# UPPCL JE 2018 Electrical question paper with Explanation 27-Aug-2018

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.

1. 18.34 kW and 7.26 kW
2. 18.34 kW and 46 kW
3. 23 kW and 23 kW
4. 25.60 kW and 23.23 kW

Given

Impedance Z = (4+ j3)Ω

Resistance R = 4 Ω

Line voltage VL = 400 V

Impedance per phase ZP

ZP = √(R2 + X2L)

ZP = √(42 + 32)

ZP = 5Ω

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

VP = VL/√3 = 400/√3 = 230.94V

IPh = Vp/Zp

IP = 230.94/5

IP = 46.19A

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

P = 3I2p Rp

P = 3 × 46.192 × 4

25602.19 W

W1 + W2 25602.19——-1

Power factor = Rp/Zp = 4/5

Cosφ = 0.7 = φ = 36.86°

Power factor of the two wattmeter is

tanØ =  √3[(W1 – W2) / (W1 + W2)]

tan36.86° = √3[(W1 – W2) ⁄ 25602.19

11086.07 = W1 – W2

W= 11086.07 + W2

Putting the value of W1  in equation 1 we get

11086.07 + W2 + W2 = 25602.19

W=  7258.06W = 7.25 kW

W1 = 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 _____

1. 12F capacitor
2. All the capacitor
3. 3F
4. 6F

The magnitude of the capacitor current IC is given by

I= V/X=  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. 1
2. 0.86 lag
4. 0.98 lag

As given in the above question

W1 = 970 W

W2 = 480 W

VL = 400 V

The power factor of the two wattmeters is

tanØ =  √3[(W1 – W2) / (W1 + W2)]

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. 1.4A
2. 2.42A
3. 24.2 A
4. 14A

As given in the above question

W1 = 970 W

W2 = 480 W

VL = 400 V

The power factor of the two wattmeters is

tanØ =  √3[(W1 – W2) / (W1 + W2)]

tanØ = √3[(970 – 480) ⁄ (970 + 480)]

Φ = 30°

Power factor = cosφ = cos30° = 0.86

Total power

W = W1 + W= 970 + 480 = 1450W

Input power

P = √3.VL.IL.cosφ

1450 = √3 × 400 × I× 0.86

IL = 2.43 A

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

1. Line voltage
2. Phase current
3. Phase voltage
4. 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’ (Vph). 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_____

1. P = 3VPh.IPh.Cosφ
2. P = 3VL.IL.Cosφ
3. P = √3VL.Iph.Cosφ
4. P = √3VPh.IPh.Cosφ

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

P = 3VPh.IPh.Cosφ

or

P = √3VL.IL.Cosφ

or

P = 3I2phRph

Where

Vph = Voltage per phase

Iph = Current per phase

VL = Line voltage

IL = Line current

Rph = Resistance per phase

Ques.77. Consider the three phase and match the following 1. A-F, B-D, C-E
2. A-F, B-E, C-D
3. A-E, B-F, C-D
4. A-D, C-F, B-E

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

P = √3VL.IL.cosφ

Reactive Power

P = √3VL.IL.sinφ

Apparent power

P = √3VL.IL

Where

VL = Line voltage

IL = Line current

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

1. 2J
2. 4J
3. 6J
4. 8J

Energy stored in the capacitor

U = CV2/2 joule

where

C = Capacitance = 1 F

V = Voltage = 4 V

U = 1 × 42/2

U = 8 Joules

Ques.79. A balanced three-phase star-connected load draws power from a 440V supply. The two connected wattmeters W1 and W2 indicate 5 kW and 1200 W respectively. Calculate the power factor of the system.

1. 0.54
2. 0.98
3. 0.70
4. 0.75

As given in the above question

W1 = 5 kW

W2 = 1200 W

VL = 440 V

The power factor of the two wattmeters is

tanØ =  √3[(W1 – W2) / (W1 + W2)]

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

Φ = 45°

Power factor = cosφ = cos45° = 0.707

Ques.80. The unit of capacitance is_____

1. Henry
2. Coulomb/Volt
3. Volts/Coulomb
4. 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.

C = Q/V = Coulomb/volt

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