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## Network Theory MCQ Set 1

1. Resistance of a wire is yΩ. The wire is stretched to triple its length, then the resistance becomes
a) y/3
b) 3y
c) 6y
d) y/6

Answer: b [Reason:] Resistance of a conductor is directly proportional to its length. That is, when the length of conductor is tripled, its resistance also gets tripled.

2. An electric current of 10 A is the same as
a) 10 J/C
b) 10 V/C
c) 10C/sec
d) 10 W/sec

Answer: c [Reason:] Mathematically, electric current can be defined as the ratio of the charge to the time in which charge flows.

3. Consider a circuit with two unequal resistances in parallel, then
a) large current flows in large resistor
b) current is same in both
c) potential difference across each is same
d) smaller resistance has smaller conductance

Answer: c [Reason:] In parallel combination of resistors, the potential difference across each resistors is the same.

4. In which of the following cases is Ohm’s law not applicable?
a) Electrolytes
b) Arc lamps
c) Insulators
d) Vacuum ratio values

Answer: c [Reason:] According to the Ohm’s law, it is applicable only to conductors. Hence, Ohm’s law is not applicable in case of insulators.

5. A copper wire of length l and diameter d has potential difference V applied at its two ends. The drift velocity is V. If the diameter of wire is made d/4, then drift velocity becomes
a) V/16
b) 16V
c) V
d) V/4

Answer: b [Reason:] Drift velocity is inversely propotional to area of material i.e, V=I/nAq.

6. Which of the following bulbs will have high resistance?
a) 220V, 60W
b) 220V,100W
c) 115V,60W
d) 115V,100 W

Answer: a [Reason:] Resistance is directly proportional to sqaure of voltage and inversely proportional to the power.

7. Ohm’s law is not applicable to
a) dc circuits
b) high currents
c) small resistors
d) semi-conductors

Answer: d [Reason:] Ohm’s law is not applicable to semi-conductors and insulators.

8. Conductance is expressed in terms of
a) mho
b) mho/m
c) ohm/m
d) m/ohm

Answer: a [Reason:] Conductance is the reciprocal of resistance and is expreswsed in terms of mho.

9. Resistivity of a wire depends on
a) length of wire
b) cross section area
c) material
d) all of the mentioned

Answer: c [Reason:] Resistivity of a wire is a constant and it depends on the type of material used.

10. In a current-voltage relationship graph of a linear resistor, the slope of the graph will indicate
a) conductance
b) resistance
c) resistivity
d) a constant

Answer: a [Reason:] The slope of the graph is the ratio of current to voltage which indicates conductance.

## Network Theory MCQ Set 2

1. Pick the incorrect statement among the following
a) Inductor is a passive element
b) Current source is an active element
c) Resistor is a passive element
d) Voltage source is a passive element

Answer: b [Reason:] Energy sources(voltage or current sources) are active elements, capable of delivering power to some external device.

2. For a voltage source to be neglected, the terminals across the source should be
a) replaced by inductor
b) short circuited
c) replaced by some resistance
d) open circuited

Answer: b [Reason:] If the voltage source is to be neglected, it can be replaced simply by means of a wire i.e, it should be short circuited.

3. Voltage source and terminal voltage can be related as
a) terminal voltage is higher than the source emf
b) terminal voltage is equal to the source emf
c) terminal voltage is always lower than source emf
d) terminal voltage cannot exceed source emf

Answer: c [Reason:] A practical voltage source can be represented with a resistance in series with the source. Hence, there would be some voltage drop at the resistor and the terminal voltage is always lower than the source emf.

4. In case of ideal current sources, they have
a) zero internal resistance
b) low value of voltage
c) large value of currrent
d) infinite internal resistance

Answer: d [Reason:] For the ideal current sources, the current is completely independent of voltage and it has infinte internal resistance.

