Final answer:
To increase the motor's power factor to 1.0 when operating on a 120 V / 60 Hz line, approximately 74.5 μF of series capacitance is required.
Step-by-step explanation:
The current drawn by the motor when attached to a 120 V / 60 Hz power line is 7.90 A. Given the average power dissipation is 850 W, we can calculate the power factor.
The apparent power 'S' (in volt-amperes) is given by:
S = V × I
= 120 V × 7.90 A
= 948 VA
Since the actual power 'P' (in watts) is 850 W, the power factor 'pf' is the ratio of P to S:
pf = P / S
= 850 W / 948 VA
≈ 0.896.
To improve the power factor to 1.0, the motor must be operating at unity power factor, indicating no phase difference between voltage and current.
This can be achieved by adding capacitance to the circuit that cancels out the effect of inductance or any existing reactance.
The required reactive power 'Q' (in vars) can be found from:
Q = S × √(1 - (pf)^2)
Q = 948 VA × √(1 - (0.896)^2)
≈ 405.3 var
The capacitive reactance 'Xc' needed is then:
Xc = V^2 / Q ≈ (120 V)^2 / 405.3 var ≈ 35.4 Ω
The series capacitance 'C' to provide this reactance at 60 Hz is:
C = 1 / (2πfXc) ≈ 1 / (2π × 60 Hz × 35.4 Ω)
≈ 74.5 μF
Therefore, we need approximately 74.5 μF of series capacitance to increase the power factor to 1.0 for the motor running on a 120 V / 60 Hz power line.