## Network Theory MCQ Set 1

1. Consider the circuit shown below. Find the equivalent Thevenin’s voltage between nodes A and B.

a) 8

b) 8.5

c) 9

d) 9.5

### View Answer

_{th}= 10×12/14 = 8.57V.

2. In the circuit shown above in question 1 find the thevenin’s resistance between terminals A and B.

a) 1

b) 2

c) 1.7

d) 2.7

### View Answer

_{th}= (12×2)/14 = 1.71Ω.

3. In the figure shown above in question 1 find the current flowing through 24Ω resistor.

a) 0.33

b) 0.66

c) 0

d) 0.99

### View Answer

4. Determine the equivalent thevenin’s voltage between terminals A and B in the circuit shown below.

a) 0.333

b) 3.33

c) 33.3

d) 333

### View Answer

_{th}=V

_{AB}=50-V=50-16.7=33.3V.

5. Find the equivalent thevenin’s resistance between terminals A and B in the circuit shown above in question 4.

a) 333

b) 33.3

c) 3.33

d) 0.333

### View Answer

_{th}, two voltage sources are removed and replaced with short circuit. The resistance at terminals AB then is the parallel combination of the 10Ω resistor and 5Ω resistor => R

_{th}=(10×5)/15=3.33Ω.

6. Determine the equivalent thevenin’s voltage between terminals A and B in the circuit shown below.

a) 5

b) 15

c) 25

d) 35

### View Answer

7. Find the equivalent thevenin’s resistance between terminals A and B in the circuit shown above in question 6.

a) 6

b) 6.25

c) 6.5

d) 6.75

### View Answer

_{th}, two voltage sources are removed and replaced with short circuit => R

_{th}=(10×6)/(10+6)+3=6.75Ω.

8. Determine the equivalent thevenin’s voltage between terminals ‘a’ and ‘b’ in the circuit shown below.

a) 0.7

b) 1.7

c) 2.7

d) 3.7

### View Answer

_{a}=(100×6)/16=37.5V, The voltage at terminal b is V

_{b}=(100×8)/23=34.7V. So the voltage across the terminals ab is V

_{ab}=V

_{a}-V

_{b}=37.5-34.7=2.7V.

9. Find the equivalent thevenin’s resistance between terminals A and B in the circuit shown above in question 8.

a) 6

b) 7

c) 8

d) 9

### View Answer

_{th}, two voltage sources are removed and replaced with short circuit => R

_{ab}=(6×10)/(6+10)+(8×15)/(8+15)=8.96≅9V.

10. Find the current through 5Ω resistor in the circuit shown above in question 8.

A) 0.1

b) 0.2

c) 0.3

d) 0.4

### View Answer

## Network Theory MCQ Set 2

1. Thevenin’s voltage is equal to the _____________ voltage across the _______________ terminals.

a) short circuit, input

b) short circuit, output

c) open circuit, output

d) open circuit, input

### View Answer

2. Consider the circuit shown below. The expression of Thevenin’s voltage (V_{Th}) is?

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

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

c) V(Z_{1})

d) V(Z_{2})

### View Answer

_{Th}= V(Z

_{2}/(Z

_{1}+Z

_{2})).

3. The value of Z_{Th} in the circuit shown in the question 2 is?

a) Z_{3}+( Z_{1}Z_{2}/(Z_{1}+Z_{2}))

b) Z_{1}+( Z_{3}Z_{2}/(Z_{3}+Z_{2}))

c) Z_{2}+( Z_{1}Z_{3}/(Z_{1}+Z_{3}))

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

### View Answer

_{Th}= Z

_{3}+(Z

_{1}Z

_{2}/(Z

_{1}+Z

_{2})).

4. In the circuit shown below, find the thevenin’s voltage across ‘ab’ terminals.

a) 48.5∠40.35⁰

b) 48.5∠-40.35⁰

c) 49.5∠-40.35⁰

d) 49.5∠40.35⁰

### View Answer

_{Th}=50∠0

^{o}×(4+j6)/((4+j6)+(3-j4) )=49.5∠40.35⁰V.

5. Find the thevenin’s impedance in the circuit shown in the question 4.

a) 4.83∠-1.13⁰

b) 5.83∠1.13⁰

c) 4.83∠1.13⁰

d) 5.83∠-1.13⁰

### View Answer

_{Th}=(j5-j4)+(3-j4)(4+j6)/(3-j4+4+j6) = 4.83∠-1.13⁰Ω.

6. Determine the Thevenin’s voltage across ‘ab’ terminals in the circuit shown below.

a) 41.86∠0⁰

b) 42.86∠0⁰

c) 43.86∠0⁰

d) 44.86∠0⁰

### View Answer

_{ab}=100∠0

^{o}×j3/(j3+j4) = 42.86∠0

^{o}V.