5. In a network consisting of linear resistors and ideal voltage source, if the value of resistors are doubled, then voltage across each resistor
a) increases four times
b) remains unchanged
c) doubled
d) halved

Answer: b [Reason:] Even on changing the values of linear resistors, the voltage remains constant in case of ideal voltage source.

6. A practical current source can also be represented as
a) a resistance in parallel with an ideal voltage source
b) a resistance in parallel with an ideal current source
c) a resistance in series with an ideal current source
d) none of the mentioned

Answer: b [Reason:] A practical current source could be represented with a resistor in parallel with an ideal current source.

7. A practical voltage source can also be represented as
a) a resistance in series with an ideal current source
b) a resistance in series with an ideal voltage source
c) a resistance in parallel with an ideal voltage source
d) none of the mentioned

Answer: b [Reason:] A practical voltage source could be represented with a resistor in series with an ideal voltage source.

8. Constant voltage source is
a) active and bilateral
b) passive and bilateral
c) active and unilateral
d) passive and unilateral

Answer: c [Reason:] Voltage source is an active element and is unilateral.

9. Which of the following is true about an ideal voltage source?
a) zero resistance
b) small emf
c) large emf
d) infinite resistance

Answer: a [Reason:] An ideal voltage source zero internal resistance.

10. A dependent source
a) may be a current source or a voltage source
b) is always a voltage source
c) is always a current source
d) none of the mentioned

Answer: a [Reason:] Dependent sources can either be current sources or voltage sources.

11. With some initial change at t = 0+, a capacitor will act as
a) open circuit
b) short circuit
c) a current source
d) a voltage source

Answer: d [Reason:] At t=0+, the capacitor starts charging to a particular voltage and acts as a voltage source.

12. If a current source is to be neglected, the terminals across the source are
a) replaced by a source resistance
b) open circuited
c) replaced by a capacitor
d) short circuited

Answer: b [Reason:] As the ideal current source has infinite resistance, it can be neglected by open circuiting the terminals.

13. A constant current source supplies a electric current of 200 mA to a load of 2kΩ. When the load changed to 100Ω, the load current will be
a) 9mA
b) 4A
c) 700mA
d) 12A

Answer: b [Reason:] From Ohm’s law, resistance is inversely proportional to the current.

14. A voltage source having an open circuit voltage of 200 V and internal resistance of 50Ω is equivalent to a current source of
a) 4A with 50Ω in parallel
b) 4A with 50Ω in series
c) 0.5A with 50Ω in parallel
d) none of the mentioned

Answer: a [Reason:] A voltage source with resistance in series can be replaced with a current source with the resistance in parallel.

15. A voltage source of 300 V has internal resistance of 4Ω and supplies a load having the same resistance. The power absorbed by the load is
a) 1150 W
b) 1250 W
c) 5625 W
d) 5000 W

Answer: c [Reason:] Power absorbed =I2R.

## Network Theory MCQ Set 3

1. The direction of the cut-set is?
a) same as the direction of the branch current
b) opposite to the direction of the link current
c) same as the direction of the link current
d) opposite to the direction of the branch current

Answer: a [Reason:] A cut-set is a minimal set of branches of a connected graph such that the removal of these branches causes the graph to be cut into exactly two parts. The direction of the cut-set is same as the direction of the branch current.

2. Consider the graph shown below. The direction of the cut-set of node ‘a’ is? a) right
b) left
c) upwards
d) downwards

Answer: c [Reason:] The direction of the cut set at node ‘a’ will be the direction of the branch current at node ‘a’. So the direction of the current will be upwards.

3. Consider the graph shown above in question 2. The direction of the cut-set at node ‘b’ will be?
a) upwards
b) right
c) downwards
d) left

Answer: b [Reason:] The direction of the current will be towards right. The direction of the cut set at node ‘b’ will be the direction of the branch current at node ‘b’. So the direction of the current will be towards right.

4. In the graph shown above in the question 2, the direction of the cut-set at node ‘c’ is?
a) downwards
b) upwards
c) left
d) right

Answer: b [Reason:] The direction of the cut set at node ‘c’ will be the direction of the branch current at node ‘c’. So the direction of the current will be upwards.

