A – 3 phase balanced load which has a power factor of 0.707 is connected to 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 watt meters are.
A – 3 phase balanced load which has a power factor of 0.707 is connected to 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 watt meters are. Correct Answer 3.94 kW and 1.06 kW
Concept:
In a two-wattmeter method, for lagging load
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
Calculation:
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 IL cos (30° + ϕ)
W1 = VL IL cos (30° + 45°) = 0.2588 VL IL
The reading of second wattmeter (W2) = VL IL cos (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
