1. Physical Science Mr. Barry

2. Series circuits have one loop through which current can flow

3. Parallel Circuits: have two or more branches through which current can flow

4. In a parallel circuit: Voltage difference is the same across all branches More current flows through branches with lower resistance Connecting wires have the most current flow

5. In your house: Parallel circuits with circuit breakers Circuit breakers have a main breaker in series with many parallel circuits. The parallel circuits each have circuit breakers. When the wires overheat, the circuit breaker switches off the current. Look at figure 18 on page 209

6. Electric Power: The rate at which electric energy is converted into another form of energy P = IV Power is measured in Watts, Current in Amperes, Voltage in Volts Appliances with heating elements usually use more electric power

7. A 9 Volt battery powers a digital clock. The current through the clock is 6.67 Amps. How much power does the clock use? P = IV P = (6.67)(9) P = 60.0 W

8. A light bulb uses 60 Watts of Power. It is plugged into a 120 Volt electrical outlet. What current flows through the bulb? P = IV I = P/V I = (60)/(120) I = 0.5 A

9. Electric Energy: The electric company charges you for the electrical energy you use, which depends on the amount of power and the length of time it is used E = Pt Electrical energy is measured in kilowatt hours (kWh), Power in kilowatts, and time in hours

10. A digital clock uses 60 Watts of power for 24 hours. How much electrical energy is used by the clock in one day? E = Pt P = 0.060 kW E = (0.060) (24) E = 1.44 kWh

11. The electrical energy used by a clock in one day is 1.44 kWh. The power company charges $0.09 per kWh. How much does it cost to run the clock for a 30-day month? Cost = E x Rate E = (1.44/day)(30days) E = 43.2 kWh all month E x Rate (43.2)($0.09) = $3.88

12. The electrical power used by a light bulb is 60 Watts. The bulb is on for approximately 7 hours each day. How much electrical energy does the light bulb use per day? E = Pt P = 0.060 kW E = (0.060) (7) E = 0.42 kWh

13. The electrical energy used by the bulb each day is 0.42 kWh. The electrical company charges $0.09 per kWh. What is the cost per year? (for this year) Cost = E x Rate E = (0.42/day)(366days) E = 153.72 kWh all year E x Rate (153.72)($0.09) = $13.83