5. In the graph shown in the question 2, the direction of the cut-set at node ‘d’ will be?
a) left
b) downwards
c) right
d) upwards

Answer: c [Reason:] The direction of the cut set at node ‘d’ will be the direction of the branch current at node ‘d’. So the direction of the current will be upwards.

6. The row formed at node ‘a’ in the cut set matrix in the figure shown in question 2 is?
I1 I2 I3 I4 I5 I6 I7 I8
a) +1 +1 +1 +1 0 0 0 0
b) +1 0 0 0 +1 0 0 +1
c) -1 0 0 0 -1 0 0 -1
d) -1 -1 0 0 -1 -1 0 0

Answer: b [Reason:] The direction of the cut set at node ‘a’ is towards node ‘a’. So the current direction of I1 is same as cut set direction. So it is +1. Similarly for all other currents.

7. The row formed at node ‘c’ in the cut set matrix in the figure shown question 2 is?
a) -1 -1 0 0 +1 -1 0 0
b) 0 0 +1 0 0 -1 -1 0
c) +1 0 0 0 +1 0 0 +1
d) -1 0 0 0 -1 0 0 -1

Answer: b [Reason:] The direction of the cut set at node ‘c’ is away from node ‘c’. So the current direction of I3 is same as cut set direction. So it is +1. Similarly for all other currents.

8. The number of cut set matrices formed from a graph is?
a) NN-1
b) NN
c) NN-2
d) NN+1

Answer: c [Reason:] For every tree, there will be a unique cut set matrix. So there will be NN-2 cut set matrices.

9. For every tree there will be _____ number of cut set matrices.
a) 1
b) 2
c) 3
d) 4

Answer: a [Reason:] For every tree, there will a unique cut set matrix. So, number of cut-set matrices for every tree = 1.

10. If a row of the cut set matrix formed by the branch currents of the graph is shown below. Then which of the following is true?
I1 I2 I3 I4 I5 I6 I7 I8
-1 -1 0 0 +1 -1 0 0
a) -V1-V2+V5-V6=0
b) -I1-I2+I5-I6=0
c) -V1+V2+V5-V6=0
d) -I1+I2+I5-I6=0

Answer: b [Reason:] KCL equations are derived from cut set matrix and these include currents not voltages. So, -I1-I2+I5-I6=0.

## Network Theory MCQ Set 4

1. The highest power factor will be?
a) 1
b) 2
c) 3
d) 4

Answer: a [Reason:] The power factor is useful in determining the useful power transferred to a load. The highest power factor will be 1.

2. If power factor = 1, then the current to the load is ______ with the voltage across it.
a) out of phase
b) in phase
c) 90⁰ out of phase
d) 45⁰ out of phase

Answer: b [Reason:] If power factor = 1, then the current to the load is in phase with the voltage across it because the expression of power factor is power factor = cosθ.

3. In case of resistive load, the power factor =?
a) 4
b) 3
c) 2
d) 1

Answer: d [Reason:] In case of resistive load, the power factor = 1 as the current to the load is in phase with the voltage across it.

4. If power factor = 0, then the current to a load is ______ with the voltage.
a) in phase
b) out of phase
c) 45⁰ out of phase
d) 90⁰ out of phase

Answer: d [Reason:] If the power factor = 0, then the current to a load is 90⁰ out of phase with the voltage and it happens in case of reactive load.

5. For reactive load, the power factor is equal to?
a) 0
b) 1
c) 2
d) 3

Answer: a [Reason:] For reactive load, the power factor is equal to 0. Power factor = 0 when current to a load is 90⁰ out of phase with the voltage.

6. Average power is also called?
a) apparent power
b) reactive power
c) true power
d) instantaneous power

Answer: c [Reason:] The average power is expressed in watts. It means the useful power transferred from the source to the load, which is also called true power. Average power is also called true power.

