# Need help on RL circuit

1. Mar 9, 2005

### andrew410

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 a component of current that is 90 degrees out of 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 this 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-Hz source is in series with a 25-mH inductor and a 20 ohm resistor. What are (a) the rms current and (b) the power factor? (c) What capacitor must be added in series to make the power factor 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?

I got part A. I need some help with part B. I know that power = (rms current)*(rms voltage)*cos(power factor). I also know that power = (rms current)^2*R. I set those two equal together and solve for the power factor, but the answer turned out wrong. Am I doing this wrong?
Any help would be great! thx! :)

2. Mar 9, 2005

### Gamma

See this web site

http://homepages.wmich.edu/~v0koppol/CH17.PPS [Broken]

Last edited by a moderator: May 1, 2017