Question

In a balanced 3-phase 400V circuit, the line current is 115.5 A. When the power is measured by two wattmeter method, one meter reads 40 kW and the other zero. What is the power factor of the load

1. 1
2. 0.5
3. 0.6
4. 0.8

Answer 1

Correct Answer - Option 2 : 0.5

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

Total reactive power in the circuit \(Q=\sqrt{3}\left( {{W}_{1}}-{{W}_{2}} \right)\)

Power factor = cos ϕ

\(\phi =\text{ta}{{\text{n}}^{-1}}\left( \frac{\sqrt{3}\left( {{W}_{1}}-{{W}_{2}} \right)}{{{W}_{1}}+{{W}_{2}}} \right)\)

Calculation:

Given that, one of the wattmeter reads zero.

W2 = 0

\(\tan \phi = \frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{\left( {{W_1} + {W_2}} \right)}}\)

\(\tan \phi = \frac{{\sqrt 3 \left( {{W_1} - 0} \right)}}{{\left( {{W_1} + 0} \right)}}\)

⇒ tan ϕ = √3

⇒ ϕ = tan-1 (√3) = 60°

Power factor = \(\cos \phi = \cos 60^\circ = \frac{1}{2} = 0.5\)