16.24
b) 17.24
c) 18.24
d) 19.24

Answer: c [Reason:] Applying nodal analysis at node ‘a’, (Va-20∠〖30o)/3+Va/(-j4)+Va/(2+j5)=0. On solving, Va = 16.27∠18.91⁰. Current in 2Ω resistor I2= Va/(2+j5)=(16.27∠18.91o)/(5.38∠68.19o )=3.02∠-49.28o. Power dissipated in 2Ω resistor P=I22 R=3.022×2= 18.24W.

7. In the circuit shown in the question 6 determine the power dissipated in 3Ω resistor.
a) 7.77
b) 8.77
c) 9.77
d) 10.77

Answer: b [Reason:] Current in 3Ω resistor I3 = (-20∠30o+16.27∠18.91o)/3=1.71∠-112o. Power dissipated in 3Ω resistor P=I32 R=1.712×3=8.77W.

8. In the circuit shown in the question 6 find the power output of the source.
a) 27
b) 28
c) 29
d) 30

Answer: a [Reason:] Total power delivered by the source is the product of voltage and current and is given by power output of the source VIcosφ = 20 x 1.71cos142⁰ = 26.95W.

9. For the circuit shown below, find the voltage across the resistance RL if RL is infinite.

a) 3
b) 2
c) 1
d) 0

Answer: d [Reason:] If RL is infinite, the voltage across it will be 0. So the voltage across the resistance RL if RL is infinite is zero.

10. Find the voltage Vab in the circuit shown question 9.
a) 21.66∠-45.02⁰
b) 20.66∠-45.02⁰
c) 21.66∠45.02⁰
d) 20.66∠45.02⁰

Answer: c [Reason:] Applying nodal analysis at node ‘a’, (Va-20∠0o)/(3+2)+(Va-20∠90o)/(j4+3)=0. On solving, we get Va = 21.66∠45.02⁰V.

Network Theory MCQ Set 3

1. Find the current flowing between terminals A and B of the circuit shown below.

a) 1
b) 2
c) 3
d) 4

Answer: d [Reason:] The magnitude of the current in the Norton’s equivalent circuit is equal to the current passing through the short circuited terminals that is I=20/5=4A.

2. Find the equivalent resistance between terminals A and B of the circuit shown below.
a) 0.33
b) 3.33
c) 33.3
d) 333

Answer: b [Reason:] Norton’s resistance is equal to the parallel combination of both the 5Ω and 10Ω resistors that is R = (5×10)/15 = 3.33Ω.

3. Find the current through 6Ω resistor in the circuit shown above.
a) 1
b) 1.43
c) 2
d) 2.43

Answer: b [Reason:] The current passing through the 6Ω resistor and the voltage across it due to Norton’s equivalent circuit is I = 4×3.33/(6+3.33) = 1.43A.

4. Find the voltage drop across 6Ω resistor in the circuit shown above.
a) 6.58
b) 7.58
c) 8.58
d) 9.58

Answer: c [Reason:] The voltage across the 6Ω resistor is V = 1.43×6 = 8.58V. So the current and voltage have same values both in the original circuit and Norton’s equivalent circuit.

5. Find the current flowing between terminals A and B.

a) 1
b) 2
c) 3
d) 4

Answer: d [Reason:] Short circuiting terminals A and B, 20-10(I1)=0, I1=2A. 10-5(I2), I2=2A. Current flowing through terminals A and B= 2+2 = 4A.

6. Find the equivalent resistance between terminals A and B
a) 3
b) 3.03
c) 3.33
d) 3.63

Answer: c [Reason:] The resistance at terminals AB is the parallel combination of the 10Ω resistor and the 5Ω resistor => R = ((10×5))/(10+5) = 3.33Ω.

7. Find the current flowing between terminals A and B obtained in the equivalent Nortan’s circuit.
a) 8
b) 9
c) 10
d) 11

Answer: d [Reason:] To solve for Norton’s current we have to find the current passing through the terminals A and B. Short circuiting the terminals a and b, I=100/((6×10)/(6+10)+(15×8)/(15+8))=11.16 ≅ 11A.

8. Find the equivalent resistance between terminals A and B obtained in the equivalent Nortan’s circuit.
a) 8
b) 9
c) 10
d) 11

Answer: b [Reason:] The resistance at terminals AB is the parallel combination of the 10Ω resistor and the 6Ω resistor and parallel combination of the 15Ω resistor and the 8Ω resistor => R=(10×6)/(10+6)+(15×8)/(15+8)=8.96≅9Ω.

