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## 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

Answer: b [Reason:] The thevenin’s voltage is equal to the open circuit voltage across the terminals AB that is across 12Ω resistor. Vth = 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

Answer: c [Reason:] The resistance into the open circuit terminals is equal to the thevenin’s resistance => Rth = (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

Answer: a [Reason:] The equivalent thevenin’s model of the circuit shown is I=8.57/(2.4+1.71)=0.33A.

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

Answer: c [Reason:] Let us find the voltage drop across terminals A and B. 50-25=10I+5I => I=1.67A. Voltage drop across 10Ω resistor = 10×1.67=16.7V. So, Vth=VAB=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

Answer: c [Reason:] To find Rth, 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 => Rth=(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

Answer: c [Reason:] Current through 3Ω resistor is 0A. The current through 6Ω resistor = (50-10)/(10+6)=2.5A. The voltage drop across 6Ω resistor = 25×6=15V. So the voltage across terminals A and B = 0+15+10 = 25V.

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

Answer: d [Reason:] To find Rth, two voltage sources are removed and replaced with short circuit => Rth=(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

Answer: c [Reason:] The voltage at terminal a is Va=(100×6)/16=37.5V, The voltage at terminal b is Vb=(100×8)/23=34.7V. So the voltage across the terminals ab is Vab=Va-Vb=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

Answer: d [Reason:] To find Rth, two voltage sources are removed and replaced with short circuit => Rab=(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

Answer: b [Reason:] The Equivalent Thevenin’s circuit for the circuit shown above is I=2.7/(8.96+5)=0.193A≅0.2A.

## 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

Answer: c [Reason:] Thevenin’s voltage is equal to open circuit voltage across output terminals not the short circuit voltage across output terminals.

2. Consider the circuit shown below. The expression of Thevenin’s voltage (VTh) is?

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

Answer: b [Reason:] Thevenin’s theorem gives us a method for simplifying a given circuit. The thevenin’s voltage is VTh = V(Z2/(Z1+Z2)).

3. The value of ZTh in the circuit shown in the question 2 is?
a) Z3+( Z1Z2/(Z1+Z2))
b) Z1+( Z3Z2/(Z3+Z2))
c) Z2+( Z1Z3/(Z1+Z3))
d) ( Z1Z2/(Z1+Z2))

Answer: a [Reason:] The thevenin’s equivalent form of any complex impedance consists of an equivalent voltage source and an equivalent impedance. The thevenin’s impedance is ZTh = Z3+(Z1Z2/(Z1+Z2)).

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

Answer: d [Reason:] Though the thevenin’s equivalent circuit is not same as its original circuit, the output current and voltage are the same in both cases. The thevenin’s voltage is equal to the voltage across (4+j6)Ω impedance. VTh =50∠0o×(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

Answer: a [Reason:] The impedance is equal to the impedance seen into the network across the output terminals. ZTh =(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

Answer: b [Reason:] The voltage across the points a and b is called thevenin’s equivalent voltage. Thevenin’s equivalent voltage Vab=100∠0o×j3/(j3+j4) = 42.86∠0o 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

Answer: c [Reason:] The impedance is equal to the impedance seen into the network across the output terminals. Zab=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

Answer: b [Reason:] The load current is the ratio of thevenin’s equivalent voltage and thevenin’s equivalent impedance. The load current IL = (42.86∠0o)/(j6.71+j5) = 3.66∠-90o 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

Answer: a [Reason:] The voltage across (-j4) Ω resistor is = (5∠90o)/((2+j2) ) (-j4)=7.07∠-45o V. The voltage across ‘ab’ = -10∠0o+5∠90o-7.07∠-45o=18∠146.31o 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

Answer: c [Reason:] The impedance is equal to the impedance seen into the network across the output terminals. Zab = 4+ (2+j6)(-j4)/(2+j2) = 11.3∠-45o Ω.

## Network Theory MCQ Set 3

1. In a balanced three-phase system-delta load, if we assume the line voltage is VRY = V∠0⁰ as a reference phasor. Then the source voltage VYB is?
a) V∠0⁰
b) V∠-120⁰
c) V∠120⁰
d) V∠240⁰

Answer: b [Reason:] As the line voltage VRY = V∠0⁰ is taken as a reference phasor. Then the source voltage VYB is V∠-120⁰.

2. In the question 1, the source voltage VBR is?
a) V∠120⁰
b) V∠240⁰
c) V∠-240⁰
d) V∠-120⁰

Answer: c [Reason:] As the line voltage VRY = V∠0⁰ is taken as a reference phasor. Then the source voltage VBR 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

Answer: c [Reason:] In a delta-connected load, the relation between line voltage and the phase voltage is line voltage = phase voltage.

4. If the load impedance is Z∠Ø, the current (IR ) is?
a) (V/Z)∠-Ø
b) (V/Z)∠Ø
c) (V/Z)∠90-Ø
d) (V/Z)∠-90+Ø

Answer: a [Reason:] As the load impedance is Z∠Ø, the current flows in the three load impedances and the current flowing in the R impedance is IR = VBR∠0⁰/Z∠Ø = (V/Z)∠-Ø.

