## 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

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2. Nodal analysis can be applied for non planar networks also.

a) true

b) false

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3. In nodal analysis how many nodes are taken as reference nodes?

a) 1

b) 2

c) 3

d) 4

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

a) 8V

b) 9V

c) 10V

d) 11V

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_{1}= (4-V)/2, I

_{2}= (V+6)/3. The nodal equation at node P will be I

_{1}+3=I

_{2}. On solving, V=9V.

5. Find the resistor value R_{1}(Ω) in the figure shown below.

a) 10

b) 11

c) 12

d) 13

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_{1}-V

_{2})/14+(V

_{1}-V

_{3})/R

_{1}. From the circuit, V

_{1}=100V, V

_{2}=15×2=30V, V

_{3}=40V. On solving, R

_{1}=12Ω.

6. Find the value of the resistor R_{2} (Ω) in the circuit shown in the question 5.

a) 5

b) 6

c) 7

d) 8

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_{1}=100V, V

_{2}=15×2=30V, V

_{3}=40V. (V

_{1}-V

_{2})/14+(V

_{1}-V

_{3})/R

_{2}=15. On solving we get R

_{2}= 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

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_{1}-(1/3)V

_{2}= 10/1. At node 2, -(1/3)V

_{1}+(1/3+1/6+1/5)V

_{2}= 2/5+5/6. On solving above equations, we get V

_{1}=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

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_{1}-(1/3)V

_{2}= 10/1. At node 2, -(1/3)V

_{1}+(1/3+1/6+1/5)V

_{2}= 2/5+5/6. On solving above equations, we get V

_{2}=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

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_{1}/10+(V

_{1}-V

_{2})/3. At node 2, (V

_{2}-V

_{1})/3+V

_{2}/5+(V

_{2}-10)/1=0. On solving the above equations, we get V

_{1}=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

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_{1}/10+(V

_{1}-V

_{2})/3. At node 2, (V

_{2}-V

_{1})/3+V

_{2}/5+(V

_{2}-10)/1=0. On solving the above equations, we get V

_{2}=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

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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

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_{a}-10∠0

^{o})/j6+V

_{a}/(-j6)+(V

_{a}-Vb)/3=0. Applying nodal analysis at node ‘b’, (V

_{b}-V

_{a})/3+V

_{b}/j4+V

_{b}/j1=0. Solving the above equations we get, V

_{a}= 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

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_{a}-10∠0

^{o})/j6+V

_{a}/(-j6)+(V

_{a}-V

_{b})/3=0. Applying nodal analysis at node ‘b’, (V

_{b}-V

_{a})/3+V

_{b}/j4+V

_{b}/j1=0. Solving the above equations we get, V

_{b}= -1.34∠-180⁰V.

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

a) (5∠0^{o})/3

b) – (5∠0^{o})/3

c) (10∠0^{o})/3

d) – (10∠0^{o})/3

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_{aa}V

_{a}+Y

_{ab}V

_{b}= I

_{1}, Y

_{ba}V

_{a}+Y

_{bb}V

_{b}= I

_{2}. We get Y

_{aa}=1/3+1/j4+1/(-j6) and the self admittance at node a is the sum of admittances connected to node a. Y

_{ab}=-(1/(-j6)). I

_{1}= (10∠0

^{o})/3=x.

5. Find the value of ‘y’ in the equation –(-1/j6)V_{a}+(1/5+1/j5-1/j6)V_{b}=y obtained from the circuit shown in the question 4.

a) (10∠30^{o})/5

b) -(10∠30^{o})/5

c) (5∠30^{o})/5

d) (-5∠30^{o})/5

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_{bb}=1/5+1/j5-1/j6 and Y

_{ab}=–(-1/j6). The mutual admittance between node b and a is the sum of the admittances between nodes b and a. I

_{2}=-(10∠30

^{o})/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

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^{o})/3+Va/(-j4)+Va/(2+j5)=0. On solving, Va = 16.27∠18.91⁰. Current in 2Ω resistor I

_{2}= Va/(2+j5)=(16.27∠18.91

^{o})/(5.38∠68.19

^{o})=3.02∠-49.28

^{o}. Power dissipated in 2Ω resistor P=I

_{2}

^{2}R=3.02

^{2}×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

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_{3}= (-20∠30

^{o}+16.27∠18.91

^{o})/3=1.71∠-112

^{o}. Power dissipated in 3Ω resistor P=I

_{3}

^{2}R=1.71

^{2}×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

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9. For the circuit shown below, find the voltage across the resistance R_{L} if R_{L} is infinite.

a) 3

b) 2

c) 1

d) 0

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_{L}is infinite, the voltage across it will be 0. So the voltage across the resistance R

_{L}if R

_{L}is infinite is zero.

