Electric Currents Topic 5 These notes were typed in association with Physics for use with the IB...

Electric CurrentsTopic 5

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

Electric Potential Difference5.1.1 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!!!!


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.


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


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

ΔV = ΔE/ q

AKA V = W/q

Where:

V= potential difference(volts)

E = Energy(Joules)

q = charge

W= work


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

electric 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 Difference5.1.3 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 Difference5.1.3 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 Difference5.1.4 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.


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


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.


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.


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


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


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***


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.


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.


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.


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?


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


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


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 ohmic if the temperature is constant. Gives a straight line on I-V graph Ohmic conductors obey “Ohm’s Law”


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


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 = I2 R = 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 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


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

Electric Currents Topic 5 These notes were typed in association with Physics for use with the IB...

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