1. Electrical Circuits
They keep the lights on!!

2. Electrical Current is how fast charge is moving in a unit of time.
I is current.
Current is measured in amperes or amps (A).

3. Remember it is charge that is moving! Electrons move very little!!

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

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

6. 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 (how much “push”) and is measured in Volts (V)
I is current and is measured in Amps (A)
R is resistance to current and is measured in Ohms (Ω)

7. Voltage is called “potential difference” and change in Voltage is called “drops”.

8. This means you have “more” charge in one area and “less” charge in another!

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

10. 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.

11. Resistance slows electricity down as it moves through a circuit
Resistance slows electricity down as it moves through a circuit. It has a symbol of R and is measured in Ohms (Ω).

12. 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.
R is directly proportional to temperature. The higher the temperature in the wire the more resistance.

13. Electrical Power
Either thermal energy or work done by an electrical device.
These equations come in handy!!

14. 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

15. Electricity and Circuits
Property
Units of Measure
Voltage (V)
Current (I )
Resistance (R)
Power (P)
Volts (V)
Amps (A)
Ohms (Ω)
Watts (W)

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

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

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

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

20. Symbols in Circuit Diagrams!
Battery
Conductor
Switch
Light Bulb or Lamp
Resistors

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
Rtotal = R1 + R2 + R3

23. Resistance on the circuit is the sum of the individual resistors2Ω 4Ω 5Ω
24 V
0 V I
Rtotal = 2Ω + 4Ω + 5Ω =
11Ω

24. Current on the circuit follows Ohm’s Law2Ω 4Ω 5Ω
24 V
0 V I
V = IR

25. Current on the circuit follows Ohm’s Law2Ω 4Ω 5Ω
24 V
0 V I
24V = I 11Ω

26. Current on the circuit follows Ohm’s Law2Ω 4Ω 5Ω
24 V
0 V I
I = 2.2A

27. This is true across any resistor as well.= 2Ω 4Ω 5Ω
24 V
0 V I
V = 2.2A 2Ω

28. There is a 4.4V drop at the first resistor.= 2Ω 4Ω 5Ω
24 V
0 V I
V = 4.4V

29. At point A the voltage has dropped.= 2Ω 4Ω 5Ω A
24 V
0 V I
24V - 4.4V = 19.6V

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

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

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

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

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

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

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

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

38. Voltage across parallel lines is always the same!2Ω I 4Ω 5Ω
2 A
Voltage across parallel lines is always the same!
The resistance of this parallel system is 1.05 Ω

39. The voltage drop can be found using V = IR2Ω I 4Ω 5Ω
2 A
The voltage drop can be found using V = IR
V = (2 A)(1.05 Ω)
V = 2.1V
When we learn current splits we will discover it is the same drop for each resistor individually!

40. The Voltage through the 4Ω and 8Ω resistors are the same!
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!

41. The Voltage through the 4Ω and 8Ω resistors are the same!
Rules to Remember
It helps to think of Voltage as the “push” through a circuit, like an invisible hand pushing charge along.
The Voltage through the 4Ω and 8Ω resistors are the same!

42. 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 = 18Ω
The Current is V = IR
9V = I 18Ω
I = 0.5 A

43. The Voltage is 9V at point 1 and 7.5V at point 2.
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

44. The Voltage is 7.5V at point 2 and 2.5V at point 3.
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

45. The Voltage is 2.5V at point 3 and 0 at point 4.
Rules to Remember
In a series circuit system, the voltage drops as current goes through each resistor.
The Voltage is 2.5V at point 3 and 0 at point 4.
V = IR
V = (0.5A)(5Ω)
V = 2.5V

46. The Voltage is 2.5V at point 3 and 0 at point 4.
Rules to Remember
To understand this it helps to think of voltage as “push”. Some of this “push” is used up as each resistor system is crossed.
The Voltage is 2.5V at point 3 and 0 at point 4.
V = IR
V = (0.5A)(5Ω)
V = 2.5V

47. Rules to Remember
In a parallel circuit system, the current splits as it enters and comes back together as it exits.
The Current through the 4Ω and 8Ω resistors are the same going in and out!

48. Rules to Remember Current is always the “same in, same out”
More Current goes through the 4Ω resistor than the 8Ω resistor!

