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 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
W1 = 11086.07 + W2
Putting the value of W1 in equation 1 we get
11086.07 + W2 + W2 = 25602.19
W2 = 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 _____
12F capacitor✓
All the capacitor
3F
6F
The magnitude of the capacitor current IC is given by
IC = V/XC = 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
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.
1.4A
2.42A✓
24.2 A
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 + W2 = 970 + 480 = 1450W
Input power
P = √3.VL.IL.cosφ
1450 = √3 × 400 × IL × 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_____
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’ (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_____
P = 3VPh.IPh.Cosφ✓
P = 3VL.IL.Cosφ
P = √3VL.Iph.Cosφ
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
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 = √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.
2J
4J
6J
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.
0.54
0.98
0.70✓
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_____
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.