5. In the question 4, the expression obtained for current (IY) is?
a) (V/Z)∠-120+Ø
b) (V/Z)∠120-Ø
c) (V/Z)∠120+Ø
d) (V/Z)∠-120-Ø

Answer: d [Reason:] As the load impedance is Z∠Ø, the current flows in the three load impedances and the current flowing in the Y impedance is IY = VYB∠120⁰/Z∠Ø = (V/Z)∠-120-Ø.

6. In the question 4, the expression obtained for current (IB) is?
a) (V/Z)∠-240+Ø
b) (V/Z)∠-240-Ø
c) (V/Z)∠240-Ø
d) (V/Z)∠240+Ø

Answer: b [Reason:] As the load impedance is Z∠Ø, the current flows in the three load impedances and the current flowing in the B impedance is IB = VBR∠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 IR . Assume the phase sequence to be RYB.
a) 44.74∠-63.4⁰A
b) 44.74∠63.4⁰A
c) 45.74∠-63.4⁰A
d) 45.74∠63.4⁰A

Answer: a [Reason:] Taking the line voltage VRY = V∠0⁰ as a reference VRY = 400∠0⁰V, VYB = 400∠-120⁰V and VBR = 400∠-240⁰V. Impedance per phase = (4+j8) Ω = 8.94∠63.4⁰Ω. Phase current IR = (400∠0o)/(8.94∠63.4o )= 44.74∠-63.4⁰A.

8. In the question 7, determine the phase current IY.
a) 44.74∠183.4⁰A
b) 45.74∠183.4⁰A
c) 44.74∠183.4⁰A
d) 45.74∠-183.4⁰A

Answer: c [Reason:] Taking the line voltage VRY = V∠0⁰ as a reference VRY = 400∠0⁰V, VYB = 400∠-120⁰V and VBR = 400∠-240⁰V. Impedance per phase = (4+j8)Ω = 8.94∠63.4⁰Ω. Phase current IY = (400∠120o)/(8.94∠63.4o )= 44.74∠-183.4⁰A.

9. In the question 7, determine the phase current IB.
a) 44.74∠303.4⁰A
b) 44.74∠-303.4⁰A
c) 45.74∠303.4⁰A
d) 45.74∠-303.4⁰A

Answer: b [Reason:] Taking the line voltage VRY = V∠0⁰ as a reference VRY = 400∠0⁰V, VYB = 400∠-120⁰V and VBR = 400∠-240⁰V. Impedance per phase = (4+j8) Ω = 8.94∠63.4⁰Ω. Phase current IB = (400∠240o)/(8.94∠63.4o)= 44.74∠-303.4⁰A.

10. Determine the power (kW) drawn by the load.
a) 21
b) 22
c) 23
d) 24

Answer: d [Reason:] Power is defined as the product of voltage and current. So the power drawn by the load is P = 3VPhIPhcosØ = 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

Answer: d [Reason:] If the system is a three-wire system, the currents flowing towards the load in the three lines must add to zero at any given instant.

2. The three impedances Z1 = 20∠30⁰Ω, Z2 = 40∠60⁰Ω, Z3 = 10∠-90⁰Ω are delta-connected to a 400V, 3 – Ø system. Determine the phase current IR.
a) (17.32-j10) A
b) (-17.32-j10) A
c) (17.32+j10) A
d) (-17.32+j10) A

Answer: a [Reason:] Taking VRY = V∠0⁰ as a reference phasor, and assuming RYB phase sequence, we have VRY = 400∠0⁰V Z1 = 20∠30⁰Ω = (17.32+j10)Ω IR = (400∠0o)/(20∠30o)=(17.32-j10) A.

3. Find the phase current IY from the question 2.
a) (10-j0) A
b) (10+j0) A
c) (-10+j0) A
d) (-10-j0) A

Answer: c [Reason:] The voltage VYB is VYB = 400∠-120⁰V. The impedance Z2 is Z2 = 40∠60⁰Ω => IY =(400∠-120o)/(40∠60o)=(-10+j0)A.

4. Find the phase current IB from the question 2.
a) (34.64-j20) A
b) (34.64+j20) A
c) (-34.64+j20) A
d) (-34.64-j20) A

Answer: d [Reason:] The voltage VBR is VBR = 400∠-240⁰V. The impedance Z3 is Z3 = 10∠-90⁰Ω => IB =(400∠240o)/(10∠-90o )=(-34.64-j20)A.

5. From the question 2, find the line current I1.
a) (-51.96-j10) A
b) (-51.96+j10) A
c) (51.96+j10) A
d) (51.96+j10) A

Answer: c [Reason:] The line current I1 is the difference of IR and IB. So the line current I1 is I1 = IR – IB = (51.96+j10) A.