49. Rules to Remember
The 0.5 A will be split between the 4Ω and 8Ω resistors!!!
8/12 of the Current goes through the 4Ω resistor!
0.5A (8/12) =
0.33 A

50. Rules to Remember
The 0.5 A will be split between the 4Ω and 8Ω resistors!!!
4/12 of the Current goes through the 8Ω resistor!
0.5A (4/12) =
0.17 A

51. Rules to Remember
The current comes back together after going through the resistors!
The Current to the battery!
0.33A A =
0.5 A

52. Rules to Remember
When you have parallel systems mixed with resistors in series, solve for the resistance of the parallel system and then treat it as part of the series.
It is very common to have series and parallel resistors mixed together on a circuit.

53. Example 2Ω 4Ω 3Ω 6Ω 2Ω
The 4Ω, 3Ω, 6Ω resistors are a parallel system in series with the other resistors!!

54. First solve for the total resistance of the parallel system!!
Example 2Ω 4Ω 3Ω 6Ω 2Ω
First solve for the total resistance of the parallel system!!

55. Example 2Ω 4Ω 3Ω 6Ω 2Ω 1 1 1 1 = + +
Rtotal 4 3 6

56. Example 2Ω 4Ω 3Ω 6Ω 2Ω 1 3 4 2 9 = + + =
Rtotal 12 12 12 12

57. Example 2Ω 4Ω 3Ω 6Ω 2Ω 1 9 12 =
Rtotal = =
1.33Ω
Rtotal 12 9

58. Now add the resistance of the parallel system to the series!!!
Example 2Ω 4Ω 3Ω 6Ω 2Ω
Now add the resistance of the parallel system to the series!!!

59. Example 2Ω 4Ω 3Ω 6Ω 2Ω
2Ω Ω + 2Ω = 5.33Ω

60. Current in Parallel

61. In is the current across the individual resistor.
Rtotal In =
Itotal Rn
In is the current across the individual resistor.
Itotal is the total current entering the parallel system.
Rtotal is the total resistance of the parallel system.
Rn is the resistance of the individual resistor.

62. Example of Parallel Current4Ω
I = 6A
I = 6A 6Ω 8Ω 1 1 1 1 = + +
Rtotal 4Ω 6Ω 8Ω

63. Example of Parallel Current4Ω
I = 6A
I = 6A 6Ω 8Ω 1 6 4 3 13 = + + =
Rtotal 24 24 24 24

64. Example of Parallel Current4Ω
I = 6A
I = 6A 6Ω 8Ω 24
Rtotal = =
1.85Ω 13
Now we can solve for the current through each resistor!

65. Rtotal In =
Itotal Rn
1.85Ω I4 = 6A 4Ω I4 =
2.77A

66. Rtotal In =
Itotal Rn
1.85Ω I6 = 6A 6Ω I6 =
1.85A

67. Rtotal In =
Itotal Rn
1.85Ω I8 = 6A 8Ω I8 =
1.38A

68. Example of Parallel Current4Ω
I = 6A
I = 6A
1.85A 6Ω
1.38A 8Ω =
2.77A +
1.85A +
1.38A 6A

69. Ammeters
They are built into circuits to give the current level at that point.
They have as little resistance as possible.
Ammeters are put in series so they do not change the current of the circuit
Remember current splits in parallel!!!

70. Galvanometers read current through magnetic fields and are obsolete.

71. Voltmeters
Voltmeters are built into circuits and give voltage at that point.
They are very good for giving the “voltage drop” of a particular device.
Voltmeters are put in parallel to the devices they read.
Remember the voltage across any two parallel lines are the same!

72. Digital Multimeters
They can read voltage, current, resistance, and even temperature.
They range from less than $10 to hundreds of dollars.
They are making ammeters and voltmeters obsolete!!

73. Fuses
A fuse is designed to melt and disrupt or break a circuit before current levels get dangerous.
This is why computers and LCD televisions should be plugged into a “surge protector” instead of straight into a wall. Modern electronics need more protection from temporarily high current levels or “surges”.

74. Fuses
A fuse is rated by the amp level at which it melts, such as a 10 amp fuse.
The power of devices is rated at a specific voltage level. A 300 W device at 110 V will have a different power at a different voltage.

