Question

In the measurement of power of balanced load by two wattmeter method in a 3-phase circuit, The readings of the wattmeters are 4 kW and 2 kW respectively, the later is being taken after reversing the connections of current coil. the power factor and reactive power of the load is

1. 0.2 & 6 kVAR
2. 0.2 & 6 √3 kVAR
3. 0.32 & 2 kVAR
4.

0.32 & 2 √3 kVAR

Answer 1

Correct Answer - Option 2 : 0.2 & 6 √3 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

\(ϕ = {\rm{ta}}{{\rm{n}}^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)\)

Calculation:

Given that, 

Reversing the connections

W1 = 4 kW, W2 = - 2 kW

Total reactive power (Q) = √3[4 - (- 2)] = 6 √3 kVAR

\(ϕ = {\rm{ta}}{{\rm{n}}^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)\)

\(ϕ = {\rm{ta}}{{\rm{n}}^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{4} - {(-2)}} \right)}}{{{4} - {2}}}} \right)\)

\(ϕ = {\rm{ta}}{{\rm{n}}^{ - 1}}\left( {\frac{{6\sqrt 3}}{{{2}}}} \right)\)

ϕ = 79.10^o

cosϕ = 0.2 (approx)

 

p.f. angle (ϕ)	p.f.(cos ϕ)	W1 [VLIL cos (30 + ϕ)]	W2 [VLIL cos (30 - ϕ)]	W = W1 + W2 [W = √3VLIL cos ϕ]	Observations
0°	1	\(\frac{{\sqrt 3 }}{2}{V_L}{I_L}\)	\(\frac{{\sqrt 3 }}{2}{V_L}{I_L}\)	√3 VLIL	W1 = W2
30°	0.866	\(\frac{{{V_L}I_L}}{2}\)	VLIL	1.5 VL IL	W1= 2W2
60°	0.5	0	\(\frac{{\sqrt 3 }}{2}{V_L}{I_L}\)	\(\frac{{\sqrt 3 }}{2}{V_L}{I_L}\)	W1 = 0
90°	0	\(\frac{{ - {V_L}{I_L}}}{2}\)	\(\frac{{{V_L}{I_L}}}{2}\)	0	W1 = -ve W2 = +ve