9. Find the current through 5Ω resistor in the circuit shown above.
a) 7
b) 8
c) 9
d) 10

Answer: a [Reason:] To solve for Norton’s current we have to find the current passing through the terminals A and B. Short circuiting the terminals a and b I=11.16×8.96/(5+8.96) = 7.16A.

10. Find the voltage drop across 5Ω resistor in the circuit shown above.
a) 33
b) 34
c) 35
d) 36

Answer: d [Reason:] The voltage drop across 5Ω resistor in the circuit is the product of current and resistance => V = 5×7.16 = 35.8 ≅ 36V.

Network Theory MCQ Set 4

1. Norton’s current is equal to the current passing through the ___________ circuited ___________ terminals.
a) short, input
b) short, output
c) open, output
d) open, input

Answer: b [Reason:] Norton’s current is equal to the current passing through short circuited output terminals not the current through open circuited output terminals.

2. The expression of Norton’s current (IN) in the circuit shown below is?

a) V/Z1
b) V/Z2
c) V(Z2/(Z1+Z2))
d) VZ1/(Z1+Z2)

Answer: a [Reason:] The Norton’s equivalent form of any complex impedance circuit consists of an equivalent current source and an equivalent impedance. The expression of Norton’s current is IN = V/Z1.

3. The expression of equivalent impedance (ZN) in the circuit shown in the question 2 is?
a) (Z1+Z2)/Z1
b) (Z1+Z2)/Z2
c) Z1Z2/(Z1+Z2)
d) Z1+Z2

Answer: c [Reason:] The impedance between the points a and b with the source replaced by a short circuit is Norton’s equivalent impedance. The Norton’s equivalent impedance is ZN = Z1Z2/(Z1+Z2).

4. Determine Norton’s equivalent current in the circuit shown below.

a) 5∠53.13⁰
b) 4∠53.13⁰
c) 4∠53.13⁰
d) 5∠-53.13⁰

Answer: d [Reason:] The current through the terminals a and b is the Norton’s equivalent current. The Norton’s equivalent current is I = (25∠0o)/(3+j4) = 5∠-53.13⁰A.

5. The Norton’s equivalent impedance in the circuit shown in the question 4.
a) 4.53∠9.92⁰
b) 4.53∠-9.92⁰
c) 5.53∠9.92⁰
d) 5.53∠-9.92⁰

Answer: a [Reason:] The Norton’s equivalent impedance is Z = (3+j4)(4-j5)/((3+j4)+(4-j5) )=4.53∠9.92oΩ. The impedance between the points a and b with the source replaced by a short circuit is Norton’s equivalent impedance.

6. Determine the Norton’s impedance seen from terminals ‘ab’.

a) 6∠90⁰
b) 7∠90⁰
c) 6∠-90⁰
d) 7∠-90⁰

Answer: c [Reason:] The impedance between the points a and b with the source replaced by a short circuit is Norton’s equivalent impedance. The Norton’s impedance is Zab=(j3)(-j2)/((j3)-(j2) )=6∠-90oΩ.

7. Find the Norton’s current passing through ‘ab’ in the circuit shown in the question 6.
a) 4.16∠126.8⁰
b) 5.16∠126.8⁰
c) 5.16∠-126.8⁰
d) 4.16∠-126.8⁰

Answer: d [Reason:] The Norton’s current is equal to the current passing through the short circuit between the points a and b. The Norton’s current is I=(10∠0o)/(3∠90o)+(5∠90o)/(2∠-90o) =4.16∠-126.8oA.

8. Find the load current in the circuit shown in the question 6.
a) 3.19∠166.61⁰
b) 3.19∠-166.61⁰
c) 4.19∠166.61⁰
d) 4.19∠-166.61⁰

Answer: b [Reason:] The load current in the circuit is given by IL = I×(6∠-90o)/(5+6∠-90o) = 3.19∠-166.61oA.

9. Determine Norton’s equivalent impedance in the circuit shown below.

a) (5+j6) Ω
b) (5-j6) Ω
c) (6+j7) Ω
d) (6-j7) Ω

Answer: a [Reason:] The impedance seen from the terminals when the source is reduced to zero is Z = (5+j6) Ω.