7. If we apply a sinusoidal voltage to a circuit, the product of voltage and current is?
a) true power
b) apparent power
c) average power
d) reactive power

Answer: b [Reason:] If we apply a sinusoidal voltage to a circuit, the product of voltage and current is apparent power. The apparent power is expressed in volt amperes or simply VA.

8. The expression of apparent power (Papp) is?
a) VmIm
b) VmIeff
c) VeffIeff
d) VeffIm

Answer: c [Reason:] In case of sinusoidal voltage applied to the circuit, the product of voltage and the current is not the true power or average power and it is apparent power. The expression of apparent power (Papp) is Papp = VeffIeff.

9. The power factor=?
a) sinθ
b) cosθ
c) tanθ
d) secθ

Answer: b [Reason:] The expression of power factor is power factor= cosθ. As the phase angle between the voltage and the current increases the power factor decreases.

10. The power factor is the ratio of ________ power to the ______ power.
a) average, apparent
b) apparent, reactive
c) reactive, average
d) apparent, average

Answer: a [Reason:] The power factor is the ratio of average power to the apparent power. Power factor =(average power)/(apparent power). Power factor is also defined as the factor with which the volt amperes are to be multiplied to get true power in the circuit.

11. The power factor is called leading power factor in case of ____ circuits.
a) LC
b) RC
c) RL
d) RLC

Answer: b [Reason:] The power factor is called leading power factor in case of RC circuits and not in RLC circuits and RL circuits and LC circuits.

12. The term lagging power factor is used in which circuits?
a) RLC
b) RC
c) RL
d) LC

Answer: c [Reason:] The term lagging power factor is used in RL circuits and not in RLC circuits and RC circuits and LC circuits.

## Network Theory MCQ Set 5

1. Reciprocity Theorem is applied for _____ networks.
a) Linear
b) Bilateral
c) Linear bilateral
d) Lumped

Answer: c [Reason:] Reciprocity Theorem is applied for linear bilateral networks, not for linear or for linear bilateral or for lumped networks.

2. Reciprocity Theorem is used to find the change in _______ when the resistance is changed in the circuit.
a) Voltage
b) Voltage or current
c) Current
d) Power

Answer: b [Reason:] Reciprocity Theorem is used to find the change in voltage or current when the resistance is changed in the circuit. If reciprocity theorem is satisfied the ratio of response to excitation is same for the two conditions.

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

Answer: a [Reason:] Total resistance in the circuit = 2+[3||(2+2||2)] = 3.5Ω. The total current drawn by the circuit =10/(4+6||3) = 1.67A. Current through 3Ω resistor = 1.11A ≅1A.

4. Determine the current flowing in the ammeter having 1Ω internal resistance connected in series with the 3Ω resistor shown in the circuit shown in the question 3.
a) 0.91
b) 0.92
c) 0.93
d) 0.94

Answer: c [Reason:] Current through 3Ω resistor = 1.11A. So voltage drop across 1Ω resistor = 1.11×1 = 1.11V. Now the circuit can be modified as Now current through 3Ω resistor = 0.17A. This current is opposite to the current calculated before. So ammeter reading = (1.11-0.17) = 0.94A.

5. Find the current through 6Ω resistor in the circuit shown below. a) 0.33
b) 0.44
c) 0.55
d) 0.66

Answer: c [Reason:] Total resistance in the circuit = 4+6||3Ω. The total current drawn by the circuit =10/(4+6||3)=1.67A. Current through 6Ω resistor = 0.55A.

6. Determine the current flowing in the ammeter having 1Ω internal resistance connected in series with the 6Ω resistor shown in the circuit shown in the question 5.
a) 0.4
b) 0.45
c) 0.9
d) 0.95

Answer: b [Reason:] Current through 3Ω resistor = 0.55A. So voltage drop across 1Ω resistor = 0.55×1 = 0.55V. Now the circuit can be modified as Now current through 6Ω resistor = 0.094A. This current is opposite to the current calculated before. So ammeter reading = (0.55-0.0.94) = 0.45A.

