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### Electric Currents

Topic 5

These notes were typed in association with Physics for use with the IB Diploma Programme by Michael Dickinson

Electric Potential Difference

5.1.1 Define electric potential difference

5.1.2 Determine the change in potential energy when a charge moves between two points at different potentials.

A man carries a bucket of water up a hill.

Pours water down slide

Turns water wheel

Turns a grind stone

Generates heat!!!

Work done by man on the water is = heat produced

Conservation of energy!!!!

Electric Potential Difference

5.1.1 Define electric potential difference

5.1.2 Determine the change in potential energy when a charge moves between two points at different potentials.

Picture an electric field that is created by to charged plates.

One on the left has a negative charge, one on the right has a positive charge. (see diagram on drawn on board) This creates an electric field.

This is a uniform field – the strength of the field is constant no matter where the charge moves.

Now place a charged particle in that field. It will feel an electric force from this field.

Electric Potential Difference

5.1.1 Define electric potential difference

5.1.2 Determine the change in potential energy when a charge moves between two points at different potentials.

Now move that charge against that force.

The charge has just done WORK!!! YAY!!!

This work done is equal to the gain in electric potential energy.

This is just like gravitational potential energy gained is equal to work done on the same mass. (DO EXAMPLE)

Work done = change in electric potential energy

W = electric potential = Fd

Fd = ΔEelec

Eqd = ΔEelec

Electric Potential Difference5.1.2 Determine the change in potential energy when a charge moves between two points at different potentials.

5.1.1 Define electric potential difference

IB Equation

ΔV = ΔE/ q

AKA V = W/q

Where:

V= potential difference(volts)

E = Energy(Joules)

q = charge

W= work

Electric Potential Difference5.1.2 Determine the change in potential energy when a charge moves between two points at different potentials.

5.1.1 Define electric potential difference

- Now lets move that charge the other direction. We’ll release it at position B.
- The electric force “pushes” it to position A and loses electric potential energy.
- It will also accelerate since it’s under a constant force.
- The increase of velocity gives the charge an increase of kinetic energy.
- Change in electric potential energy = change in kinetic energyelectric potential = KE

Eqd = ½ mv2

v2 = 2Eqd/m

v =

Electric Potential Difference

As the charge moves between the platesthe “energy difference per unit charge” remains constant (potential difference).

Potential Difference is voltage(V)

Potential Difference(voltage) = the energy per unit charge

V = (ΔEelec/ q)

Vq= ΔEelec

Vq = KE = ½ mv2

Electric Potential Difference

Potential Difference(voltage) = the energy per unit charge

V = (ΔEelec / q)

Unit is J/ c

Think of it as how much energy the charge is going to get as it zips from one plate to the other.

Electric Potential Difference

5.1.3 Define the electronvolt.

Electrical energy = potential difference x charge

Eelec = Vq

This gives us a unit of Joules. A Joule is such a large quantity of energy when looking at the energy carried by electrond, that a different unit is sometimes used.

1 electronvolt = 1 volt x charge on 1 electron

1eV = 1.6 x 10-19 J

Electric Potential Difference

5.1.3 Define the electronvolt.

IB Formal Definition and Equation

Electronvolt – the energy gained by an electron accelerated through a potential difference of 1 volt.

Ve = ½ mv2

Electric Potential Difference

5.1.4 Solve problems involving electric potential difference.

An electron moves from a negative to a positive plate when a potential difference of 50Vols applied to the plates. If the plates are separated by a distance of 5mm, calculate:

a) the strength of the uniform field between the plates

b) the force on the electron.

c) the loss of electric potential energy as the electron moves between the plates.

d) the gain in kinetic energy for the electron

e) the speed of the electron as it reaches the positive plate.

Electric Potential Difference

5.1.4 Solve problems involving electric potential difference.

Solutions

a) E = V/d

E = 50/0.005

E = 1000 V/m

b) F = Eq

F = 10000 x 1.6 x 10-19

F = 1.6 x 10-15N

c) ΔEelec = Vq

ΔEelec = 50 x 1.6 x 10-19

Δeelec = 8 x 10-18J or 50eV

Electric Potential Difference

5.1.4 Solve problems involving electric potential difference.

Solutions

d) Loss of Ep = gain in Ek

Δek = 8 x 10-18J or 50 eV

e) Δek = 8 x 10-18J

½ mv2 = 8 x 10-18J

√2 x 8 x 10-18

v = 9.11 x 10-31

v = 4.19 x 106 m/s

Electric current and resistance

5.1.5 Define electric current.

New Diagram ***See board***

Conducting material has a “lattice structure” – orderly rows and columns. Because of the lattice structure electrons can freely move from one atom to the next.

