Ques.71. A three-phase star-connected balanced load of (4+ j3)Ω per phase is connected across three-phase, 50 Hz, 400V AC supply. If the two wattmeter method is used to determine input power, find each wattmeter reading.

18.34 kW and 7.26 kW✓

18.34 kW and 46 kW

23 kW and 23 kW

25.60 kW and 23.23 kW

Given

Impedance Z = (4+ j3)Ω

Resistance R = 4 Ω

Line voltage V_{L} = 400 V

Impedance per phase Z_{P}

Z_{P} = √(R^{2} + X^{2}_{L})

Z_{P} = √(4^{2} + 3^{2})

Z_{P} = 5Ω

In star connected system line voltage is equal to the phase voltage

V_{P} = V_{L}/√3 = 400/√3 = 230.94V

I_{Ph} = V_{p}/Z_{p}

I_{P} = 230.94/5

I_{P} = 46.19A

The power consumed by the three-phase delta connected circuit is

P = 3I^{2}_{p} R_{p}

P = 3 × 46.19^{2} × 4

25602.19 W

W_{1} + W_{2} = 25602.19——-1

Power factor = Rp/Zp = 4/5

Cosφ = 0.7 = φ = 36.86°

Power factor of the two wattmeter is

tanØ = √3[(W_{1} – W_{2}) / (W_{1} + W_{2})]

tan36.86° = √3[(W_{1} – W_{2}) ⁄ 25602.19

11086.07 = W_{1} – W_{2}

W_{1 }= 11086.07 + W_{2}

Putting the value of W_{1 }in equation 1 we get

11086.07 + W_{2} + W_{2} = 25602.19

W_{2 }= 7258.06W = 7.25 kW

W_{1} = 18344.13 W = 18.34 kW

Ques.72. Three capacitors of 3F, 6F, and 12F are connected in parallel across an AC source. The maximum current will pass through the _____

12F capacitor✓

All the capacitor

3F

6F

The magnitude of the capacitor current I_{C} is given by

I_{C }= V/X_{C }= V/1/2.π.f.C = 2.π.f.C.V

or

Charged in the capacitor

Q = C.V

where Q = IT

∴ I = CV/Q

Hence from the above relation, it is clear that the current is directly proportional to the voltage and capacitor.

Therefore the capacitance of higher value will be drawn the maximum current i.e 12 F

Ques.73. The balanced load of a delta connection is powered by a three-phase balanced 400 V, 50 Hz AC power supply. The readings of the two power meters are 970 W and 480 W respectively. Each phase load consists of a series of resistors and inductors. Calculate the power factor.

1

0.86 lag✓

0.86 lead

0.98 lag

As given in the above question

W_{1} = 970 W

W_{2} = 480 W

V_{L} = 400 V

The power factor of the two wattmeters is

tanØ = √3[(W_{1} – W_{2}) / (W_{1} + W_{2})]

tanØ = √3[(970 – 480) ⁄ (970 + 480)]

Φ = 30°

Power factor = cosφ = cos30° = 0.86

Since the load is inductive in nature hence the power factor will be lagging

0.86 lag

Ques.75. The balanced load of a delta connection is powered by a three-phase balanced 400 V, 50 Hz AC power supply. The readings of the two power meters are 970 W and 480 W respectively. Each phase load consists of a series of resistors and inductors. Calculate the line current.

1.4A

2.42A✓

24.2 A

14A

As given in the above question

W_{1} = 970 W

W_{2} = 480 W

V_{L} = 400 V

The power factor of the two wattmeters is

tanØ = √3[(W_{1} – W_{2}) / (W_{1} + W_{2})]

tanØ = √3[(970 – 480) ⁄ (970 + 480)]

Φ = 30°

Power factor = cosφ = cos30° = 0.86

Total power

W = W_{1} + W_{2 }= 970 + 480 = 1450W

Input power

P = √3.V_{L}.I_{L}.cosφ

1450 = √3 × 400 × I_{L }× 0.86

I_{L} = 2.43 A

Ques.75. In a three-phase system, the current passing through any two lines of the supply is called as_____

Line voltage

Phase current

Phase voltage

Line current✓

The potential difference between any two lines of supply is called line voltage and the current passing through any two line is called line current.

The potential difference between any one of the three lines and the neutral conductor or earth is called ‘phase voltage’ (V_{ph}). and the current passing through any branch of the three-phase load is called phase current.

Ques.76. The power in a the three-phase circuit is given by the equation_____

P = 3V_{Ph}.I_{Ph}.Cosφ✓

P = 3V_{L}.I_{L}.Cosφ

P = √3V_{L}.I_{ph}.Cosφ

P = √3V_{Ph}.I_{Ph}.Cosφ

The power in three-phase AC circuit is given by the following relation

P = 3V_{Ph}.I_{Ph}.Cosφ

or

P = √3V_{L}.I_{L}.Cosφ

or

P = 3I^{2}_{ph}R_{ph}

Where

V_{ph} = Voltage per phase

I_{ph} = Current per phase

V_{L} = Line voltage

I_{L} = Line current

R_{ph} = Resistance per phase

Ques.77. Consider the three phase and match the following

A-F, B-D, C-E

A-F, B-E, C-D✓

A-E, B-F, C-D

A-D, C-F, B-E

In three phase circuit, the active power in the circuit is given by the equation

P = √3V_{L}.I_{L}.cosφ

Reactive Power

P = √3V_{L}.I_{L}.sinφ

Apparent power

P = √3V_{L}.I_{L}

Where

V_{L} = Line voltage

I_{L} = Line current

Ques.78. When a 4-V EMF is applied across a 1-F capacitor, it will store_____of energy.

2J

4J

6J

8J✓

Energy stored in the capacitor

U = CV^{2}/2 joule

where

C = Capacitance = 1 F

V = Voltage = 4 V

U = 1 × 4^{2}/2

U = 8 Joules

Ques.79. A balanced three-phase star-connected load draws power from a 440V supply. The two connected wattmeters W_{1} and W_{2} indicate 5 kW and 1200 W respectively. Calculate the power factor of the system.

0.54

0.98

0.70✓

0.75

As given in the above question

W_{1} = 5 kW

W_{2} = 1200 W

V_{L} = 440 V

The power factor of the two wattmeters is

tanØ = √3[(W_{1} – W_{2}) / (W_{1} + W_{2})]

tanØ = √3[(5000 – 1200) ⁄ (5000 + 1200)]

Φ = 45°

Power factor = cosφ = cos45° = 0.707

Ques.80. The unit of capacitance is_____

Henry

Coulomb/Volt✓

Volts/Coulomb

Ohms

Capacitance is the ratio of the change in an electric charge in a system to the corresponding change in its electric potential. The SI unit of capacitance is the farad (symbol: F). A 1-farad capacitor, when charged with 1 coulomb of electrical charge, has a potential difference of 1 volt between its plates.