7. Find the Thevenin’s impedance across ‘ab’ terminals in the circuit shown in the question 6.

a) j4.71

b) j5.71

c) j6.71

d) j7.71

### View Answer

_{ab}=j5+ (j4)(j3)/j7 = j6.71Ω.

8. Determine the load current across j5Ω in the circuit shown in the question 2.

a) 3.66∠90⁰

b) 3.66∠-90⁰

c) 4.66∠90⁰

d) 4.66∠-90⁰

### View Answer

_{L}= (42.86∠0

^{o})/(j6.71+j5) = 3.66∠-90

^{o}A.

9. Determine the thevenin’s voltage in the circuit shown below.

a) 18∠146.31⁰

b) 18∠-146.31⁰

c) 19∠146.31⁰

d) 19∠-146.31⁰

### View Answer

^{o})/((2+j2) ) (-j4)=7.07∠-45

^{o}V. The voltage across ‘ab’ = -10∠0

^{o}+5∠90

^{o}-7.07∠-45

^{o}=18∠146.31

^{o}V.

10. Find the Thevenin’s impedance in the circuit shown in the question 9.

a) 11.3∠45⁰

b) 12.3∠45⁰

c) 11.3∠-45⁰

d) 12.3∠-45⁰

### View Answer

_{ab}= 4+ (2+j6)(-j4)/(2+j2) = 11.3∠-45

^{o}Ω.

## Network Theory MCQ Set 3

1. In a balanced three-phase system-delta load, if we assume the line voltage is V_{RY} = V∠0⁰ as a reference phasor. Then the source voltage V_{YB} is?

a) V∠0⁰

b) V∠-120⁰

c) V∠120⁰

d) V∠240⁰

### View Answer

_{RY}= V∠0⁰ is taken as a reference phasor. Then the source voltage V

_{YB}is V∠-120⁰.

2. In the question 1, the source voltage V_{BR} is?

a) V∠120⁰

b) V∠240⁰

c) V∠-240⁰

d) V∠-120⁰

### View Answer

_{RY}= V∠0⁰ is taken as a reference phasor. Then the source voltage V

_{BR}is V∠-240⁰.

3. In a delta-connected load, the relation between line voltage and the phase voltage is?

a) line voltage > phase voltage

b) line voltage < phase voltage

c) line voltage = phase voltage

d) line voltage >= phase voltage

### View Answer

4. If the load impedance is Z∠Ø, the current (I_{R} ) is?

a) (V/Z)∠-Ø

b) (V/Z)∠Ø

c) (V/Z)∠90-Ø

d) (V/Z)∠-90+Ø

### View Answer

_{R}= V

_{BR}∠0⁰/Z∠Ø = (V/Z)∠-Ø.

5. In the question 4, the expression obtained for current (I_{Y}) is?

a) (V/Z)∠-120+Ø

b) (V/Z)∠120-Ø

c) (V/Z)∠120+Ø

d) (V/Z)∠-120-Ø

### View Answer

_{Y}= V

_{YB}∠120⁰/Z∠Ø = (V/Z)∠-120-Ø.

6. In the question 4, the expression obtained for current (I_{B}) is?

a) (V/Z)∠-240+Ø

b) (V/Z)∠-240-Ø

c) (V/Z)∠240-Ø

d) (V/Z)∠240+Ø

### View Answer

_{B}= V

_{BR}∠240⁰/Z∠Ø = (V/Z)∠-240-Ø.

7. A three phase, balanced delta connected load of (4+j8) Ω is connected across a 400V, 3 – Ø balanced supply. Determine the phase current I_{R} . Assume the phase sequence to be R_{YB}.

a) 44.74∠-63.4⁰A

b) 44.74∠63.4⁰A

c) 45.74∠-63.4⁰A

d) 45.74∠63.4⁰A

### View Answer

_{RY}= V∠0⁰ as a reference V

_{RY}= 400∠0⁰V, V

_{YB}= 400∠-120⁰V and V

_{BR}= 400∠-240⁰V. Impedance per phase = (4+j8) Ω = 8.94∠63.4⁰Ω. Phase current I

_{R}= (400∠0

^{o})/(8.94∠63.4

^{o})= 44.74∠-63.4⁰A.

8. In the question 7, determine the phase current I_{Y}.

a) 44.74∠183.4⁰A

b) 45.74∠183.4⁰A

c) 44.74∠183.4⁰A

d) 45.74∠-183.4⁰A

### View Answer

_{RY}= V∠0⁰ as a reference V

_{RY}= 400∠0⁰V, V

_{YB}= 400∠-120⁰V and V

_{BR}= 400∠-240⁰V. Impedance per phase = (4+j8)Ω = 8.94∠63.4⁰Ω. Phase current I

_{Y}= (400∠120

^{o})/(8.94∠63.4

^{o})= 44.74∠-183.4⁰A.