10. Find the voltage V_{ab} 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

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^{o})/(3+2)+(Va-20∠90

^{o})/(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

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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

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3. Find the current through 6Ω resistor in the circuit shown above.

a) 1

b) 1.43

c) 2

d) 2.43

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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

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

a) 1

b) 2

c) 3

d) 4

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_{1})=0, I

_{1}=2A. 10-5(I

_{2}), I

_{2}=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

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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

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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

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9. Find the current through 5Ω resistor in the circuit shown above.

a) 7

b) 8

c) 9

d) 10

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10. Find the voltage drop across 5Ω resistor in the circuit shown above.

a) 33

b) 34

c) 35

d) 36

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## 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

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2. The expression of Norton’s current (I_{N}) in the circuit shown below is?

a) V/Z_{1}

b) V/Z_{2}

c) V(Z_{2}/(Z_{1}+Z_{2}))

d) VZ_{1}/(Z_{1}+Z_{2})

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_{N}= V/Z

_{1}.

3. The expression of equivalent impedance (Z_{N}) in the circuit shown in the question 2 is?

a) (Z_{1}+Z_{2})/Z_{1}

b) (Z_{1}+Z_{2})/Z_{2}

c) Z_{1}Z_{2}/(Z_{1}+Z_{2})

d) Z_{1}+Z_{2}

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_{N}= Z

_{1}Z

_{2}/(Z

_{1}+Z

_{2}).

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

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^{o})/(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

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^{o}Ω. 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

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_{ab}=(j3)(-j2)/((j3)-(j2) )=6∠-90

^{o}Ω.

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

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^{o})/(3∠90

^{o})+(5∠90

^{o})/(2∠-90

^{o}) =4.16∠-126.8

^{o}A.

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

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_{L}= I×(6∠-90

^{o})/(5+6∠-90

^{o}) = 3.19∠-166.61

^{o}A.

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

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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

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## 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 I_{0} and the voltage at the capacitor is V_{0}. The inductor is represented by a transform impedance _________ in series with a voltage source __________

a) Ls, L V_{0}

b) Ls, LI_{0}

c) 1/Ls, LI_{0}

d) 1/Ls, L V_{0}

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_{0}. It is represented by a transform impedance Ls in series with a voltage source L V

_{0}.

2. In the circuit shown in question 1, the capacitor is replaced by a transform impedance of __________ with an initial voltage ___________

a) 1/Cs, V_{0}/S

b) 1/Cs, I_{0}/S

c) Cs, I_{0}/S

d) Cs, V_{0}/S

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_{0}across it. It is represented by a transform impedance of 1/Cs with an initial voltage V

_{0}/S.

3. The value of the total voltage after replacing the inductor and capacitor in question 1 is?

a) V_{1}(S)-LI_{0}-V_{0}/S

b) V_{1}(S)+LI_{0}-V_{0}/S

c) V_{1}(S)+LI_{0}+V_{0}/S

d) V_{1}(S)-LI_{0}+V_{0}/S

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_{1}(S)+LI

_{0}-V

_{0}/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

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5. The current flowing in the circuit in question 1 is?

a) (V_{1}(S)-LI_{0}-V_{0}/S)/( R+LS+1/CS)

b) (V_{1}(S)-LI_{0}+V_{0}/S)/( R+LS+1/CS)

c) (V_{1}(S)+LI_{0}+V_{0}/S)/( R+LS+1/CS)

d) (V_{1}(S)+LI_{0}-V_{0}/S)/( R+LS+1/CS)

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_{1}(S)+LI

_{0}-V

_{0}/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

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_{1}(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)

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_{1}(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)

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_{1}(s) = 1/(s+4) and now the impedance of capacitor is 1/2s. So the series combination of the last elements is Z

_{2}(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) (4s^{2}-19s+4)/(6s-8)

b) (4s^{2}+19s-4)/(6s+8)

c) (4s^{2}+19s-4)/(6s-8)

d) (4s^{2}+19s+4)/(6s+8)

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_{2}(s) =1/2s+1/(s+4)=(3s+4)/2s(s+4) , the admittance parallel combination of the last elements is Y

_{2}(s) = 1/2+2s(s+4)/( 3s+4)=(4s

^{2}+19s+4)/(6s+8).

10. Obtain the transform impedance of the network shown in question 6.

a) (6s-8)/( 4s^{2}+19s-4)

b) (6s+8)/( 4s^{2}+19s+4)

c) (6s+8)/( 4s^{2}-19s+4)

d) (6s-8)/( 4s^{2}+19s+4)

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_{2}(s) = (4s

^{2}+19s+4)/(6s+8). So the transform impedance of the network is Z (s) = 1 / Y

_{2}(s) = (6s+8)/( 4s

^{2}+19s+4).