If a potential difference(voltage) is applied to the ends of a conducting material then the “free electrons” feel an electrical force.

All of them feel the same force.

Just like F=ma says, this electrical force causes the electrons to speed up and accelerate.

Just like gravitational force causes you to accelerate toward earth.

Electric current and resistance

5.1.5 Define electric current.

As they move they occasionally collided with the lattice structure.

The more they accelerate the more collisions they have.

Eventually, there are so many collisions that the free electrons cannot continue to accelerate.

This maximum velocity is called – Drift velocity.

Drift velocity is similar in concept to terminal velocity.

Electric current and resistance

5.1.5 Define electric current.

Which direction do the electrons move?

From the negative terminal to the positive terminal, the source of the potential difference.

Electric current - is this movement of electrons.

Convention is different!!! - electric current was originally thought to be the movement of positively charged particles and we are now stuck with this convention. For this reason, conventional current moves from the positive terminal to the negative. Sorry

Electric current and resistance

5.1.5 Define electric current.

IB Formula/Definition

Current is defined as the amount of electric charge passing a pint in a circuit in unit time.

Current = charge/time

I = Δq/ Δt

1 Ampere = 1coulomb / 1second

Electric current and resistance

5.1.5 Define electric current.

Practice problem 2

5 amps flow in an electric circuit. How many coulombs of charge pass a point in the circuit in 15 seconds? How many electrons is this equal to?

Answer: 75 Coulombs, 4.69 x 1020 electrons

***1 coulomb = 6.25 x 1018 electrons***

Electric current and resistance

5.1.5 Define electric current.

- What kind of current flows

though your cell phone?

- Direct current(DC) is an electric current that continually flows in one direction.
- Alternating current(AC) is a current with a regular change of change of direction.
- Electrons oscillate back and forth in the conducting wires.

Electric current and resistance

5.1.6 Define resistance.

Remember the concept of drift velocity

Some of that kinetic energy is conducted to the lattice in the form of heat

This is called resistance, R.

Measured in Ohms, Ω

IB Formula and Definition

Resistance = voltage / current

R = V/I

Resistance – the ratio of the voltage across a conductor to the current flowing through it.

Electric current and resistance

5.1.6 Define resistance.

Practice 3

When 6 volts is applied to the ends of a resistor, 3 amps of current flows. Calculate the value of the resistor.

Practice 4

If 12 volts is applied to the ends of a 60Ω resistor, calculate the current which will flow.

Practice 5

0.4 amps flows through a 125Ω resistor, calculate the potential difference across the ends of the resistor.

Electric current and resistance

5.1.6 Define resistance.

- Some conductors allow electrons to flow without very much hindrance. – good conductors
- Ex. Copper, aluminum, silver, gold.
- Resistors
- In some conductors the free flow of electrons is made difficult.
- The lattice structure is irregular or disjointed.
- Ex. Wire, light bulb, fan, ect.
- Think of conductors as a tunnel the charge has to move through.
- Resistors – have a muddy floor
- Good conductors – have paved floor
- If we make the tunnel smaller or longer it will also make it harder for the charges to move through the tunnel.

Electric current and resistance

5.1.6 Define resistance.

- Resistance of wire can be changed.
- Material
- Length
- Cross-sectional area
- Temperature
- Which will have a lower/high resistance for each to the above factors?

Electric current and resistance

5.1.6 Define resistance.

- Material
- Depends on each substance.
- Length
- If the length doubles then the resistance doubles R ∝ L
- Cross-sectional area
- If the CSA doubles then the resistance halves R ∝ 1/CSA
- Temp
- Higher temp, means more atomic vibrations, means more collisions between electron and lattice.

