1. Sources of Electricity
1. Friction
2. Electromagnetism
3. Photovoltaic (Solar)
4. Peizoelectric
5. Thermoelectric
6. Chemical

2. 2. Electromagnetism Hydro Electric Nuclear Power
Nuclear Power

3. Fossil Fuels
Wood, Coal, LNG, Oil, etc.
Same as Nuclear except that the source of
Heat to produce steam is different

4. 3. Photovoltaic (Solar)
4. Peizoelectric

5. 5. Thermoelectric
Thermocouple

6. Chemical
Voltaic Cell

12. Electric Current (I)
The rate of flow of electric charge measured
in units of Amperes (A)
Electric Potential Difference or Voltage (V)
The work done per unit charge measured in
units of Volts (V)

13. Electric Power (P)
The rate at which work or energy is used
measured in units of watts (w)
P is power in watts
W is work or energy in Joules (J)
t is time in seconds (s)
A more familiar version of the power formula
connects power with current and voltage
P is power in watts
V potential difference in volts (V)
I is current in amperes (A)

14. e.g. 1. How much energy does a 60 w light bulb use in 10 s?

15. e. g. 2. If a light bulb uses 200 J of energy every 5 s,
e.g. 2. If a light bulb uses 200 J of energy every 5 s, how many watts would it be?

16. e.g. 3. What current would flow through a 60 w light bulb that is connected to a 120 V outlet?

17. e.g. 4. What would the power rating be of a hairdryer that has a current of 10 A passing through it when it is plugged into a 120 V outlet?

18. Cost of Electric Power (P)
In NB, Homeowners pay about 9 cents for every kwh
(kilo-watt-hour) that they use.
A kilo-watt-hour is a modified version of a Joule that
is used because a Joule is a very small amount of
energy.

19. e.g. 5. How many kilo-watt-hours would a 100 w light use in 5 hours?
kwh is a measure of work or energy so
we are looking for “W”, but the units must be in kw and hours.
Convert watts to kilowatts by dividing by 1000.

20. If each kwh costs 9 cents then multiply by 0.5 kwh

21. Electrical Resistance
A measure of how good a conductor an object is
is called Resistance.
It is the ratio of the Potential Difference to the
Current
This is called Ohm’s Law named after
George Ohm who discovered it.
V is potential difference in volts (V)
I is current in ampere (A)
R is the resistance in ohms (Ω)
This is the most common form of the equation

22. e.g. 6. If a current of 0.5 A passes through a light bulb when it is connected to a 120 V outlet, find the resistance of the light bulb.

23. e.g. 7. What current would pass through a lamp having a resistance of 60 Ω if it is connected to a 120 V outlet?

24. e.g. 8. Find the resistance of a 40 w if it is plugged into a 120 V outlet.
Find the current first
Then find resistance

28. 6400 +/- 5%

29. /- 10%

30. 3300 +/- 5%

31. Pass out problem sheet DC Electricity
Do questions 1-14
Use the following formulae or or or
If you are calculating the cost, find the Work (W). make sure that Power (watts) is converted to kilowatts (move decimal three places to left and that time is in hours. Then multiply by the cost per kwh.

46. Resistors in Series
Resistors are connected in series when there is one conducting path through them. i.e. the same current must pass through each.
e.g. Assume that a 6 Ω resistor is connected to a 12 V battery, find the current.

47. e.g. 2 If two 6 Ω resistors are connected in series to a 12 V battery, find the current through each.
Since the voltage has to be “shared” between each resistor, we must find the total resistance first. In series the total resistance is just the sum of the resistances. The formula would be:

48. Find the current and voltages through two 6 Ω
resistors connected in series to a 12 V battery? V I R 1 6 2 A 12 6 1 6 1
Since the current that goes
through RA must be the same
as the current through R1 and R2 1 12
Note that the voltage V1 and V2 add up to the
total Voltage

49. In general, in series:
In series, Current is Constant

50. Find the current and voltages through three resistors,
2 Ω, 4 Ω, and 6 Ω, connected in series to a 12 V battery V I R 1 2 4 3 6 A 2 1
RA = 2+4+6=12 4 1 6 1 1 12 12

51. Resistors in Parallel:
If there is more than one path through two or more resistors they are said to be in parallel:
Consider the original question
that had one 6 Ω resistor connected
to the 12 V battery. If a second 6 Ω
is connected in parallel, there are two
paths for the electrons to travel.
When any electron reaches the point where the path splits, it can only go one way. Whichever way it goes, it can only go through one resistor on its way around the circuit.
Since each of the resistors will have electrons pass through them, the total current has to split upon reaching the point where the paths split.

52. If the current through R1 is calculated using Ohm’s law
I1 = 2 A
Since the current split the total current must be
If the current through R2 is calculated using Ohm’s law it would also give a current of 2 A
IT = I1 + I2 + …
Since each resistor gets 12 V
VT = V1 = V2 = …
Finally, since the current has increased when a second resistor is added in series, the resistance must have decreased, therefore

53. In general, in parallel:
Note that the voltage is constant in parallel (in series the current was constant)

54. V I R 1 6 2 A 12 2 12 2 12 4 3

55. Resistors and Series and Parallel
Find the currents and voltages
through the resistors. The resistance
of each is given in the table below V I R 1 2 6 3 12 A B 4 2 8
1.3 8
0.7 8 2 4 12 2 6

56. Assign the following questions from the problem sheet
#15-17, 19-25

57. In general, in series:
In general, in parallel: