1. Power in Electrical Systems Power in Electrical Systems

2. Power in Electrical Systems : Objectives:   Explain the relationship between power, current, and voltage in electrical systems.   Explain the relationship between power, current, and resistance in electrical systems.   Calculate energy usage in kilowatt-hours.   Solve problems involving power in electrical systems : Objectives:   Explain the relationship between power, current, and voltage in electrical systems.   Explain the relationship between power, current, and resistance in electrical systems.   Calculate energy usage in kilowatt-hours.   Solve problems involving power in electrical systems

3. Power in Electrical Systems Voltage causes charges to move in circuits. This movement creates work. How fast the charge does work is called Power. Voltage causes charges to move in circuits. This movement creates work. How fast the charge does work is called Power.

4. Power in Electrical Systems In electrical systems, power is measured in watts (W), current is measured in amperes (A), and potential difference is measured in volts (V). 1 Watt = 1 Ampere x 1 Volt 1 W = 1 A·V In electrical systems, power is measured in watts (W), current is measured in amperes (A), and potential difference is measured in volts (V). 1 Watt = 1 Ampere x 1 Volt 1 W = 1 A·V

5. Power in Electrical Systems Power = Current x Voltage P = I x V Power = Current x Voltage P = I x V

6. Power in Electrical Systems A circuit in the power supply for a high- powered laser has a current of 15 amps when the voltage across the circuit is 110 volts. What is the power? P = I x V P = 15 A x 110 V P = 1,650 W or 1.65 kW A circuit in the power supply for a high- powered laser has a current of 15 amps when the voltage across the circuit is 110 volts. What is the power? P = I x V P = 15 A x 110 V P = 1,650 W or 1.65 kW

7. Power in Electrical Systems When a circuit resists changes in the charge flow, there are 3 equations that are used: P = I x V P = I 2 R P = V 2 /R When a circuit resists changes in the charge flow, there are 3 equations that are used: P = I x V P = I 2 R P = V 2 /R

8. Power in Electrical Systems A current of 2.2 A has a resistance of 5 ohms. What is the power? We have current (I) and resistance (R), which equation? P = I 2 R P = 2.2 2 x 5 P = 24.2 W A current of 2.2 A has a resistance of 5 ohms. What is the power? We have current (I) and resistance (R), which equation? P = I 2 R P = 2.2 2 x 5 P = 24.2 W

9. Power in Electrical Systems A current of 2.2 A has a Voltage of 110 V, what is the power? We have current (I) and Voltage (V), which equation? P = I x V P = 2.2 x 110 P = 242 W A current of 2.2 A has a Voltage of 110 V, what is the power? We have current (I) and Voltage (V), which equation? P = I x V P = 2.2 x 110 P = 242 W

10. Power in Electrical Systems A Voltage of 110 V, has a resistance of 5 ohms. What is the power? We have voltage (V) and Resistance (R), what equation? P = V 2 /R P = 110 2 /5 P = 2,420 W or 2.42 kW A Voltage of 110 V, has a resistance of 5 ohms. What is the power? We have voltage (V) and Resistance (R), what equation? P = V 2 /R P = 110 2 /5 P = 2,420 W or 2.42 kW

11. Power in Electrical Systems Power companies sell energy, not Power. When you use an electrical device you pay for the energy used by the device. Power companies sell energy, not Power. When you use an electrical device you pay for the energy used by the device.

12. Power in Electrical Systems Equation for energy and power: Energy = Power x change in time E = P x ∆t Equation for energy and power: Energy = Power x change in time E = P x ∆t

13. Power in Electrical Systems An electric company charges $0.085 per kWh of electric energy consumption. What is the cost of operating a TV set for one month (30 days) if the set is operated an average of 7.5 hours per day? The TV set draws 1.5 A when connected to a 120 V outlet. E = P x t P = I x V = (1.5)(120) = 180 W or 0.18 kW t = (30 days)(7.5 h/d) = 225 h E = (0.18 kW)(225 h) = 40.5 kWh Cost = ($0.085/kWh)(40.5 kWh) = $3.44 An electric company charges $0.085 per kWh of electric energy consumption. What is the cost of operating a TV set for one month (30 days) if the set is operated an average of 7.5 hours per day? The TV set draws 1.5 A when connected to a 120 V outlet. E = P x t P = I x V = (1.5)(120) = 180 W or 0.18 kW t = (30 days)(7.5 h/d) = 225 h E = (0.18 kW)(225 h) = 40.5 kWh Cost = ($0.085/kWh)(40.5 kWh) = $3.44