Suppose you manage a factory that uses many electric motors. The motors create a

Question

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 with the voltage 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 30.0-mH inductor and a 16.0- resistor.

(a) What is the rms current?
???? A

(b) What is the power factor?


(c) What capacitor must be added in series to make the power factor equal to 1?
???? µF

(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?
??? V

Explanation / Answer

Impedence Z^2 = R^2 + XL^2

Xl = wL = 2pi fL

XL = 2*3.14 * 60 * 0.03 = 11.3 ohms

s Z^2 = 16^2 +11.3^2

Z = 19.58 ohms

part A :

Irms = Vrms/Z

Irms = 120/19.58 = 6.12 Amps

---------------------------------

POwer Factor Cos theta = R/Z = 16/19.58 = 0.817

---------------------------------

if power factor = 1

then R = Z

if R= Z, XL - Xc = 0 or XL = Xc

Xc= 1/2pifC = 11.3

C = 1/(2 *3.14 * 60 * 11.3)

C = 234.8 uF

----------------------------------------

Vrms = IRmms * Z

Vrms = 6.12 * 16

Vrms = 98 V

so at 98 Volts, with capaciatcne C, gives same power

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