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Three Phase AC Circuits MCQs With Explanatory Answers

Three Phase AC Circuits  (MCQs  With Explanatory Answers)

Three Phase AC Circuits MCQ’s with explanation. For explanatory answer, click on the toggle button labeled as “check explanatory answer”.

Q1. Power in a Three Phase Circuit = _________.

1. P = 3 VPh IPh CosФ
2. P = √3 VL IL CosФ
3. Both 1 & 2.
4. None of The Above

Total Power in a Three Phase Circuit,
P = 3 x Power per Phase,
P = 3 x VPh IPh CosФ
P = 3 VPh IPh CosФ…………(1)

[For a Delta Connection]

[VPh = VL and IPh = IL/√3.]

then putting the values in eq …..(1)
P = 3 x VL x ( IL/√3) x CosФ
P = √3 x√3 x VL x ( IL/√3) x CosФ …{ 3 = √3x√3 }
P = √3 x VLx IL x CosФ ….Ans.

Also
[For Star Connection]

[VPh = VL/√3 and IPh = IL] Putting the values again in eq…….(1)
P = 3 x (VL/√3 ) x IL x CosФ
P = √3 x√3 x (VL/√3 ) x IL x CosФ …{ 3 = √3x√3 }
P = √3 x VL x IL x CosФ ….Ans.

Q2. A polyphase system is generated by______?

1. Having two or more generator windings separated by equal electrical angle.
2. Having generator windings at equal distances
3. None of the above
4. A and C

Answer: 1. Having two or more generator windings separated by equal electrical angle.

A generator having two or more electrical windings which are separated by equal electrical angle generates a polyphase electrical system. The electrical angle or displacement depends upon the number of windings or phases. For example, in a three-phase electrical system, the generated voltages are separated from each other by 120° degrees.

Q3. In a three phase AC circuit, the sum of all three generated voltages is _______ ?

1. Infinite ()
2. Zero (0)
3. One (1)
4. None of the above

Three phase voltages are generated by having an alternator with three armature windings such that each winding is displaced from the other by 120 degrees. When these windings are placed in a rotating magnetic field or rotated in a stationary magnetic field, electromotive force is generated in each coil, of same magnitude and direction. Consider the below diagram

Figure : 3 Phase AC Waveforms

As seen EMF generated in coil R-R1 is eR, which is the reference in this case. EMF generated in coil Y-Y1 is eY which is 120° degrees ahead of eR and EMF generated in coil B-B1 is eB which is 240° degrees ahead of eR.

Therefore the voltage equations are as given below;

eR=Em sin⁡ wt
eY= Em sin⁡ (wt – 120°)
eB= Em sin⁡ (wt – 240) = Em sin ⁡(wt + 120°)
Adding all three equations, we get
eR+eY+eB = Em (sin ⁡wt + sin⁡ (wt – 120°) + sin ⁡(wt + 120°) )
= Em (sin ⁡wt + sin⁡ wt cos⁡ 120° – cos⁡wt sin⁡ 120° + sin⁡ wt cos ⁡120°+cos ⁡wt sin⁡ 120°)=0
i.e, eR+eY+eB = 0

Hence, sum of all three voltages is zero.

Q4. For a star connected three phase AC circuit ———

1. Phase voltage is equal to line voltage and phase current is three times the line current
2. Phase voltage is square root three times line voltage and phase current is equal to line current
3. Phase voltage is equal to line voltage and line current is equal to phase current
4. None of the above

Answer: 2.  Phase voltage is square root three times line voltage and phase current is equal to line current

A star connected AC circuit is achieved by connecting each end of the winding to a common point known as neutral point and leaving the other end of each winding free. While voltage across each coil is the phase voltage, potential difference between each free end is the line voltage.

Consider the circuit below;

Now as said above, phase voltages are equal

Hence, VNR = VNY = VNB = Vph

Therefore, line voltage,

VRY = √3 VPH

Since the line conductor is in series with the phase winding, same current will flow through the line conductor as through the phase windings, hence phase current is equal to phase current.

Q5. In a three phase, delta connection ——-

1. line current is equal to phase current
2. Line voltage is equal to phase voltage
3. None of the above
4. Line voltage and line current is zero

Answer: 2.  Line voltage is equal to phase voltage

A delta connected AC circuit is achieved by connecting the start end of a winding to the finish end of another winding such that all three windings form a mesh. Since each end of the windings forms the line connection, voltage across each winding is equal to the potential difference between the corresponding lines taken from that winding. Hence the phase voltage is equal to the line voltage.

Q6. For a star connection network, consuming power of 1.8kW and power factor 0.5, the inductance and resistance of each coil at a supply voltage of 230 Volts, 60 Hz is ______?

