Correct Answer : Option (C) : 6350W, 3650W
Explanation :
Let reading of one Wattmeter = W1
Reading of second Wattmeter = W2
Input Power, P = W1+W2 = √3VLILcosφ = 10000………………1
Power Factor, cosφ = 0.9
Phase angle, φ = 25.8 degrees
Therefore, W1 = VLILcos(30-φ) = 0.99VLIL = 6350W
W2 = VLILcos(30+φ) = 0.56VLIL = 3650W