5) (Based on Jackson, ch 22, Q 24, 29) An induction motor draws 10.0A at a 0.8la

Question

5) (Based on Jackson, ch 22, Q 24, 29) An induction motor draws 10.0A at a 0.8lagging Power Factor from a 208 VAC, 60 Hz source: a) What value of capacitance must be connected across the motor to increase the Power Factor to 0.96lagging? b) What will the Power Factor be if a 100Mu F, capacitor is connected in parallel with the motor instead of the value you determined in a)? c) What are the current draws from the 208V source (magnitude) for each of the 3 cases above (i.e. uncompensated, Fp = .96, and with a 100MuF capacitor)? d) What are the power savings (i.e. % I^R? loss savings) at a .96 Power Factor?

Explanation / Answer

INITIAL CURRENT = 10(0.8-0.6j) = 8-6j

For 0.96 power factor

Current = 10(0.96 - 0.28j) = 9.6-2.8j

The difference of reactive components = (6-2.8)j = 3.2j

This is supplied by capacitor, so

V/(wc) = 3.2

w = 2 *3.14 * 60 = 376.8

C = 208/(376.8*3.2) = 0.172505307F

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