Question

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 and the need for two components of current by installing a capacitor between the power line and your factory. But, you need to convince the owners of the factory to spend the funds to purchase and install this capacitor. You decide to make a presentation to the owners, using a simple RL circuit as a demonstration device. In your demonstration circuit, you represent the power company with a 120 V (rms), 60.0 Hz source. This source is in series with a series combination of a 21.0 mH inductor and a 20.0 ? resistor. This combination represents the inductive and resistive loads for your factory. (a) To impress the owners, you calculate for them the power factor for the circuit and show that it is not equal to 1. power factor = 0.911 (b) You then determine the capacitance (in µF) of a capacitor that will bring the power factor to 1. (c) Demonstrate to the owners the percentage of increased power delivered to the factory. (P_new - P_old) / P_old x 100% = %

          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 and the need for two components of current by installing a capacitor between the power line and your factory. But, you need to convince the owners of the factory to spend the funds to purchase and install this capacitor. You decide to make a presentation to the owners, using a simple RL circuit as a demonstration device. In your demonstration circuit, you represent the power company with a 120 V (rms), 60.0 Hz source. This source is in series with a series combination of a 21.0 mH inductor and a 20.0 ? resistor. This combination represents the inductive and resistive loads for your factory.
(a) To impress the owners, you calculate for them the power factor for the circuit and show that it is not equal to 1.
power factor = 0.911
(b) You then determine the capacitance (in µF) of a capacitor that will bring the power factor to 1.
(c) Demonstrate to the owners the percentage of increased power delivered to the factory.
(P_new - P_old) / P_old x 100% = %

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 and the need for two components of current by installing a capacitor between the power line and your factory. But, you need to convince the owners of the factory to spend the funds to purchase and install this capacitor. You decide to make a presentation to the owners, using a simple RL circuit as a demonstration device. In your demonstration circuit, you represent the power company with a 120 V (rms), 60.0 Hz source. This source is in series with a series combination of a 21.0 mH inductor and a 20.0 ? resistor. This combination represents the inductive and resistive loads for your factory.
(a) To impress the owners, you calculate for them the power factor for the circuit and show that it is not equal to 1.
power factor = 0.911
(b) You then determine the capacitance (in µF) of a capacitor that will bring the power factor to 1.
(c) Demonstrate to the owners the percentage of increased power delivered to the factory.
(Pnew - Pold) / Pold x 100% = %

Added by Amanda G.

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