1. 0.1H, 8 Ohms
2. 0.5H, 10 Ohms
3. 0.3H, 7.4 Ohms
4. 1H, 7 Ohms

Given values are:

Line voltage, VL = 230 V

Line frequency, f = 60 Hz

Power Factor, cosφ = 0.5

Power consumed = P = 1800 Watts = 3 VL x IL x cosφ

Hence, line current, IL = 9 Amperes

Since it is a star connection, phase current = line current = 9 Amperes

Phase Voltage, Vph = VL/3 = 132.8 Volts

Phase Impedance, Zph = Vph/Iph = 14.7 Ohms

Now, Power Factor = Resistance/Impedance

Hence, Resistance of Coil = Impedance X Power Factor = 7.4 Ohms

Substituting values, we get Reactance of coil = 12.7 Ohms

Thus, inductance of coil , L = 0.03H

Q7. For a three-phase delta connected load, fed from a star connected network, the power transferred to the load is _____?

1. 3 kW
2. 4.7 kW
3. 5 kW
4. 7 kW

Given values:

Star Connected phase voltage, VPH = 230 Volts

Phase load resistance, RPHLd = 20 Ohms

Phase load reactance, XPHLd= 40 Ohms

Star connected line voltage, VL = VPHs = 398.37 Volts

For the delta connected load, Phase Voltage, VPHLd = VL = 398.37 Volts

Therefore, current through each phase of load, IPHLd = VPHLd / ZPHLd = 8.9 Amperes

Line current for delta connected load, IL =  3 IPHLd = 15.41 Amperes

Power Factor, pfs = RphLd/ ZPHLd = 0.44

Thus, the power supplied to load then, PL =  VL IL pfs = 4.7 KW

Q8. In a three phase AC circuit, power is measured using a Wattmeter.

1. True
2. False

Power is measured using a Wattmeter which consists of two coils – Current coil, connected in series with the load, carrying the load current and Voltage coil, connected in parallel to the load.

Q9. For a polyphase system, the number of Wattmeter required to measure power is equal to ——

1. Number of wires
2. One less than number of wires
3. Number of phases
4. None of the above

Answer: 2.  One less than number of wires

The number of Wattmeter required to measure power in a polyphase system is determined using Blondell’s theorem. According to this, the number of Wattmeter required is equal to one less than the number of wires in the circuit. For example, in a three phase, four wire system (Star network), the number of Wattmeter required is three.

Q10.  For the below star connected network of equal resistances, if the Wattmeter reading is 5kW and ammeter reading is 25 Amperes, the power factor, resistance and inductance are __________  respectively.

1. 1.5 Ohms, 0.1H
2. 0.866, 8 Ohms, 0.02H
3. 5.10 Ohms, 0.01H
4. 4. Ohms, 0.02H

Answer: 2.  0.866, 8 Ohms, 0.02H

Given

Line Voltage, VL = 400 Volts

Frequency, f = 60 Hz

Line Current, IL = 25 Amperes

Power per phase, Pph = 5kW

Phase Voltage, Vph = VL/3^1/2 = 230.9 Volts

Phase current, Iph = 25 Amperes

Hence, power factor, cosφ = Pph/VphIph = 0.866

Impedance, Zph = Vph/Iph = 9.236 Ohms

Resistance, R = Zphcosφ = 8 Ohms

Putting the values in the below equation, Reactance, X =  Therefore, Inductance, L = 0.02H

Q11. For a three phase, three wire system, the two Wattmeter read 4000 Watts and 2000 Watts respectively. The power factor when both meters give direct reading is _______ ?

1. 1
2. 0.5
3. 0.866
4. 0.6

Reading of Wattmeter 1, W1 = 4000 Watts

Reading of Wattmeter 2, W2 = 2000 Watts

Phase angle;

Power Factor,  = 0.866

Q12. For a balanced three phase, three wire system with input power of 10kW, at 0.9 power factor, the readings on both wattmeter are ————– respectively

1. 7kW, 3kW
2. 6350W, 3650W
3. 5000W, 5000W
4. 7600W, 1200W

Let reading of one Wattmeter = W1

Reading of second Wattmeter = W2

Input Power, P = W1+W2 = VL IL cosφ = 10 kW ……………… (1)

Power Factor, cos φ = 0.9

Phase angle, φ = 25.8 degrees …… (i.e. Cos -1 = 09 = 25.8°)

Therefore,

W1 = VL IL cos (30 – φ) = 0.99VL IL = 6350W

W2 = VL IL cos (30 + φ) = 0.56 VL IL = 3650W

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