6. From the question 2, find the line current I2.
a) (-27.32+j10) A
b) (27.32+j10) A
c) (-27.32-j10) A
d) (27.32-j10) A

Answer: a [Reason:] The line current I2 is the difference of IY and IR. So the line current I2 is I2 = IY – IR = (-27.32+j10) A.

7. From the question 2, find the line current I3.
a) (24.646+j20) A
b) (-24.646+j20) A
c) (-24.646-j20) A
d) (24.646-j20) A

Answer: c [Reason:] The line current I3 is the difference of IB and IY. So the line current I3 is I3 = IB – IY = (-24.646-j20) A.

8. In the question 2 find the power in the R phase.
a) 6628
b) 6728
c) 6828
d) 6928

Answer: d [Reason:] The term power is defined as the product of square of current and the impedance. So the power in the R phase = 202 x 17.32 = 6928W.

9. In the question 2 find the power in the Y phase.
a) 1000
b) 2000
c) 3000
d) 4000

Answer: b [Reason:] The term power is defined as the product of square of current and the impedance. So the power in the Y phase = 102 x 20 = 2000W.

10. In the question 2 find the power in the B phase.
a) 0
b) 1
c) 3
d) 2

Answer: a [Reason:] The term power is defined as the product of square of current and the impedance. So the power in the B phase = 402 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)

Answer: a [Reason:] The transfer function is defined as the s-domain ratio of the laplace transfrom of the output to the laplace transfrom of the input. The transfer function of a system having the input as X(s) and output as Y(s) is H (s) = Y(s)/X(s).

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/(S2 LC+RCS+1)
b) H(s)=SC/(S2 LC-RCS+1)
c) H(s)=SC/(S2 LC+RCS+1)
d) H(s)=SC/(S2 LC+RCS-1)

Answer: c [Reason:] If the current is defined as the response signal of the circuit, then the transfer function H (s) = I/V = 1/(R+sL+1/sC) = sC/(s2Lc+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/(S2 LC-RCS+1)
b) H(s)=1/(S2 LC+RCS+1)
c) H(s)=1/(S2 LC+RCS-1)
d) H(s)=1/(S2 LC-RCS-1)

Answer: b [Reason:] If the voltage across the capacitor is defined as the output signal of the circuit, the transfer function is H(s) = Vo/V = (1/sC)/(R+sL+1/sC)=1/(S2LC+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)/(S22 )
b) X(S)=A(Scos Ø+ω sinØ)/(S22 )
c) X(S)=A(Scos Ø+ω sinØ)/(S22 )
d) X(S)=A(Scos Ø-ω sinØ)/(S22 )

Answer: d [Reason:] We use the transfer function to relate the study state response to the excitation source. And we had assumed that x (t) = A cos(ωt + φ). On expanding and taking the laplace transform we get X(s) = AcosØs/(s22 )-AsinØs/(s22) = A(Scos Ø-ω sinØ)/(S22 ).

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)/(S22 )
b) Y(s)=H(s) A(Scos Ø+ω sinØ)/(S22 )
c) Y(s)=H(s) A(Scos Ø-ω sinØ)/(S22 )
d) Y(s)=H(s) A(Scos Ø+ω sinØ)/(S22 )

Answer: c [Reason:] We had the equation Y(s) = H(s)X(s) to find the steady state solution of y(t). The s-domain expression for the response for the input is Y(s)=H(s) A(Scos Ø-ω sinØ)/(S22 ).

6. On taking the partial fractions for the response in the question 4, we get?
a) Y(s)=k1/(s-jω)+(k1)/(s+jω)+Σterms generated by the poles of H(s)
b) Y(s)=k1/(s+jω)+(k1)/(s+jω)+Σterms generated by the poles of H(s)
c) Y(s)=k1/(s+jω)+(k1)/(s-jω)+Σterms generated by the poles of H(s)
d) Y(s)=k1/(s-jω)+(k1)/(s-jω)+Σterms generated by the poles of H(s)

Answer: a [Reason:] By taking partial fractions, Y(s)=k1/(s-jω)+(k1)/(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 k1 in the question 6 is?
a) 1/2 H(jω)Ae
b) H(jω)Ae-jØ
c) H(jω)Ae
d) 1/2 H(jω)Ae-jØ

Answer: d [Reason:] The first two terms on the right hand side of Y(s) determine the steady state response. k1=H(s) (A(scosØ-ωsin⁡Ø))/( 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)|ejθ (ω)
d) H(jω)=ejθ (ω)

Answer: c [Reason:] In general, H(jω) is a complex quantity, thus H(jω) = |H(jω)|ejθ(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 k1 in the question 6 considering θ (ω) is?
a) |H(jω)|ej[θ (ω)+Ø]
b) A/2|H(jω)|ej[θ (ω)+Ø]
c) |H(jω)|e-j[θ (ω)+Ø]
d) A/2 |H(jω) | e-j[θ (ω)+Ø]

Answer: b [Reason:] The expression of k1 becomes K1 = A/2|H(jω)|ej[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)].