75. Calculating Power

76. Calculating Power
When using the P = VI equation the voltage used is the “voltage drop” or push “used up” at that particular resistor.
The current used is the current through the resistor.
Same with resistance, the Ω value of that resistor.
Only exception is when asked the power on the circuit.

77. What is the resistance of the two 4Ω in parallel?
A Power Example 4Ω 4Ω 2Ω
20 V
0 V I
What is the resistance of the two 4Ω in parallel?
Rp = 2Ω

78. What is the resistance on the circuit?
A Power Example 4Ω 4Ω 2Ω
20 V
0 V I
What is the resistance on the circuit?
Rt = 4Ω

79. What is the current on the circuit?
A Power Example 4Ω 4Ω 2Ω
20 V
0 V I
What is the current on the circuit?
I = 5A

80. What is the voltage drop on the 2Ω resistors?
A Power Example 4Ω 4Ω 2Ω
20 V
0 V
I = 5A
What is the voltage drop on the 2Ω resistors?
V = 10V

81. What is the voltage drop on the parallel 4Ω resistors?
A Power Example 4Ω 4Ω 2Ω
V = 10V
20 V
0 V
I = 5A
What is the voltage drop on the parallel 4Ω resistors?
V = 10V

82. What is the current on the 2Ω resistor?
A Power Example 4Ω 5A 4Ω
V = 10V 2Ω
V = 10V
20 V
0 V
I = 5A
What is the current on the 2Ω resistor?

83. What is the current on the parallel 4Ω resistors?
A Power Example
2.5A 4Ω 5A
2.5A 4Ω
V = 10V 2Ω
V = 10V
20 V
0 V
I = 5A
What is the current on the parallel 4Ω resistors?

84. Now calculate Power using P = VI
A Power Example
2.5A 4Ω 5A
2.5A 4Ω
V = 10V 2Ω
V = 10V
20 V
0 V
I = 5A
Now calculate Power using P = VI
P for 2Ω = 10V 5A = 50W

85. Now calculate Power using P = VI
A Power Example
2.5A 4Ω 5A
2.5A 4Ω
V = 10V 2Ω
V = 10V
20 V
0 V
I = 5A
Now calculate Power using P = VI
P for 4Ω = 10V 2.5A = 25W

86. Now calculate Power using P = RI2
A Power Example
2.5A 4Ω 5A
2.5A 4Ω
V = 10V 2Ω
V = 10V
20 V
0 V
I = 5A
Now calculate Power using P = RI2
P for 2Ω = 2Ω (52)A = 50W

87. Now calculate Power using P = RI2
A Power Example
2.5A 4Ω 5A
2.5A 4Ω
V = 10V 2Ω
V = 10V
20 V
0 V
I = 5A
Now calculate Power using P = RI2
P for 4Ω = 4Ω (2.52)A = 25W

88. Now calculate Power using P = (V2) /R
A Power Example
2.5A 4Ω 5A
2.5A 4Ω
V = 10V 2Ω
V = 10V
20 V
0 V
I = 5A
Now calculate Power using P = (V2) /R
P for 2Ω = (102)V / 2Ω = 50W

89. Now calculate Power using P = (V2) /R
A Power Example
2.5A 4Ω 5A
2.5A 4Ω
V = 10V 2Ω
V = 10V
20 V
0 V
I = 5A
Now calculate Power using P = (V2) /R
P for 4Ω = (102)V / 4Ω = 25W

90. See how they are all the same?
A Power Example
2.5A 4Ω 5A
2.5A 4Ω
V = 10V 2Ω
V = 10V
20 V
0 V
I = 5A
See how they are all the same?

91. Resistance can be found with the equation
Power Changes
An electrical device is rated at 500 W at 220 V. What will the power of the device be if it is switched to a 110 V power source? Assume that its resistance is not variable.
Resistance can be found with the equation
P = V2/R.
500W = 2202V/R.
R= 48400V/500W.
R= 96.8Ω.

92. Current can be found with the equation
Power Changes
Now we need to find the current when the device is plugged in to the 110V source.
Current can be found with the equation
V = I R.
110V = I 96.8Ω.
I = 110V / 96.8Ω
I = 1.14A.

93. Power Changes
Now we can find the power of the device plugged into the 110V outlet.
Current can be found with the equation
P = V I
P = 110V 1.14A
P = 125W

94. The End