11. In electrodynamometer type wattmeters, pressure coil inductance produces an error which is:
Constant irrespective of power factor
Higher at low power factors of load
Zero at leading power factors of load
Lower at low power factors of load
Answer.2. Higher at low power factors of load
Explanation:
Pressure coil Inductance error in Electrodynometer wattmeter
In an ideal dynamo-meter type wattmeter the current in the pressure coil is in phase with the applied voltage. But practically pressure coil of a wattmeter has inductance and thus current in it will lag behind the applied voltage.
If there was no inductance, the current in the pressure coil will be in phase with the applied voltage.
That means in the absence of inductance in the pressure coil of the wattmeter, it will read correctly at all power factors and frequencies.
The wattmeter will read high when the load power factor is lagging, as in that case, the effect of pressure coil inductance is to reduce the phase angle between load current and pressure coil current. Hence the wattmeter will read high, which is a very serious error.
The wattmeter will read low when the power factor is leading, in that case, the effect of pressure coil inductance is to increase the phase angle between load current and pressure coil current. Hence the wattmeter will read low.
Hence pressure coil inductance produces an error that is high at low power factors of the load and low at higher factors of the load.
12. A 400 V three-phase 50 Hz balanced source is supplying power to a balanced three-phase load. Line current flowing through the load is 5A at a power factor angle of 30 degrees lagging. Reading of two wattmeters used to measure the load power are:
1000 W, 2000 W
2000 W, 4000 W
2000 W, 3000 W
1000 W, 3000 W
Answer.1. 1000 W, 2000 W
Explanation:
In a two-wattmeter method,
The reading of first wattmeter (W1) = VL IL cos (30 + ϕ)
The reading of second wattmeter (W2) = VL IL cos (30 – ϕ)
Total power in the circuit (P) = W1 + W2 = 3 Vph Iph cos ϕ
Given:
VL = 400 V, IL = 5 A, ϕ = 30°
W1 = 400 × 5 × cos (30 + 30) = 1000 W
W2 = 400 × 5 × cos (30 – 30) = 2000 W
13. A 3 phase balanced load that has a power factor of 0.707 is connected to a balanced supply. The power consumed by the load is 5 kW. The power is measured by the two-wattmeter method. The readings of the two wattmeters are
3.94 kW and 1.06 kW
2.50 kW and 2.50 kW
5.00 kW and 0.00 kW
2.96 kW and 2.04 kW
Answer.1. 3.94 kW and 1.06 kW
Explanation:
Given,
Total power consumed by load W1 + W2= 5 kW
Power factor cos ϕ = 0.707
⇒ ϕ = cos-1 (0.707) = 45°
The reading of first wattmeter (W1) = VL ILcos (30° + ϕ)
W1 = VL IL cos (30° + 45°) = 0.2588 VL IL
The reading of second wattmeter (W2) = VL ILcos (30° – ϕ)
W2 = VL IL cos (30° – 45°) = 0.966 VL IL
The total power consumed by the load is
W1 + W2 = 5 kW = (0.2588 + 0.966) VL IL
⇒ VL IL = 4.082 kW
Therfore the reading in each wattmeter is
W1 = 0.966 VL IL = 0.966 × (4.082 kW)
W1 = 3.94 kW
W2 = 0.2588 VL IL = 0.2588 × (4.082 kW)
W2 = 1.06 kW
14. In a 3-phase system, the two-wattmeter method is used to measure the power. If one of the wattmeters shows a negative reading and the other shows a positive reading, and the magnitude of the readings are not the same, then what will be the power factor (p.f.) of the load?
0.0 < p.f. < 0.5
0.5 < p.f. < 1.0
1
0.5
Answer.1. 0.0 < p.f. < 0.5
Explanation:
The reading of two wattmeters can be expressed as
W1 = VLILcos(30 + φ) W2 = VLILcos(30 − φ)
(i) When PF is unity ( φ = 0°)
W1 = VLILcos30° W2 = VLILcos30°
Both wattmeters read equal and positive reading i.e upscale reading
(ii) When PF is 0.5 (φ = 60°)
W1 = VLILcos90° = 0 W2 = VLILcos30°
Hence total power is measured by wattmeter W2 alone
(iii) When PF is less than 0.5 but greater than 0 i.e ( 90° > φ > 60°)
The wattmeter W2 reads positive (i.e.upscale) because for the given conditions (i.e. ( 90° > φ > 60°), the phase angle between voltage and current will be less than 90°. However, in wattmeter W1, the phase angle between voltage and current shall be more than 90° and hence the wattmeter gives negative (i.e. downscale) reading.
Wattmeter cannot show negative reading as it has only a positive scale. An indication of negative reading is that the pointer tries to deflect in a negative direction i.e. to the left of zero. In such a case, reading can be converted to positive by interchanging either pressure coil connections or by interchanging current coil connections. Remember that interchanging connections of both the coils will have no effect on wattmeter reading.
