1. 2. Please answer ALL PARTS of these QUESTIONS with COMPLETE STEPS, only then

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Please answer ALL PARTS of these QUESTIONS with COMPLETE STEPS, only then I will give you a 5-star rating.

Suppose you manage a factory that uses many electric motors. The motors create a large inductive load to the electric power line as well as a resistive load. The electric company builds an extra-heavy distribution line to supply you with two components of current: one that is 90° out of phase withthevoltage and another that is in phase with the voltage. The electric company charges you an extra fee for "reactive volt-amps" in addition to the amount you pay for the energy you use. You can avoid the extra fee by installing a capacitor between the power line and your factory. The following problem models this solution. In an RL circuit, a 120-V (rms), 60.0-Hz source is in series with a 24.0-mH inductor and a 21.0-n resistor. (a) What is the rms current? (b) What is the power factor? (c) What capacitor must be added in series to make the power factor equal to 1? (d) To what value can the supply voltage be reduced if the power supplied is to be the same as before the capacitor was installed?

Explanation / Answer

(A) R = 21 ohm

XL = 2 pi f L = 2 x pi x 60 x 0.0240

XL = 9.05 ohm

Z = sqrt(R^2 + XL^2) = 22.9 ohm

Irms = Vrms / Z = 5.25 A

(B) cos(phi) = R/Z = 0.918

(c) xC = XL

1 / 2 pi C = 2 pi f L

C = 1 / (4 pi^2 x 60^2 x 0.024)

C = 293 x 10^-6 F = 293 uF

(d) P1 = Vrms Irms cos(phi) = 578.4

now P = V^2 / R

578.4 = V^2 / 21
V = 110 Volt

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