Correct Answer - Option 3 : 17.32 kvar
Concept:
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
Total reactive power in the circuit \(Q = \sqrt 3 \left( {{W_1} - {W_2}} \right)\)
Power factor = cos ϕ
\(\phi = {\rm{ta}}{{\rm{n}}^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)\)
Calculation:
Given that,
W1 = 20 kW, W2 = 10 kW
Total reactive power (Q) = √3(20 – 10) = 17.42 kvar
