Circuit electricity

Atomic structure

Atoms are composed of protons (+),

electrons (-) and neutrons. The nucleus

contains the protons and neutrons and the electrons surround the

nucleus.

Atomic structure

The outer layer of electrons in a metal is

incomplete which allows them to pass from atom to atom

Atomic structure

Because electrons can pass from atom to

atom. charge can pass through a conducting

material such as a metal.

Metals are conductors

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

Current

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

The Ampere (Named after Andre Marie Ampere)

QI = ---- t

The Ampere is a measure of how much

electrical current is flowing and is measured

in units of amps

I = amps Q = charge (in coulombs)and t = time ( in seconds)

Potential Difference or Voltage

Potential difference, or voltage, is the electrical

potential energy per coulomb of charge.

EV = ---- Q

Alessandro Volta

V = voltage E = energy in Joules Q = charge (in

coulombs)

Resistance

Resistance is a measure of opposition to the flow of charge and is measured in

ohms () VI = ---- R

Georg Ohm

I = current V = voltage R = resistance in ohms

Ohms Law (three versions)

VI = ---- R

V = IR VR = ---- I

Electrical Power

Power is the rate of using energy in joules per

second

P = E tor E = Pxt

Electrical Power

From previous slides we know that

EV = ---- Q

QI = ---- t

and

Electrical PowerCombine the two and

cancel the Q from each EV = ---- Q

QI = ---- t

X

Leaving E/t so electrical power is P = V x I

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

Kirchoff’s Laws

I1

Kirchoff’s first LawThe total current flowing into a junction is the total current flowing out of the circuit

I2

I3I1 = I2 + I3

Kirchoff’s Second Law

1. The sum of the potentialdifferences around an

electricalcircuit equals the supply

voltage.

Resistors in series The total resistance is

found by simply

adding the resistance

of each R1 + R2 +R3

etc

Resistors in series The supply voltage (pd) is shared across the resistors. The voltage across each

depends on the resistance of

each

Resistors in seriesThe 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)

Resistors in parallel

The total resistance is calculated as

below

Resistors in parallelThe current in a parallel

circuit is shared

between each resistor. (The amount in each depends

on the resistance)

Resistors in parallel The supply voltage (pd) across each resistor is

the same as the supply

voltage

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

Example 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.

Example

R1 = 50Ω, R2 = 100Ω and supply voltage = 12V.

Total resistance = R1 + R2 =150ΩTotal (supply) current V/I = 12/150 =0.08 amps.Voltage across R1 = 50 x 0.08 = 4 voltsVoltage across R2 = 100 x 0.08 = 8 voltsPower loss in R1 = V x I = 4 x 0.08 = 0.32 Watts

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.

Example1/Total resistance = 1/100 +1/1000.= 10/1000 + 1/1000 = 11/1000 Total resistance = 1000/11 =90.9Ω

ExampleTotal current = V/R = 12/90.9 = 0.132 ampsCurrent through R1, V/R1 = 12/100 = 0.12 ampsCurrent through R2, V/R2 = 12/1000 = 0.012 amps

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

Example Resistance of the parallel resistors

1/total = 1/100 +1/100=2/100

Total resistance = 100/2= 50Ω

Total resistance in circuit = 200+50 =

250Ω Current = V/R

=12/250=0.048 amps

Example

Voltage across R1I x R

=0.048 x 200= 9.6 volts

Example Voltage across R1 & R2V = I x R

0.048 x 50= 2.4 volts

Current through R2I = V/R

=2.4/100=0.024amps

Example

Power dissipatedP = V x I

2.4 x 0.024= 0.058 watts