1. Circuit electricity

2. Atomic structure
Atoms are composed of protons (+), electrons (-) and neutrons. The nucleus contains the protons and neutrons and the electrons surround the nucleus.

3. Atomic structure
The outer layer of electrons in a metal is incomplete which allows them to pass from atom to atom

4. Atomic structure
Because electrons can pass from atom to atom. charge can pass through a conducting material such as a metal.
Metals are conductors

5. These are called Insulators
Atomic structure
Some materials such as rubber and plastic have complete outer layers of electrons so they cannot pass from atom to atom. Charge cannot pass through these materials.
These are called Insulators

6. Conventional current flows from positive to negative
Because it is the electrons which move from atom to atom in reality negative charge flows from negative to positive.
This has the same effect as positive charge moving from positive to negative
Conventional current flows from positive to negative

7. The Ampere (Named after Andre Marie Ampere)
The Ampere is a measure of how much electrical current is flowing and is measured in units of amps
Q I = t
I = amps Q = charge (in coulombs)
and t = time ( in seconds)

8. E V = ---- Q Potential Difference or Voltage
Alessandro Volta
Potential difference, or voltage, is the electrical potential energy per coulomb of charge.
E V = Q
V = voltage E = energy in Joules Q = charge (in coulombs)

9. I = current V = voltage R = resistance in ohms
Georg Ohm
Resistance is a measure of opposition to the flow of charge and is measured in ohms ()
V I = R
I = current V = voltage R = resistance in ohms

10. Ohms Law (three versions)
V I = R
V = IR
V R = I

11. Power is the rate of using energy in joules per second
Electrical Power
Power is the rate of using energy in joules per second
P = E t
or E = Pxt

12. From previous slides we know that
Electrical Power
From previous slides we know that
E V = Q
Q I = t
and

13. Electrical Power Combine the two and cancel the Q from each
E V = Q
Q I = t X
Leaving E/t so electrical power is P = V x I

14. Electrical Power Equation variations
P = V x I
P = I2R
P = V2/R
These were obtained by using Ohm’s law to substitute for V and I

15. Kirchoff’s Laws I2 I1 I3 I1 = I2 + I3 Kirchoff’s first Law
The total current flowing into a junction is the total current flowing out of the circuit I2 I1 I3
I1 = I2 + I3

16. Kirchoff’s Second Law 1. The sum of the potential
differences around an electrical
circuit equals the supply
voltage.

17. The total resistance is found by simply adding the resistance of each
Resistors in series
The total resistance is found by simply adding the resistance of each
R1 + R2 +R3 etc

18. Resistors in series
The supply voltage (pd) is shared across the resistors. The voltage across each depends on the resistance of each

19. Resistors in series
The current in a series circuit is the same all the way round the circuit
(as per Kirchoff’s first Law). Current flowing into the resistor is the same as the current flowing out of the resistor)

20. The total resistance is calculated as below
Resistors in parallel
The total resistance is calculated as below

21. Resistors in parallel
The current in a parallel circuit is shared between each resistor. (The amount in each depends on the resistance)

22. Resistors in parallel
The supply voltage (pd) across each resistor is the same as the supply voltage

23. Combined resistors
To calculate the total resistance of the circuit calculate the parallel set first and treat it as a single resistor in series with the other resistor

24. Example R1 = 50Ω, R2 = 100Ω and supply voltage = 12V.
From the following diagram determine:
a) Total resistance.
b) Total (supply) current.
c) Voltage across each resistor.
d) Power loss in resistor R1.
R1 = 50Ω, R2 = 100Ω and supply voltage = 12V.

25. Example R1 = 50Ω, R2 = 100Ω and supply voltage = 12V.
Total resistance = R1 + R2 =150Ω
Total (supply) current V/I = 12/150 =0.08 amps.
Voltage across R1 = 50 x 0.08 = 4 volts
Voltage across R2 = 100 x 0.08 = 8 volts
Power loss in R1 = V x I = 4 x 0.08 = 0.32 Watts
R1 = 50Ω, R2 = 100Ω and supply voltage = 12V.

26. R1 100Ω, R2 = 1kΩand supply voltage = 12V.
Example
From the following diagram determine:
a) Total resistance.
b) Total (supply) current.
c) Current through each resistor.
R1 100Ω, R2 = 1kΩand supply voltage = 12V.

27. Example
1/Total resistance = 1/100 +1/1000.= 10/ /1000 = 11/1000
Total resistance = 1000/11 =90.9Ω

28. Example Total current = V/R = 12/90.9 = 0.132 amps
Current through R1, V/R1 = 12/100 = 0.12 amps
Current through R2, V/R2 = 12/1000 = amps

29. Example From the diagram below, determine:
a) The total resistance, and the supply current.
b) The voltage across the R1 resistor.
c) The current through R2 , and the power dissipated in it.
R1 = 200Ω R2 and R3 are both 100Ω and the supply voltage is 12 volts

30. Example Resistance of the parallel resistors 1/total = 1/100 +1/100
=2/100
Total resistance = 100/2
= 50Ω
Total resistance in circuit = = 250Ω
Current = V/R
=12/250
=0.048 amps

31. Example
Voltage across R1
I x R
=0.048 x 200
= 9.6 volts

32. Example Voltage across R1 & R2 V = I x R 0.048 x 50 = 2.4 volts
Current through R2
I = V/R
=2.4/100
=0.024amps

33. Example
Power dissipated
P = V x I
2.4 x 0.024
= watts