9. In the question 7, determine the phase current I_{B}.

a) 44.74∠303.4⁰A

b) 44.74∠-303.4⁰A

c) 45.74∠303.4⁰A

d) 45.74∠-303.4⁰A

### View Answer

_{RY}= V∠0⁰ as a reference V

_{RY}= 400∠0⁰V, V

_{YB}= 400∠-120⁰V and V

_{BR}= 400∠-240⁰V. Impedance per phase = (4+j8) Ω = 8.94∠63.4⁰Ω. Phase current I

_{B}= (400∠240

^{o})/(8.94∠63.4

^{o})= 44.74∠-303.4⁰A.

10. Determine the power (kW) drawn by the load.

a) 21

b) 22

c) 23

d) 24

### View Answer

_{Ph}I

_{Ph}cosØ = 24kW.

## Network Theory MCQ Set 4

1. If the system is a three-wire system, the currents flowing towards the load in the three lines must add to ___ at any given instant.

a) 1

b) 2

c) 3

d) zero

### View Answer

2. The three impedances Z_{1} = 20∠30⁰Ω, Z_{2} = 40∠60⁰Ω, Z_{3} = 10∠-90⁰Ω are delta-connected to a 400V, 3 – Ø system. Determine the phase current I_{R}.

a) (17.32-j10) A

b) (-17.32-j10) A

c) (17.32+j10) A

d) (-17.32+j10) A

### View Answer

_{RY}= V∠0⁰ as a reference phasor, and assuming RYB phase sequence, we have V

_{RY}= 400∠0⁰V Z

_{1}= 20∠30⁰Ω = (17.32+j10)Ω I

_{R}= (400∠0

^{o})/(20∠30

^{o})=(17.32-j10) A.

3. Find the phase current I_{Y} from the question 2.

a) (10-j0) A

b) (10+j0) A

c) (-10+j0) A

d) (-10-j0) A

### View Answer

_{YB}is V

_{YB}= 400∠-120⁰V. The impedance Z

_{2}is Z

_{2}= 40∠60⁰Ω => I

_{Y}=(400∠-120

^{o})/(40∠60

^{o})=(-10+j0)A.

4. Find the phase current I_{B} from the question 2.

a) (34.64-j20) A

b) (34.64+j20) A

c) (-34.64+j20) A

d) (-34.64-j20) A

### View Answer

_{BR}is V

_{BR}= 400∠-240⁰V. The impedance Z

_{3}is Z

_{3}= 10∠-90⁰Ω => I

_{B}=(400∠240

^{o})/(10∠-90

^{o})=(-34.64-j20)A.

5. From the question 2, find the line current I_{1}.

a) (-51.96-j10) A

b) (-51.96+j10) A

c) (51.96+j10) A

d) (51.96+j10) A

### View Answer

_{1}is the difference of I

_{R}and I

_{B}. So the line current I

_{1}is I

_{1}= I

_{R}– I

_{B}= (51.96+j10) A.

6. From the question 2, find the line current I_{2}.

a) (-27.32+j10) A

b) (27.32+j10) A

c) (-27.32-j10) A

d) (27.32-j10) A

### View Answer

_{2}is the difference of I

_{Y}and I

_{R}. So the line current I

_{2}is I

_{2}= I

_{Y}– I

_{R}= (-27.32+j10) A.

7. From the question 2, find the line current I_{3}.

a) (24.646+j20) A

b) (-24.646+j20) A

c) (-24.646-j20) A

d) (24.646-j20) A

### View Answer

_{3}is the difference of I

_{B}and I

_{Y}. So the line current I

_{3}is I

_{3}= I

_{B}– I

_{Y}= (-24.646-j20) A.

8. In the question 2 find the power in the R phase.

a) 6628

b) 6728

c) 6828

d) 6928

### View Answer

^{2}x 17.32 = 6928W.

9. In the question 2 find the power in the Y phase.

a) 1000

b) 2000

c) 3000

d) 4000

### View Answer

^{2}x 20 = 2000W.

10. In the question 2 find the power in the B phase.

a) 0

b) 1

c) 3

d) 2

### View Answer

^{2}x 0 = 0W.

## Network Theory MCQ Set 5

1. The transfer function of a system having the input as X(s) and output as Y(s) is?

a) Y(s)/X(s)

b) Y(s) * X(s)

c) Y(s) + X(s)

d) Y(s) – X(s)

### View Answer

2. In the circuit shown below, if current is defined as the response signal of the circuit, then determine the transfer function.

a) H(s)=C/(S^{2} LC+RCS+1)

b) H(s)=SC/(S^{2} LC-RCS+1)

c) H(s)=SC/(S^{2} LC+RCS+1)

d) H(s)=SC/(S^{2} LC+RCS-1)

### View Answer

^{2}Lc+RCs+1) where I corresponds to the output Y(s) and V corresponds to the input X(s).

