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Three Phase Two Wattmeter Method-2

A 3-phase bal...

Question

A 3-phase balanced load which has a power factor of 0.707 is connected to a balanced supply. The power consumed by the load is 5 kW. The power is measured by the two-wattmeter method. The readings of the two wattmeters are

A

3.94 kW and 1.06 kW

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B

2.50 kW and 2.50 kW

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C

5.00 kW and 0.00 kW

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D

2.96 kW and 2.04 kW

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Solution

The correct option is A 3.94 kW and 1.06 kW
$p . f . = 0.707, ϕ = c o s^{- 1} (0.707) = 45^{\circ}$
$P = 5 k W$
$W_{1} + W_{2} = 5 . . . (i)$
$t a n ϕ = \frac{\sqrt{3} (W_{1} - W_{2})}{(W_{1} + W_{2})}$
$(W_{1} + W_{2}) \times 1 = \sqrt{3} (W_{1} - W_{2})$
$W_{1} - W_{2} = \frac{5}{\sqrt{3}} . . . (i i)$
Form equation (i) and (ii),
$W_{1} = 3.94 k W$
$W_{2} = 1.06 k W$

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