Electric current and resistance

5.1.7 Apply the equation for resistance in the form R = ρ L/A

Now we can take the factors, “material, length, and CSA” and combine them we get a new equation:

IB Equation

R = ρ L/A

Electric current and resistance

5.1.7 Apply the equation for resistance in the form R = ρ L/A

Here is how we got there….

1st Length - doubles then the resistance doubles R ∝ L

2nd CSA doubles then the resistance halves R ∝ 1/CSA

Combine those to get R ∝ L / A or R = L / A

R = resistance

L = length

A = cross sectional area

Electric current and resistance

5.1.7 Apply the equation for resistance in the form R = ρ L/A

Here is how we got there….

Last we add in resistivity. (ρ) “rho”

Resistivity is a property associated with actual material something is made of.

Unit of ohm meter = Ωm

Resistance is a quantity for a specific component or part.

So we get … R = ρ (L / A)

Electric current and resistance

5.1.8 State Ohm’s Law.

IB Formula and Definition

Ohm’s Law sates that the current flowing through a resistor is proportional to the voltage across it (as long as the temperature remains constant).

Voltage = Current x Resistance

V = IR

http://www.youtube.com/watch?v=zYS9kdS56l8&kw=electricity&ad=6599373247&feature=pyv&lr=1

Electric current and resistance

5.1.9 Compare ohmic and non-ohmic behavior.

- Three different types of conductors
- Ohmic
- Filament Lamp
- Diode

Electric current and resistance

5.1.9 Compare ohmic and non-ohmic behavior.

- Resistor is “ohmic” if the current flowing is proportional to the voltage across its ends.
- Metals are ohmicif the temperature is constant.
- Gives a straight line on I-V graph
- Ohmic conductors obey “Ohm’s Law”

Electric current and resistance

5.1.9 Compare ohmic and non-ohmic behavior.

The filament lamp gets hot as there is an creasing voltage. This means the resistance increases with higher voltages.

Electric current and resistance

5.1.9 Compare ohmic and non-ohmic behavior.

- Diodes
- Lets very little ore no current flow until voltage reaches a threshold.
- 0.6V is very common
- Once the threshold is met, there is essentially zero risistance and a large amount of current can flow.
- Act as a switch, only allowing current to flow in one direction.
- Can turn AC in to DC

Electric current and resistance

5.1.10 Derive and apply expressions for electrical power dissipation in resistors

5.1.11 Solve problems involving potential difference, current and resistance.

- Hot battery demo***
- When a current flows in a resistor electrical potential energy is converted into heat energy.
- Where does this heat come from?
- We can solve for the amount of power, P, that is dissipated in the resistor.
- Remember…. power = energy per second, P = ΔE/t

Electric current and resistance

5.1.10 Derive and apply expressions for electrical power dissipation in resistors

5.1.11 Solve problems involving potential difference, current and resistance.

Remember…. power = energy per second, P = ΔE/t

See work on board.

Units for electrical power are Joules per second, J/s or Watts, W

IB Formula

P = VI = I2R = V2 / R

Electric current and resistance

Practice 6

When 20 volts is applied to the ends of a resistor, 0.5 Amps of current flows. Calculate the power dissipated in the resistor.

Answer: 10 Watts

Practice 7

If 12 volts is applied to the ends of a 40Ω resistor, calculate the power dissipated in the resistor.

Answer: 3.6 Watts

Practice 8

0.4 Amps flows through a 125Ω resistor, calculate the power dissipated in the resistor.

Answer: 20 Watts

Electric current and resistance

Practice 9

A resistor of resistance 12Ω has a current of 2.0A flowing through it. How much energy is generated in the resistor in one minute?

Answer: 2.9 x 103J

Electric current and resistance

Electric devices are usually rated according to the power they use. A light bulb rated as 60W at 220V means that it will dissipate 60W when a potential difference of 220V is applied across its ends. If the potential difference across its ends is anything other than 220V, the power dissipated will be different from 60W

Practice 10

A light bulb rated as 60W at 220V has a potential difference of 110V across its ends. Find the power dissipated in this light bulb.

Answer: 15W

Electric current and resistance

Practice

A resistor of resistance 12Ω has a current of 2.0A flowing through it. How much energy is generated in the resistor in one minute?

Answer: 2.9 x 103J

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