3. In the circuit shown in question 2, if voltage across the capacitor is defined as the output signal of the circuit, then the transfer function is?

a) H(s)=1/(S^{2} LC-RCS+1)

b) H(s)=1/(S^{2} LC+RCS+1)

c) H(s)=1/(S^{2} LC+RCS-1)

d) H(s)=1/(S^{2} LC-RCS-1)

### View Answer

_{o}/V = (1/sC)/(R+sL+1/sC)=1/(S

^{2}LC+RCS+1).

4. Let us assume x (t) = A cos(ωt + φ), then the Laplace transform of x (t) is?

a) X(S)=A(Scos Ø-ω sinØ)/(S^{2}-ω^{2} )

b) X(S)=A(Scos Ø+ω sinØ)/(S^{2}+ω^{2} )

c) X(S)=A(Scos Ø+ω sinØ)/(S^{2}-ω^{2} )

d) X(S)=A(Scos Ø-ω sinØ)/(S^{2}+ω^{2} )

### View Answer

^{2}+ω

^{2})-AsinØs/(s

^{2}+ω

^{2}) = A(Scos Ø-ω sinØ)/(S

^{2}+ω

^{2}).

5. The s-domain expression for the response for the input mentioned in question 4 is?

a) Y(s)=H(s)A(Scos Ø-ω sinØ)/(S^{2}-ω^{2} )

b) Y(s)=H(s) A(Scos Ø+ω sinØ)/(S^{2}+ω^{2} )

c) Y(s)=H(s) A(Scos Ø-ω sinØ)/(S^{2}+ω^{2} )

d) Y(s)=H(s) A(Scos Ø+ω sinØ)/(S^{2}-ω^{2} )

### View Answer

^{2}+ω

^{2}).

6. On taking the partial fractions for the response in the question 4, we get?

a) Y(s)=k_{1}/(s-jω)+(k_{1}^{‘})/(s+jω)+Σterms generated by the poles of H(s)

b) Y(s)=k_{1}/(s+jω)+(k_{1}^{‘})/(s+jω)+Σterms generated by the poles of H(s)

c) Y(s)=k_{1}/(s+jω)+(k_{1}^{‘})/(s-jω)+Σterms generated by the poles of H(s)

d) Y(s)=k_{1}/(s-jω)+(k_{1}^{‘})/(s-jω)+Σterms generated by the poles of H(s)

### View Answer

_{1}/(s-jω)+(k

_{1}

^{‘})/(s+jω)+Σterms generated by the poles of H(s). The first two terms result from the complex conjugate poles of the deriving source.

7. The value of k_{1} in the question 6 is?

a) 1/2 H(jω)Ae^{jØ}

b) H(jω)Ae^{-jØ}

c) H(jω)Ae^{jØ}

d) 1/2 H(jω)Ae^{-jØ}

### View Answer

_{1}=H(s) (A(scosØ-ωsinØ))/( s+jω)|s=jω = 1/2 H(jω)Ae

^{-jØ}.

8. The relation between H (jω) and θ (ω) is?

a) H(jω)=e^{-jθ (ω)}

b) H(jω)=|H(jω)|e^{-jθ (ω)}

c) H(jω)=|H(jω)|e^{jθ (ω)}

d) H(jω)=e^{jθ (ω)}

### View Answer

^{jθ(ω)}where |H(jω)| is the magnitude and the phase angle is θ(ω) of the transfer function vary with frequency ω.

9. The value of k_{1} in the question 6 considering θ (ω) is?

a) |H(jω)|e^{j[θ (ω)+Ø]}

b) A/2|H(jω)|e^{j[θ (ω)+Ø]}

c) |H(jω)|e^{-j[θ (ω)+Ø]}

d) A/2 |H(jω) | e^{-j[θ (ω)+Ø]}

### View Answer

_{1}becomes K

_{1}= A/2|H(jω)|e

^{j[θ (ω)+Ø]}. We obtain the steady state solution for y(t) by taking the inverse transform ignoring the terms generated by the poles of H(s).

10. The final steady state solution for y (t) in the question 4 is?

a) A|H(jω) |cos[ωt+Ø+ θ (ω)].

b) A|H(jω) |cos[ωt-Ø+ θ (ω)].

c) A|H(jω) |cos[ωt-Ø- θ (ω)].

d) A|H(jω) |cos[ωt+Ø- θ (ω)].

### View Answer

_{ss}(t) = A|H(jω)|cos[ωt+Ø+ θ (ω)] which indicates how to use the transfer function to find the steady state response of a circuit.