Electricity

1. Electricity And Electrical Circuits!! Core 5.4 & 5.5

2. Physics knows its electron flow but often use conventional current.

3. Direction of the current is taken to be opposite of the electron motion.

4. Electrical Current is the amount of charge that moves through wire in a unit of time. I is current. Current is measured in amperes. 1 A = C/s ΔQ I = Δt ΔQ is change in electrical charge.

5. George Ohm discovered that current across a metallic conductor is proportional to the potential difference (Volts) across it. I ∝ V

6. Deviation from Ohm’s Law • Not all materials obey Ohm’s Law. • The filament of a light bulb has increasing resistance as current increases. • It only obeys Ohm’s Law at low temperatures.

7. Factors Affecting Resistance R is directly proportional to length of the wire. The longer the wire the more resistance. R is inversely proportional to cross-sectional area of the wire. The “thicker” the wire the less resistance. L R ∝ A

8. Ohm’s Law is that different metals have different resistance to electrical flow independent of Voltage. V = IR V is voltage or how much electricity and is measured in Volts (V) I is current or Coulombs/Second and is measured in Amps (A) R is resistance to current and is measured in Ohms (Ω)

9. There must be a potential drop from one end to another to have current. V = IR V is created by having more electrical charge at one end and less electrical charge at another end.

10. Electrical Power • Either thermal energy or work done by an electrical device.

11. Electrical Power If any two of voltage, current, and resistance are known then power can be calculated. P = V I P = R I2 V2 P = R

12. Electrical Circuits • In a battery, work is done to push electrons from the negative to the positive terminal. • This is the electromotive force or emfof the battery. • ε = emf = W/q • Remember q is electrical charge and W is work. • Remember emf is the potential difference or voltage of the battery.

13. + I ˗ r ε 0 Potential drops by Ir The area inside the dashed lines is the battery. The electrons are being pushed from the negative to the positive terminals of the battery. The current is going the opposite way that the electrons are being pushed.

14. + I ˗ r ε 0 Potential drops by Ir The chemicals and other materials inside the battery cause an internal resistance. The current internal to the battery also takes away from the emf, which is across the positive and negative terminals.

15. + I ˗ r ε 0 Potential drops by Ir The realized voltage of the battery is V=ε-Ir As electrons make it from the positive to negative terminal emf is reduced, also reducing the voltage. This is how batteries “go out”.

16. The current travels from the positive to negative terminal.

17. Simple Circuits Here are examples of both a series and parallel circuit.

18. Simple Circuits The left circuit is in series, if one bulb goes out the entire circuit is disrupted.

19. Simple Circuits The right circuit is in parallel. If one bulb goes out the other will remain on.

20. Simple Circuits All circuits must be complete from the positive to negative terminal of the battery.

21. Resistance in Series R1 R2 R3 24 V 0 V I Rtotal = R1 + R2 + R3

22. Resistance on the circuit is the sum of the individual resistors R1 R2 R3 24 V 0 V I Rtotal = R1 + R2 + R3

23. Resistance on the circuit is the sum of the individual resistors 2Ω 4Ω 5Ω 24 V 0 V I Rtotal = 2Ω + 4Ω + 5Ω = 11Ω

24. Current on the circuit follows Ohm’s Law 2Ω 4Ω 5Ω 24 V 0 V I V = IR

25. Current on the circuit follows Ohm’s Law 2Ω 4Ω 5Ω 24 V 0 V I 24V = I 11Ω

26. Current on the circuit follows Ohm’s Law 2Ω 4Ω 5Ω 24 V 0 V I I = 2.2A

27. This is true across any resistor as well. R1 = 2Ω 4Ω 5Ω 24 V 0 V I V = I1R1

28. This is true across any resistor as well. R1 = 2Ω 4Ω 5Ω 24 V 0 V I 24V = I12Ω

29. This is true across any resistor as well. R1 = 2Ω 4Ω 5Ω 24 V 0 V I I1 =12A

30. Resistance in Parallel R3 R2 R1 24 V 0 V I 1 1 1 1 = + + Rtotal R1 R2 R3

31. Resistance in Parallel R3 R2 R1 24 V 0 V I R1= 2Ω 1 1 1 1 R2= 4Ω = + + Rtotal R1 R2 R3 R3= 5Ω

32. Resistance in Parallel R3 R2 R1 24 V 0 V I R1= 2Ω 1 1 1 1 R2= 4Ω = + + Rtotal 2Ω 4Ω 5Ω R3= 5Ω

33. Resistance in Parallel R3 R2 R1 24 V 0 V I R1= 2Ω 1 R2= 4Ω = + 0.5 + 0.25 0.2 Rtotal R3= 5Ω

34. Resistance in Parallel R3 R2 R1 24 V 0 V I R1= 2Ω 1 R2= 4Ω = 0.95 Rtotal R3= 5Ω

35. Resistance in Parallel R3 R2 R1 24 V 0 V I R1= 2Ω R2= 4Ω Rtotal = 1.05 Ω R3= 5Ω

36. Resistance in Parallel R3 R2 R1 24 V 0 V I Remember the Golden Rule: In parallel resistors resistance on the circuit must be less than that of the lowest resistor!

37. This is due to the property current has to take the path of least resistance R3 R2 R1 24 V 0 V I R1= 2Ω R2= 4Ω R3= 5Ω

38. How to solve this???

39. Solve each parallel system individually.

40. Then add them together!!

41. R1 thru R6 are all 10Ω 10Ω 10Ω 10Ω 10Ω 10Ω 10Ω

42. R4 and R5 add to 20Ω 10Ω 20Ω 10Ω 10Ω 10Ω

43. R4 and R5 are in parallel to R6 and resistance comes out to 6.7Ω 6.7Ω 10Ω 10Ω 10Ω

44. R2 and R3 add to 20Ω 10Ω 20Ω 20Ω 10Ω

45. R2 and R3 are in parallel to R1 and resistance comes out to 6.7Ω 6.7Ω 6.7Ω

46. The two sets of resistors are in parallel to each other and add up to 13.4Ω. 6.7Ω 13.4Ω 6.7Ω

47. Rules to Remember • In a parallel circuit system, the voltage is the same in all parallel paths. The Voltage through the 4Ω and 8Ω resistors are the same!

48. Rules to Remember • To find current on the circuit the resistance across the entire circuit must be known as well as voltage out of the battery. The Resistance is 3+10+5 = 18Ω The Currentis V= IR 9V= I 18Ω I = 0.5 A

49. Rules to Remember • In a series circuit system, the voltage drops as current goes through each resistor. The Voltage is 9V at point 1 and 7.5V at point 2. V = IR V = (0.5A)(3Ω) V = 1.5V

50. Rules to Remember • In a series circuit system, the voltage drops as current goes through each resistor. The Voltage is 7.5V at point 2 and 2.5V at point 3. V = IR V = (0.5A)(10Ω) V = 5V