Significant Figures Feedback: Your answer 90.476 Vwas either rounded differently

Question

Significant Figures Feedback: Your answer 90.476 Vwas either rounded differently or used a differe than required for this part. If you need this result for any later calculation in this item, keep all the digits submitting your answer. Significant Figures Feedback: Your answer 10.770 12 was either rounded differently or used a differ* than required for this part. If you need this result for any later calculation in this item, keep all the digits submitting your answer. A motor attached to a 120 V/60 Hz power line draws an 8.40 A current. Its average energy dissipation is 760 W.

Explanation / Answer

If you add series capacitance, the voltage drop across the capacitor may decrease/increase the voltage across the motor to a level that the motor may not work. Power factor correction is typically done with the capacitor in parallel with the motor.

Apparent Power
S = V_S I = 120V * 8.40A = 1008 VA

Power Factor
pf = P / S = 760 W / 1008VA = 0.753 lagging

Reactive Power
Q_L = ( S² - P² )= ( (1008 VA)² - (760W)² ) = 662VAR

This is where it gets a little dicey for me (as an engineer). I'd do this with powers and parallel capacitor so that the voltage is known.

Inductive Reactance
Q_L = I² X_L
X_L = Q_L / I² = 662.12VAR / (8.4A)² = 9.38

To bring the pf = 1, leading capacitive reactive power must equal lagging inductive reactive power:
X_C = X_L = 9.38

=== Capacitance+
C = 1 / (2 f X_C) = 1 / (2 * * 60Hz * 9.38) = 283F

That's the answer. New current will be (Z = R, since X_L = X_C - sort of like resonance):
I = V_S / R = 120V / 10.8 = 11.1A

V_motor = I Z_motor = 11.1A * 14.1 = 157V

So motors typically handle 10% variation in supply voltage.

37V / 120V * 100% = +30.8% increase, at increased current, which means motor will/may overheat.

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