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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.
1. 3.94 kW and 1.06 kW
2. 2.50 kW and 2.50 kW
3. 5.00 kW and 0.00 kW
4. 2.96 kW and 2.04 kW

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Correct Answer - Option 1 : 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

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