Discussion
During the measurement of power by two-wattmeter method, the total input power, to a three-phase, 440 V motor running at power factor of 0.8, was found to be 25 kW. The readings of wattmeters will be :
[Given : cos−1(0.8) = 36.87°, sin (36.87) = 0.6]
Answer: C
The readings are 17.9 kW and 7.1 kW.
In the two-wattmeter method the sum of the readings is the total power, and their difference is related to it through the power factor angle:
\[ \begin{aligned} W_{1} + W_{2} &= 25 \\ W_{1} - W_{2} &= \frac{(W_{1} + W_{2})\tan\varphi}{\sqrt{3}} \\ &= \frac{25 \times 0.75}{1.732} \approx 10.83 \end{aligned} \]
Here \(\tan\varphi = 0.6/0.8 = 0.75\). Solving, \(W_{1} \approx 17.9\) kW and \(W_{2} \approx 7.1\) kW.
Equal readings of 12.5 kW occur only at unity power factor. The pairs 20.2 and 4.8, and 10.9 and 14.1, add up to 25 kW but have the wrong difference for a power factor of 0.8.
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