1. Ch. 20 - Electricity I. Electric Charge  Static Electricity  Conductors  Insulators  Electroscope

2. A. Static Electricity  Static Electricity the net accumulation of electric charges on an object  Electric Field force exerted by an e - on anything that has an electric charge opposite charges attract like charges repel

3. A. Static Electricity  Static Discharge the movement of electrons to relieve a separation in charge

4. B. Conductors  Conductor material that allows electrons to move through it easily e - are loosely held ex: metals like copper and silver

5. C. Insulators  Insulator material that doesn’t allow electrons to move through it easily e - are tightly held ex: plastic, wood, rubber, glass

6. D. Electroscope  Electroscope instrument that detects the presence of electrical charges leaves separate when they gain either a + or - charge

7. Ch. 20 - Electricity II. Electric Current  Circuit  Potential Difference  Current  Resistance  Ohm’s Law

8. A. Circuit  Circuit closed path through which electrons can flow

9. A. Potential Difference  Potential Difference (voltage) difference in electrical potential between two places large separation of charge creates high voltage the “push” that causes e - to move from - to + measured in volts (V)

10. B. Current  Current flow of electrons through a conductor depends on # of e - passing a point in a given time measured in amperes (A)

11. C. Resistance  Resistance opposition the flow of electrons electrical energy is converted to thermal energy & light measured in ohms () Copper - low resistance Tungsten - high resistance

12. C. Resistance  Resistance depends on… the conductor wire thickness  less resistance in thicker wires wire length  less resistance in shorter wires temp - less resistance at low temps

13. E. Ohm’s Law  Ohm’s Law V = I × R V: potential difference (V) I: current (A) R: resistance (  ) Voltage increases when current increases. Voltage decreases when resistance increases.

14. E. Ohm’s Law  A light bulb with a resistance of 160  is plugged into a 120-V outlet. What is the current flowing through the bulb? GIVEN: R = 160  V = 120 V I = ? WORK : I = V ÷ R I = (120 V) ÷ (160  ) I = 0.75 A I V R

15. Ch. 20 - Electricity III. Electrical Circuits  Circuit components  Series circuits  Parallel circuits  Household circuits

16. A. Circuit Components A - batteryC - light bulb B - switchD - resistor

17. B. Series Circuits  Series Circuit current travels in a single path  one break stops the flow of current current is the same throughout circuit  lights are equal brightness each device receives a fraction of the total voltage  get dimmer as lights are added

18. C. Parallel Circuits  Parallel Circuits current travels in multiple paths  one break doesn’t stop flow current varies in different branches  takes path of least resistance  “bigger” light would be dimmer each device receives the total voltage  no change when lights are added

19. D. Household Circuits  Combination of parallel circuits too many devices can cause wires to overheat  Safety Features: fuse - metal melts, breaking circuit circuit breaker - bimetallic strip bends when hot, breaking circuit

20. Ch. 20 - Electricity IV. Measuring Electricity  Electrical Power  Electrical Energy

21. A. Electrical Power  Electrical Power rate at which electrical energy is converted to another form of energy P = I × V P: power (W) I: current (A) V: potential difference (V)

22. A. Electrical Power  A calculator has a 0.01-A current flowing through it. It operates with a potential difference of 9 V. How much power does it use? GIVEN: I = 0.01 A V = 9 V P = ? WORK : P = I · V P = (0.01 A) (9 V) P = 0.09 W I P V

23. B. Electrical Energy  Electrical Energy energy use of an appliance depends on power required and time used E = P × t E: energy (kWh) P: power (kW) t: time (h)

24. B. Electrical Energy  A refrigerator is a major user of electrical power. If it uses 700 W and runs 10 hours each day, how much energy (in kWh) is used in one day? GIVEN: P = 700 W = 0.7 kW t = 10 h E = ? WORK : E = P · t E = (0.7 kW) (10 h) E = 7 kWh P E t