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Three Phase Two Wattmeter Method-2

A symmetrical...

Question

A symmetrical three-phase three-wire RYB system is connected to a balanced delta-connected load. The RMS values of the line current and line-to-line voltage are 10 A and 400 V respectively. The power in the system is measured using the two wattmeter method. The first wattmeter connected between R-line and Y-line reads zero. The reading of the second wattmeter (connected between B-line and Y-line) in watt is
3464.1

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Solution

The correct option is A 3464.1
As the load is balanced and one of the watt-meter read zero power factor is 0.5.

Reading of the other wattmeter will be

$W_{1} = V_{L} I_{L} c o s (30 - ϕ)$
$W_{2} = V_{L} I_{L} c o s (30 + ϕ)$

when,
$c o s ϕ = 0.5$

$W_{2} = V_{L} I_{L} c o s (30 + 60)$

$= V_{L} I_{L} c o s 90 = 0$

$W_{1} = 10 \times 400 c o s (30 - 60) = 3464.1 W$

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