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A synchronous...

Question

A synchronous motor improves the power factor of a 300kW load from 0.6 lagging to 0.8 lagging and simultaneously the motor carries a load of 100 kW. The power factor at which the motor operates is ______
0.7

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Solution

The correct option is A 0.7
Load, $P_{1} = 300 k W, m o t o r l o a d = 100 k W$
P.f. of load, $c o s ϕ_{1} = 0.6 l a g g i n g$
P.f. of load, $c o s ϕ_{2} = 0.8 l a g g i n g$
Combined load $P = P_{1} + P_{2} = 300 + 100 = 400 k W$
Leading kVAR taken by the motor
$= P_{1} t a n ϕ_{1} - P_{2} t a n ϕ_{2}$
$= 300 t a n (c o s^{- 1} 0.6) - 400 t a n (c o s^{- 1} 0.8)$
= 400 - 300 = 100 kVAR
Rating of the motor $= \sqrt{(100)^{2} + (100)^{2}} = 141.42 k V A$
$P . f o f m o t o r, c o s ϕ_{m} = \frac{100}{141.42} = 0.707$ leading

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