1. Chapter 19 Review Current and Resistance

2. 1. A current of 2 amps flows for 30 seconds. How much charge is transferred?

3. I = q/t It = q 2 x 30 = q 60 C = q

4. 2. A charge of 10 coulombs is transmitted over a time of 20 seconds. What is the current over this period?

5. I = q/t I = 10 C/20 s I = 0.5 Amps

6. 3. When you flip a switch to turn on a light, how long does it take the electric field in the wire to develop?

7. Close to the speed of light.

8. 4. What is drift speed? Why is it so much slower than the speed of the electric field?

9. Drift speed is the actual speed the electrons are moving in the conductor. It is much slower because the electrons must follow a zig-zag path.

10. 5. What is the difference in AC and DC?

11. AC is alternating current. In AC the direction of the current changes rapidly. DC is direct current. In DC the current moves in the same direction all the time.

12. 6. If the current in a wire is 10 A and the resistance is 12 Ω, what is the potential difference across the circuit?

13. V = IR V = 10 x 12 V = 120 V

14. 7. A motor of resistance 5 Ω carries a current of 12 A. What is the potential difference provided by the voltage source?

15. V = IR V = 12 x 5 V = 60 V

16. 8. A light bulb with a resistance of 144 Ω is in a 120 V circuit. What is the current in the circuit?

17. V =IR 120 = I(144) 0.833 A = I

18. 9. A 12 V battery is connected to a heating element. If the current is 0.5 A, what is the resistance of the heating element?

19. V =IR 12 = 0.5 R 24 Ω = R

20. 10. What four factors affect the resistance of a conducting wire and what effect do they have?

21. 1. Length - the longer the wire, the higher the resistance 2. Cross section – the “fatter” the wire, the lower the resistance 3. Material – some materials are better conductors than others 4. Temperature – the higher the temperature, the higher the resistance

22. 11. What is a superconductor? What is critical temperature?

23. A superconductor is a material that will conduct electricity with zero resistance. The critical temperature is the temperature below which a superconductor has zero resistance. (These critical temperatures tend to be VERY low.)

24. 12. If the resistance in a circuit is doubled while the voltage is held constant, how does this change the power output of the resistor?

25. By V =IR, doubling the R while V is held constant would cut I in half. By P = IV, if I is cut in half while V is held constant, the power output P is cut in half.

26. 13. If the voltage in a circuit is doubled while the resistance is held constant, how does this change the power output of the resistor?

27. By V = IR, doubling V while R is held constant would cause I to double. By P = IV, if I is doubled and V is doubled, power output P would be quadrupled!

28. 14. If the current in a circuit is doubled while the resistance is held constant, how does this change the power output of the resistor?

29. By V = IR, doubling I while R is held constant would cause V to double. By P = IV, if I is doubled and V is doubled, power output P would be quadrupled!

30. 15. If a circuit with a current of 15 A has a potential difference of 120 V, what is the power output of the circuit?

31. P = IV P = 15 x 120 P = 1800 W

32. 16. If a resistance of 240 Ω is in a 120 V circuit, what is the power output of the resistance?

33. V = IR 120 = I(240) I = 0.5 A P = IV P = 0.5 x 120 P = 60 W

34. 17. A 1500 W hair dryer is plugged into a 110 V outlet. What is the current of the circuit? What is the resistance of the hair dryer?

35. P = IV 1500 = I(110) I = 13.6 A V = IR 110 = 13.6 R 8.1 Ω = R

36. 18. A device uses 4 A when plugged into a 120 V outlet. What is the total cost of running this device for 5 hours if the cost of energy is $0.12 per kilowatt·hour?

37. P = IV P = 4 x 120 P = 480 W 480 W / 1000 = 0.480 kW 0.480 kW x 5 hours = 2.4 kWh 2.4 kWh x 12 cents = 28.8 = 29 cents

38. 19. A device uses 6 A when plugged into a 120 V outlet. What is the total cost of running this device for 2 hours if the cost of energy is $0.11 per kilowatt·hour?

39. P = IV P = 6 x 120 P = 720 W 720 W / 1000 = 0.720 kW 0.720 kW x 2 hours = 1.44 kWh 1.44 kWh x 11 cents = 15.84 = 16 cents