Two-wattmeter method star load

Given: $W_1=2100$ W, $W_2=1550$ W, $V_L=400$ V, star connection.

(i) Per-phase average power. Total power is the sum of the readings:

$$P = W_1+W_2 = 3650\ \text{W}, \qquad P_{ph}=\frac{3650}{3}=1216.67\ \text{W}$$

(ii) Per-phase reactive power. Total reactive power is $\sqrt3\,(W_1-W_2)$:

$$Q=\sqrt3(2100-1550)=952.63\ \text{VAR}, \qquad Q_{ph}=\frac{952.63}{3}=317.54\ \text{VAR}$$

(iii) Power factor.

$$\tan\phi=\frac{Q}{P}=\frac{952.63}{3650}=0.261\ \Rightarrow\ \phi=14.63^\circ$$ $$\cos\phi=0.9676\ \text{(lagging, for an inductive load)}$$

(iv) Phase impedance. Line current from $P=\sqrt3 V_L I_L\cos\phi$:

$$I_L=\frac{3650}{\sqrt3\times400\times0.9676}=5.445\ \text{A}=I_{ph}$$ $$V_{ph}=\frac{400}{\sqrt3}=230.94\ \text{V}, \qquad Z_{ph}=\frac{230.94}{5.445}=42.41\ \Omega$$

Then $R_{ph}=Z\cos\phi=41.04\ \Omega$ and $X_{ph}=Z\sin\phi=10.71\ \Omega$, so $Z_{ph}=41.04+j10.71\ \Omega$.

Answer: (i) 1216.67 W, (ii) 317.54 VAR, (iii) 0.9676 lagging, (iv) 42.41 Ω ($41.04+j10.71\ \Omega$).