Question

A 3-phase balanced R-L load is connected to a 3-phase balanced symmetrical 400 V, 50 Hz, RYB sequence mains as shown in figure. The wattmeter W reads 1.732 kW. Then,

• A. the active power in the load is 1.732 kW
• B. the reactive power of the load is 1.732 kVAr
• C. the active power in the load is 3 kW
• D. the reactive power of the load is 3 kVAr

Answer 1

The correct option is B the reactive power of the load is 1.732 kVAr

Wattmeter reading,

$B=\left|V_{PC}\right|\left|I_{CC}\right|cos\varphi$

Now,

$V_{PC}=V_B=V_Y-V_B$

$I_{CC}=I_R$


For phasor diagram,

$\varphi =(90-\theta )$

So,

Wattmeter reading,

$W=\left|V_{YB}\right|\left|I_R\right|cos\varphi$

$\Rightarrow W=\sqrt{3}VIcos(90-\theta )$

$\Rightarrow W=\sqrt{3}VIsin\theta =Q$

So, meter reads reactive power of the load.