10. Find the Norton’s current in the circuit shown in the question 9.
a) 40∠30⁰
b) 40∠-30⁰
c) 30∠30⁰
d) 30∠-30⁰

Answer: c [Reason:] The Norton’s current is equal to the current passing through the short circuit between the points a and b. So the current passing through the short circuited terminals ‘a’ and ‘b’ is I = 30∠30⁰A.

Network Theory MCQ Set 5

1. In the circuit shown below, switch K is moved from position to position 2 at time t = 0. At time t = 0-, the current in the inductor is I0 and the voltage at the capacitor is V0. The inductor is represented by a transform impedance _________ in series with a voltage source __________

a) Ls, L V0
b) Ls, LI0
c) 1/Ls, LI0
d) 1/Ls, L V0

Answer: a [Reason:] The inductor has an initial current I0. It is represented by a transform impedance Ls in series with a voltage source L V0.

2. In the circuit shown in question 1, the capacitor is replaced by a transform impedance of __________ with an initial voltage ___________
a) 1/Cs, V0/S
b) 1/Cs, I0/S
c) Cs, I0/S
d) Cs, V0/S

Answer: a [Reason:] The capacitor has an initial voltage V0 across it. It is represented by a transform impedance of 1/Cs with an initial voltage V0/S.

3. The value of the total voltage after replacing the inductor and capacitor in question 1 is?
a) V1(S)-LI0-V0/S
b) V1(S)+LI0-V0/S
c) V1(S)+LI0+V0/S
d) V1(S)-LI0+V0/S

Answer: b [Reason:] The current I(s) is given as the total transform voltage in the circuit divided by the total transform impedance. The value of the total voltage after replacing the inductor and capacitor is V (s) = V1(S)+LI0-V0/S.

4. The value of the total impedance after replacing the inductor and capacitor in question 1 is?
a) R-LS-1/CS
b) R-LS+1/CS
c) R+LS+1/CS
d) R+LS-1/CS

Answer: c [Reason:] The value of the total impedance after replacing the inductor and capacitor is Z (s) = R+LS+1/CS. By knowing the V(s) and Z(s) we can calculate I(s) as I(s) is given as the total transform voltage in the circuit divided by the total transform impedance.

5. The current flowing in the circuit in question 1 is?
a) (V1(S)-LI0-V0/S)/( R+LS+1/CS)
b) (V1(S)-LI0+V0/S)/( R+LS+1/CS)
c) (V1(S)+LI0+V0/S)/( R+LS+1/CS)
d) (V1(S)+LI0-V0/S)/( R+LS+1/CS)

Answer: d [Reason:] The current I(s) is given as the total transform voltage in the circuit divided by the total transform impedance. The current flowing in the circuit is I (s) = V(s)/I(s) =(V1(S)+LI0-V0/S)/( R+LS+1/CS).

6. Obtain the admittance of the last two elements in the parallel combination after transformation in the circuit shown below.

a) 1+s
b) 2+s
c) 3+s
d) 4+s

Answer: d [Reason:] The term admittance is defined as the inverse of impedance. The admittance of capacitor is 1/s and the admittance of resistor is 1/4 mho. So the admittance of the last two elements in the parallel combination is Y1(s) = 4 + s.

7. The impedance of the last two elements in the parallel combination after transformation in the circuit shown in question 6 is?
a) 1/(s+4)
b) 1/(s+3)
c) 1/(s+2)
d) 1/(s+1)

Answer: a [Reason:] The impedance of resistor is 4Ω and the impedance of capacitor is s. So the impedance of the last two elements in the parallel combination is Z1(s) = 1/(s+4).

8. The series combination of the last elements after replacing 1/s and 1/4Ω with 1/(S+4) in the question 6 is?
a) (3s+4)/2s(s-4)
b) (3s-4)/2s(s-4)
c) (3s+4)/2s(s+4)
d) (3s-4)/2s(s+4)

Answer: c [Reason:] We got the impedance of last two elements in parallel combination as Z1(s) = 1/(s+4) and now the impedance of capacitor is 1/2s. So the series combination of the last elements is Z2(s) =1/2s+1/(s+4)=(3s+4)/2s(s+4).

9. Determine the admittance parallel combination of the last elements after replacing with (3s+4)/2s(s+4) in the question 6 is?
a) (4s2-19s+4)/(6s-8)
b) (4s2+19s-4)/(6s+8)
c) (4s2+19s-4)/(6s-8)
d) (4s2+19s+4)/(6s+8)