1. Electricity
Chapter 19

2. Electric Current
Current is the rate at which electric charges move through a given area.
SI unit is the Ampere or Amp.
1 A = 1 C/s
I = ΔQ/t
Current = charge / time

3. Example problem
The current in a light bulb is A. How long does it take for a total charge of 1.67 C to pass a point in the wire?
ΔQ = 1.6 C I = A t= ?
I = ΔQ/t
t = ΔQ/I
t= 1.6C/0.835A
t= 2.00s

4. Homework
P695 #1,2,4

5. Electric Current
The moving charges that make up a current can be positive or negative.
Conventional Current assumes that current is made of positive charges flowing out of the positive terminal, through the circuit and into the negative terminal of the source.
This was the convention chosen during the discovery of electricity. They were wrong!

6. Electric Current
In most conductors, such as copper, the protons are fixed in the nucleus, so current is really caused by moving electrons. This is called Electron Flow.
Electron Flow says that electrons flow out of the negative terminal, through the circuit and into the positive terminal of the source.

7. Electric Current
In fact, it makes no difference which way current is flowing as long as it is used consistently. The direction of current flow does not affect what the current does.
Our book always uses conventional current.

8. Electric Current
Batteries maintain electric current by converting chemical energy into electrical energy.
Generators convert mechanical energy into electrical energy.

9. AC/DC There are two kinds of current:
Direct current is where charges are always moving in the same direction.
Batteries produce direct current because the positive and negative terminals always stay the same.

10. AC/DC
Alternating current is where the charges change the direction of flow constantly.
Power plants supply alternating current to homes and businesses by using giant electromagnets to change positive and negative terminals.
In the US current alternates (changes direction) 60 times every second while in Europe, current alternates 50 times every second.

11. Resistance
Resistance- The opposition to the flow of current in a conductor
R = V/I
Resistance = Potential difference/Current
SI unit – ohm Symbol-  (omega)

12. Resistance
Ohmic materials- Materials that have a constant resistance over a wide range of potential differences
Non-ohmic- Materials that do not function according to Ohm’s law

13. Resistance
Resistance depends on length, cross-sectional area, material and temperature.
Length: short = ↓ R; long = ↑ R
Area: skinny = ↑R; wide = ↓R
Material: insulator = ↑R; conductor = ↓R
Temperature: hot = ↑R; cold = ↓R

14. Resistance
Resistance is important in controlling the amount of current in a circuit.
If the voltage is constant, resistance is the only way to adjust the current.
Change the material of the wires, or add resistors to the circuit.

15. Example Problem
The resistance of a steam iron is 19.0 Ω. What is the current in the iron when it is connected across a potential difference of 120V?
R= 19.0 Ω V= 120V I= ?
R=V/I
I=V/R
I=120V/19.0 Ω
I= 6.32 A

16. Homework
P 703 #2,4,5

17. Electric Power
Electric power is the rate of conversion of electrical energy
Formula for Electric Power:
P = IV
Electric power = current X potential difference

18. Electric Power Because P= IV and V=IR we can also say;
P= IV = I(IR) = I2R
P = I2R
Or, because I = V/R, we can also say:
P = IV = (V/R)V = V2/R
P=V2/R

19. Electric Power
An electric space heater is connected across a 120 V outlet. The heater dissipates 1320 W of power in the form of electromagnetic radiation and heat. Calculate the resistance of the heater.
P = V2/R R = V2/P
R = 1202/1320
R = 10.9 Ω

20. Homework
P 710 #1 - 4

21. Electric Power
Power companies measure energy not power, using the kilowatt-hour as the unit
One kilowatt-hour = the energy delivered in 1 hour at the constant rate of 1 kW.
To convert between kWh and the SI unit of Joule:
1 kWh = 3.6 X 106 J

22. Example Problem
How much does it cost to operate a W light bulb for 24 h if electrical energy costs $0.080 per kWh?
P= 100W = kW; t= 24 h
Energy = Pt = kW*24 h = 2.4 kWh
Cost = 2.4 kWh*$0.080 = $0.19

23. Homework
P 712 #1-2