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# Circuit electricity - PowerPoint PPT Presentation

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.

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## PowerPoint Slideshow about ' Circuit electricity' - cerise

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Presentation Transcript

### Circuit electricity

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

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

Because electrons can pass from atom to atom. charge can pass through a conducting material such as a metal.

Metals are conductors

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

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)

The Ampere is a measure of how much electrical current is flowing and is measured in units of amps

QI = ---- t

I = amps Q = charge (in coulombs)

and t = time ( in seconds)

Alessandro Volta

Potential difference, or voltage, is the electrical potential energy per coulomb of charge.

EV = ----Q

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

Georg Ohm

Resistance is a measure of opposition to the flow of charge and is measured in ohms ()

VI = ---- R

I = current V = voltage R = resistance in ohms

VI = ---- R

V = IR

VR = ----I

Power is the rate of using energy in joules per second

P = E

t

or E = Pxt

From previous slides we know that

EV = ---- Q

QI = ---- t

and

Combine the two and cancel the Q from each

EV = ---- Q

QI = ---- t

X

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

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

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

Kirchoff’s Second Law

1. The sum of the potential

differences around an electrical

circuit equals the supply

voltage.

The total resistance is found by simply adding the resistance of each

R1 + R2 +R3etc

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

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)

Resistors in series

Resistors in parallel round the circuit

The total resistance is calculated as below

Resistors in parallel round the circuit

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

Resistors in parallel round the circuit

• The supply voltage (pd) across each resistor is the same as the supply voltage

Combined resistors round the circuit

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 round the circuit

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 = 100Ωand supply voltage = 12V.

Example round the circuit

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 = 100Ωand supply voltage = 12V.

Example round the circuit

From the following diagram determine:

a) Total resistance.

b) Total (supply) current.

c) Current through each resistor.

R1 100Ω, R2 = 1kΩand supply voltage = 12V.

Example round the circuit

1/Total resistance = 1/100 +1/1000.= 10/1000 + 1/1000 = 11/1000

Total resistance = 1000/11 =90.9Ω

Example round the circuit

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 = 0.012 amps

Example round the circuit

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 round the circuit

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 round the circuit

Voltage across R1

I x R

=0.048 x 200

= 9.6 volts

Example round the circuit

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

Example round the circuit

Power dissipated

P = V x I

2.4 x 0.024

= 0.058 watts