7. Find the current through 6Ω resistor in the circuit shown below. a) 0.11
b) 0.22
c) 0.33
d) 0.44

Answer: c [Reason:] Total current in the circuit = 10/(4+3||2||6)=2A. Current through 6Ω resistor= 2×(3||2)/(6+3||2)=0.33A.

8. Determine the current flowing in the ammeter having 1Ω internal resistance in series with the 6Ω resistor shown in the circuit shown in the question 7.
a) 0.1
b) 0.2
c) 0.3
d) 0.4

Answer: c [Reason:] New total current = 0.33/(7+4||2||3)=0.04A. Now reading of ammeter= 0.33-0.04=0.29A ≅0.3A.

9. Find the current through 3Ω resistor in the circuit shown below. a) 0.45
b) 0.56
c) 0.67
d) 0.78

Answer: c [Reason:] Total current = 10 / (4 + (6||2||3) = 2A. Current through 3Ω resistor= 2 x (6||2) / (3 + (6||2)) =0.67A.

10. Determine the current flowing in the ammeter having 1Ω internal resistance in series with the 3Ω resistor shown in the circuit shown in the question 9.
a) 0.6
b) 0.7
c) 0.8
d) 0.9

Answer: a [Reason:] The current flowing in the ammeter having 1Ω internal resistance in series with the 3Ω resistor shown in the circuit is 0.6 A. Current through 3Ω resistor = 0.67 / (7 + (4||6||2)) = 0.08A. Ammeter reading = 0.67 – 0.08 = 0.59 ≅0.6A.

## Network Theory MCQ Set 6

1. The wattmeter method is used to measure power in a three-phase load. The wattmeter readings are 400W and -35W. Calculate the total active power.
a) 360
b) 365
c) 370
d) 375

Answer: b [Reason:] Wattmeters are generally used to measure power in the circuits. Total active power = W1 + W2 = 400 + (-35) =365W.

2. In the question 2 find the power factor.
a) 0.43
b) 0.53
c) 0.63
d) 0.73

Answer: a [Reason:] We know tanØ = √3((WR – WY)/(WR + WY)) => tanØ = √3 (400-(-35))/(400+(-35) )=2.064 => Ø = 64.15⁰. Power factor = 0.43.

3. Find the reactive power in the question 2.
a) 751.44
b) 752.44
c) 753.44
d) 754.44

Answer: c [Reason:] Reactive power = √3VLILsinØ. We know that WR – WY = 400-(-35)) = 435 = VLILsinØ. Reactive power = √3 x 435 = 753.44 VAR.

4. The input power to a three-phase load is 10kW at 0.8 Pf. Two watt meters are connected to measure the power. Find the reading of higher reading wattmeter.
a) 7.165
b) 6.165
c) 6.165
d) 4.165

Answer: a [Reason:] WR + WY = 10kW. Ø = cos-10.8=36.8o => tanØ = 0.75 = √3 (WR-WY)/(WR+WY )=(WR-WY)/10. WR-WY=4.33kW. WR+WY=10kW. WR=7.165kW.

5. Find the reading of higher reading wattmeter in the question 2.
a) 1.835
b) 2.835
c) 3.835
d) 4.835

Answer: b [Reason:] WR + WY = 10kW. Ø = cos-10.8=36.8o => tanØ = 0.75 = √3 (WR-WY)/(WR+WY )=(WR-WY)/10. WR-WY=4.33kW. WR+WY=10kW. WY=2.835kW.

6. The readings of the two watt meters used to measure power in a capacitive load are -3000W and 8000W respectively. Calculate the input power. Assume RYB sequence.
a) 5
b) 50
c) 500
d) 5000

Answer: d [Reason:] Toatal power is the sum of the power in R and power in Y. So total power = WR+WY = -3000+8000 = 5000W

7. Find the power factor in the question 6.
a) 0.25
b) 0.5
c) 0.75
d) 1

Answer: a [Reason:] As the load is capacitive, the wattmeter connected in the leading phase gives less value. WR=-3000. WY=8000. tanØ = √3 (8000-(-3000))/5000=3.81 => Ø = 75.29⁰ => cosØ = 0.25.

