Question

A single-phase load is connected between R and Y terminals of a 220 V, symmetrical, 3-phase, 4 wire systems with phase sequence RYB. A watt-meter is connected in the system. The watt-meter will read (pf = 0.8 lagging)?

(a) – 795 W

(b) – 168 W

(c) + 597 W

(d) + 795 W

I have been asked this question in homework.

I want to ask this question from Advanced Problems on Measurement of Power in Polyphase Circuit in chapter Measurement of Power and Related Parameters of Electrical Measurements

Answer 1

The correct choice is (b) – 168 W

Easiest explanation: VRY = 220∠30°

IRY = \(\frac{220∠30°}{100∠36.86°}\) = 2.2∠-6.68° A

Power = VBN IRY = cos∅ = \(\frac{220}{(3)^{0.5}}\)  × 2.2 cos (126.86°)

= -167.6 W ≈ – 168 W.