15. Which power factor results in an equal reading of both wattmeters in the two wattmeter method?
Half
Zero
60°
Unity
Answer.4. Unity
Explanation:
The reading of two wattmeters can be expressed as
W1 = VLILcos(30 + φ) W2 = VLILcos(30 − φ)
(i) When PF is unity ( φ = 0°)
W1 = VLILcos30° W2 = VLILcos30°
Both wattmeters read equal and positive reading i.e upscale reading
16. The power of the system, a three-phase 10 kVA load with a power factor of 0.342, is measured by the two-wattmeter method. The readings of two wattmeters are W1 and W2. What can be said about these readings?
When the power factor is changed from lagging to leading, the reading of W1 and W2 increase.
When power factor is changed from lagging to leading, the reading of W1 and W2 get interchanged.
When power factor is changed from lagging to leading, the reading of W1 and W2 decrease.
When power factor is changed from lagging to leading, the reading of W1 and W2 attain negative sign, always.
Answer.2. When power factor is changed from lagging to leading, the reading of W1 and W2 get interchanged.
Explanation:
The reading of first wattmeter (W1) = VL IL cos (30 + ϕ)
The reading of second wattmeter (W2) = VL IL cos (30 – ϕ)
Hence, When the power factor is changed from lagging to leading, the reading of W1 and W2 get interchanged.
17. The wattmeter reads
Instantaneous power
Average power
Maximum power
None of the above
Answer.2. Average power
Explanation:
A wattmeter is an electrical instrument that is used to measure the electric power of any electrical circuit. The power measured by each wattmeter varies from instant to instant. Due to the inertia of the moving system, the wattmeter measures the average power. The sum of the two wattmeter readings gives the total 3-phase power.
A single wattmeter can also measure the average power in a three-phase system that is balanced. The total power is three times the reading of that one wattmeter. However, two or three single-phase wattmeters are necessary to measure power if the system is unbalanced.
18. A three-phase 500 V motor load has a power factor of 0.4. Two wattmeters connected to measure the input read 20 kW and 10 kW. Find the reactive power (Q)
51.96 kVar
10 kVar
17.42 kVar
30 kVar
Answer.3. 17.42 kVar
Explanation:
Total reactive power in the circuit in two-wattmeter
Q = √3(W1 − W2)
Given that,
W1 = 20 kW, W2 = 10 kW
Total reactive power (Q) = √3(20 – 10) = 17.42 kvar
19. In the two-wattmeter method of measurement of three-phase power of a balanced load, if both wattmeters indicate the same reading, then the power factor of the load is:
0.5 lag
< 0.5 lag
Unity
> 0.5 lag
Answer.3. Unity
Explanation:
The reading of two wattmeters can be expressed as
W1 = VLILcos(30 + φ) W2 = VLILcos(30 − φ)
(i) When PF is unity ( φ = 0°)
W1 = VLILcos30° W2 = VLILcos30°
Both wattmeters read equal and positive reading i.e upscale reading
20. When two wattmeters are used to measure the power of a 3 – φ balanced circuit and one wattmeter reads negative, it means the angle of lag is
0°
30°
60°
Above 60°
Answer.4. Above 60°
Explanation:
The reading of two wattmeters can be expressed as
W1 = VLILcos(30 + φ) W2 = VLILcos(30 − φ)
When PF is less than 0.5 but greater than 0 i.e ( 90° > φ > 60°)
The wattmeter W2 reads positive (i.e.upscale) because for the given conditions (i.e. ( 90° > φ > 60°), the phase angle between voltage and current will be less than 90°. However, in wattmeter W1, the phase angle between voltage and current shall be more than 90° and hence the wattmeter gives a negative (i.e. downscale) reading.
Wattmeter cannot show negative reading as it has only a positive scale. An indication of negative reading is that the pointer tries to deflect in a negative direction i.e. to the left of zero. In such a case, reading can be converted to positive by interchanging either pressure coil connections or by interchanging current coil connections. Remember that interchanging connections of both the coils will have no effect on wattmeter reading.
21. The power of n-phase unbalanced circuit can be measured by using a minimum of _________.
n wattmeters
(n + 1) wattmeters
(n – 2) wattmeters
(n – 1) wattmeters
Answer.4. (n – 1) wattmeters
Explanation:
According to Blondel’s theorem:
When a system contains ‘N’ number of phases & ‘N+1’ wires, then ‘N’ number of watt-meters is required to calculate the total power of the system.
When a system contains ‘N’ number of phases & ‘N’ wires, then the ‘N-1’ number of watt-meters is required to calculate the total power of the system.
In a 3-phase balanced and unbalanced star-connected system, two watt-meters are sufficient to measure the total power.
But if a 3-phase star connected system is balanced then only one wattmeter is sufficient to measure the total power.
