Maximizing Power Efficiency: Solving for Power Factor in RL Circuit

[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! :)

 

[Gamma]

See this web site

http://homepages.wmich.edu/~v0koppol/CH17.PPS

 

[thepatientmental]

In order to solve for the power factor in an RL circuit, we need to first understand what it represents. Power factor is a measure of how efficiently the electrical power is being used in a circuit. In this case, the inductive load from the electric motors is causing the current to lag behind the voltage, resulting in a low power factor.

To solve for the power factor, we can use the formula: power factor = cos(theta), where theta is the angle between the voltage and current. In this case, the angle is 90 degrees, since the current is 90 degrees out of phase with the voltage.

Using the formula, we can calculate the power factor to be 0.707. This means that only 70.7% of the supplied power is being used effectively, while the remaining 29.3% is being lost due to the inductive load.

To improve the power factor to 1, we need to add a component that will create a capacitive load, which will counteract the inductive load. This can be achieved by connecting a capacitor in series with the circuit. The value of the capacitor needed to achieve a power factor of 1 can be calculated using the formula: C = 1/ (2*pi*f*R), where C is the capacitance, f is the frequency (60 Hz in this case), and R is the resistance (20 ohms). Plugging in the values, we get a capacitance of approximately 13 microfarads.

Finally, to determine the new supply voltage that will result in the same power being supplied, we can use the formula: V2 = V1 * (power factor2 / power factor1), where V1 is the original supply voltage (120 V), and power factor1 and power factor2 are the original and desired power factors, respectively. Plugging in the values, we get a new supply voltage of approximately 169 V.

In conclusion, by installing a capacitor in the circuit to improve the power factor, we can avoid paying extra fees for reactive volt-amps and also reduce the supply voltage while maintaining the same power supply. This not only saves costs for the factory but also results in a more efficient use of electrical power.