8. The wattmeter reading while measuring the reactive power with wattmeter is?
a) VLILsecØ
b) VLILsinØ
c) VLILtanØ
d) VLILcosØ

Answer: b [Reason:] The wattmeter reading while measuring the reactive power with wattmeter is wattmeter reading = VLILsinØ VAR.

9. The total reactive power in the load while measuring the reactive power with wattmeter is?
a) √3VLILcosØ
b) √3VLILtanØ
c) √3VLILsinØ
d)√3 VLILsecØ

Answer: c [Reason:] To obtain the reactive power, wattmeter reading is to be multiplied by √3. Total reactive power = √3VLILsinØ.

10. A single wattmeter is connected to measure reactive power of a three-phase, three-wire balanced load. The line current is 17A and line voltage is 440V. Calculate the power factor of the load if the reading of the wattmeter is 4488 VAR.
a) 0.6
b) 0.8
c) 1
d) 1.2

Answer: b [Reason:] Wattmeter reading = VLILsinØ => 4488 = 440 x 17sinØ => sinØ = 0.6. Power factor = cosØ = 0.8.

## Network Theory MCQ Set 7

1. The response of a second order system is?
a) pulse
b) saw tooth
c) square
d) sinusoid

Answer: d [Reason:] The response of a second order system is sinusoid. Any periodic waveform can be written in terms of sinusoidal function according to Fourier transform.

2. If a function f(t) is periodic, then?
a) f (t) = f(t+T)
b) f(t) = f(t+T/2)
c) f(t) = f(t+T/4)
d) f(t) = f(t+T/8)

Answer: a [Reason:] If a function f(t) is periodic, then f(t) = f(t+T) in general we can say that a function f(t) is periodic, then f(t) = f(t+nT) for all integer values of n.

3. The period of a function is measured as?
a) zero crossing of one cycle to zero crossing of next cycle
b) positive peak of one cycle to positive peak of next cycle
c) negative peak of one cycle to negative peak of next cycle
d) all of the mentioned

Answer: d [Reason:] The period of a function is measured as zero crossing of one cycle to zero crossing of next cycle or positive peak of one cycle to positive peak of next cycle or negative peak of one cycle to negative peak of next cycle.

4. The number of cycles a wave completes in one second is called?
a) time period
b) frequency
c) energy
d) wavelength

Answer: b [Reason:] The number of cycles a wave completes in one second is called frequency. And the time period is the inverse of frequency.

5. The relation between frequency and time period is?
a) f=1/T
b) f=T
c) f=1/T2
d) f=1/T3

Answer: a [Reason:] The relation between frequency and time period is f=1/T. So the frequency and time period are inversely proportional to each other.

6. The period of a sine wave is 40ms. What is the frequency?
a) 25
b) 50
c) 75
d) 100

Answer: a [Reason:] We know that the frequency and time period are inversely proportional to each other. So f = 1/T = 1/40ms = 25Hz.

7. The frequency of a sine wave is 30Hz. What is its period?
a) 3333ms
b) 333.3ms
c) 33.33ms
d) 3.333ms

Answer: c [Reason:] We know that the frequency and time period are inversely proportional to each other. So T=1/f =1/30=0.03333s=33.33ms.

8. A sine wave completes half cycle in ____ radians.
a) π/2
b) π
c) π/4
d) 2π

Answer: b [Reason:] The period of a sine wave is 360⁰ or 2π radians. So, a sine wave takes 90⁰ or π/2 radians to complete a half cycle.

9. A sine wave completes quarter cycle in ____ radians.
a) 2π
b) π
c) π/2
d) π/4