Physics II PHY 202/222 Electricity

1. Physics IIPHY 202/222Electricity 452 South Anderson Road Rock Hill, SC 29730 www.yorktech.com

2. Electricity – Test 4 • Beiser Chapters 23-26 • Multiple Choice – Odd • Supplementary Problems – Every Other Odd (1,not3,5,not7…) • Browne Chapter 20-25 for PHY 222 • 20: 3,13 21: 8 22: 5 • 24: 7 25: 9

3. Chapter 23 – Electricity Beiser p.266

4. Electric Charge • Positive charge from protons • Negative charge from electrons • Measured in Coulombs (C) • e = ±1.6 x 10-19 C Like charges repel. Unlike charges attract. _ _ _ + + + Coulomb’s Law Beiser p.266

5. Coulomb’s Law Example _ + Find the force of attraction between a ball with a charge of +0.2 C and a ball with a charge of -0.3 C if they are separated by .5 m. Beiser p.266,7

6. F23 r13=.8 m _ + + F13 Q3=.03 C Q1=-0.04 C r23=.6 m Q2=+0.02 C Superposition of Electric Forces Find the electric force on Q3 from the other charges. Fnet 350 Beiser p.269

7. Electric Fields If you had a “small positive test charge” and placed it near other charges, it would experience a force at every point in space. Mapping these lines of force show the electric field. Measured in N/C or V/m. Beiser p.269,270

8. Potential Difference • The amount of work needed to move a charge of 1C from one point to another. • Measured in volts (V) • 1V = 1J/C Beiser p.266

9. Chap 23 - Summery

10. 23:6,8

11. q1 q2 r 23.10

12. q1 23.12 q2 q x 40 - x 40 cm

13. - - - - + + + + F 23.14

14. + + + - - - 23.16 100 V

15. 23.18 - - - - + + + + F

16. 23.20

17. Chapter 24 – Electric Current Beiser p.277

18. Current • The flow of charge • Measured in Amperes (or Amps), A • 1 A = 1 C/s Beiser p.277,8

19. Direction of Current • A complete circuit is needed for electrons to travel. • Electrons actually travel from negative terminal of battery to positive. • Current is said to go from positive to negative. Direction of current is positive to negative + V = 12 V Mr. Electron sez: I’m going this way!

20. Conductors & Insulators • Conductors – a material through which current flows easily • Insulator - a material through which current will not (generally) flow • Resistor – a material through which current flows with some difficulty • Semiconductor - a material that is sometimes a conductor sometimes a resistor • Superconductor – a material that carries current effortlessly with no loss Beiser p.277

21. Resistance • A measure of the opposition to current in a given material • Measured in Ohms (Ω) • 1 Ω = 1 V/A • A resistor is a device with resistance Resistor color band example red yellow blue is 2 4 6 so 24x106Ω R = 24x106 Ω Beiser p.278

22. Ohm’s Law • Relates current, voltage and resistance. • It takes more voltage to push current through a high resistance material Beiser p.278

23. Power • The rate at which work is done to maintain current. or • The rate at which a current at a voltage can do work • Measured in Watts (W) • 1 W = 1 J/s Beiser p.282

24. 24.2

25. 24.4

26. 24.6

27. 24.8

28. 24.10

29. 24.12

30. 24.14

31. 24.16

32. 24.18

33. 24.20

34. Chapter 25 – Direct Current Circuits Beiser p.288

35. R1 = 100 Ω R2 = 300 Ω R3 = 500 Ω Mr. Electron sez: Uuuugggh, I gotta make it through all three resistors! Resistors in Series Current “has to fight it’s way through” R1 thenR2 and then R3 Add resistors in series. RTotal = R1,+R2 + R3 = 100 Ω + 300 Ω + 500 Ω = 900 Ω Beiser p.288

36. R1 = 100 Ω R2 = 300 Ω Resistors in Parallel • Current “can choose” to go through R1 or R2, so total resistance is less that either individual resistor. • Use formula to get total resistance TI-83 keystrokes: Mr. Electron sez: Whee! I can go either way. That makes it easy for me/hard for you! Beiser p.290

37. R1 = 100 Ω R3 = 100 Ω R2 = 300 Ω R4 = 100 Ω R5 = 300 Ω R1 R2 and 3 R4 and 5 R6 = 300 Ω R7 = 500 Ω RTotal R7 R6 R1 to 5 R6 to 7 Mr. Electron sez: looks like fun! Combinations of Resistors 1) Add the parallel resistors 2) Add the series resistors RTotal = 191 Ω 3) Add the parallel resistors Beiser p.293

38. EMF & Internal Resistance • Batteries have a small internal resistance so that V = Ve – Ir or Terminal Voltage = emf – potential drop due to internal resistance • The total internal resistance of batteries in series is the sum of the individual internal resistances. • The total voltage of batteries in series is the sum of the individual batteries. Beiser p.294

39. Batteries in Series • The total voltage of batteries in series is the sum of the individual batteries. • The total internal resistance is the sum of the individual internal resistances Beiser p.295

40. Batteries in Parallel • Batteries in parallel should always have the same voltage, so that back currents don’t flow through the weaker batteries and waste power. • The total voltage of batteries in parallel is the voltage of any of the batteries. • The total internal resistance is added like resistors in parallel. Beiser p.295

41. I 3 R1 = 100 Ω R3 = 300 Ω I 1 A V = 12 V I 2 R2 = 500 Ω Kirchhoff’s Rules • The sum of the currents into any point is equal to the sum of the current from that point, • The sum of the voltage around a loop is zero. Beiser p.298

42. I 3 R1 = 100 Ω R3 = 300 Ω I 1 A V = 12 V I 2 R2 = 500 Ω #4 #5 Kirchhoff Example 1 • Step 1 – pick a point where all the legs of the circuit come together. • Step 2 – pick a direction that you think current will flow in each leg and label each leg as I1, I2 … • Step 3 – The current into point A = the current out of point A I1=I2+ I3 • Step 4 – Trace a complete circuit and add the voltages – batteries increase and resistors decrease – see #4 12V –I1R1 – I2R2 = 0 (If moving against the current reverse the signs) • Step 5 – Trace another path – see #5 12V –I1R1 – I3R3 = 0 • Solve the three simultaneous equations Beiser p.299-302

43. Solving by Substitution Method Substitute both into eq.1 I1= I2+ I3 12 –I1R1 – I2R2 = 0 12 –I1R1 – I3R3 = 0 Solve eq.2 for I2, and eq.3 for I3. Substitute into eq.2 and eq.3

44. Solving by matrices On TI-83+

45. 25.2

46. 25.4

47. 25.6

48. 25.8

49. 25.10

50. 25.12