AS Physics Unit 1 4 Electric Current - Animated Science

AS Physics Unit 1 4 Electric Current

Mr D Powell

Mr Powell 2009 Index

Chapter Map


4.1 Current & Charge

Specification link-up 3.1.3: Charge, current and potential difference What is an electric current? How can we calculate the charge flow in a circuit? What are charge carriers?

When electrons move through a wire we call it an electrical current. It has a simple definition which leads to 1A = 2 x10-7Nm-1 or 1A = 1Cs-1

Units of Electrical Current

When electrons move through a wire we call it an electrical current. The electrons move as there is a potential difference.

The larger the p.d. the higher the current flow or Coulombs per second.

1A = 1Cs-1

A simple graph of this process would be where a steady

current has flowed for 20s seconds;

The number of Coulombs of charge that have flowed is 100C

Current Flow Basics

The area under the graph can be found by simple calculus and integration;

Current Flow Calculus

ItItItItQ

dtIQ

IdtdQ

so

dt

dQI

t

QI

t

QI

ItQ

tt

t

0

0

0

,0

NB: calculus not required for exam but good simple example for Mathematicians!

This process whereby ions exchange electrons through a molten liquid or dissolved solid is also a way in which a “current” flows. Also a PD between ground and cloud causing a spark to move.

As charge carriers are moving. Hence we can say that energy converted is;

E = QV E = 5C x 3000V = 15000J or E = VIt E = 3V x 2A x 3s E = 18J

GCSE Background – Electrolysis / Static

Make a note of these…

Potential Difference (chemical cells)

One way of separating charges is a chemical cell;

1) Two chemical pastes are separated 2) An anode is formed at one end (positive) & a

cathode at the other (negative) 3) When a wire is connected charges try and equalise

pushing electrons through the wire. 4) Energy is lost inside the cell in the process.

(Internal Resistance) 5) This can also be done with liquids i.e. a Daniell cell

as shown on the right but obviously not a lot of good for transportation in an ipod!

HSW - Extension


Categories of Carriers....

When a voltage is applied across a conductor delocalised electrons are attracted to the positive terminal.

Electrons are firmly attached to atoms and can not be moved through the conductor.

The number of Charge carriers available for conduction increases with

temperature . Resistance decreases with temperature.

Superconductors – conduct massively below a certain temperature. & Ions in solution can also be charge carriers.

Materials can be classified by how the conduct or don’t conduct charge carriers such as electrons….

Insulator

Conductor

Semi-conductor


Total charge charge on an electron 1.6 x 10 -19 C Number of charge carriers

Q=I∆t

Q =Ne Or

I∆t=Ne

Conduction of electrons Summary...

Electrical conduction: 1. is the movement of charge carriers, e.g. electrons

2. the rate of flow of charge is called the current and is measured in

amperes or amps

Make a note of this…


Problem: A wire carries a current of 5mA for 15 mins. i) calculate the charge passing through the wire in this time, Ii) calculate the number of electrons passing through the wire each second

Worked problem… Make a note of this…


Problem: A wire carries a current of 5mA for 15 mins. i) calculate the charge passing through the wire in this time, Ii) calculate the number of electrons passing through the wire each second

Worked Problem Part II Make a note of this…


Summary Questions


Or try this…. Q = ne So sub in Q = It It = ne It/e = n

Summary Questions


Summary Questions

The Basics

Extension

Detailed Example…. Galvanic cell (also called voltaic cell) uses chemical reaction to produce electrical energy (flow of electrons).

CuSO4(aq)

ZnSO4(aq)

Key points 1. Metal electrodes can be placed in a solution

of same metal and neg ions. Either more metal ions are produced or metal in solution plates the electrode.

2. Copper from solution ends up on anode. Zinc ends up in solution in this case.

3. The salt bridge provides a ready supply of neg ions but blocks the metallic ions.

4. If solutions are 1 molar the e.m.f produced in total is 1.10V

5. You can measure this by using a voltmeter and a larger resistance in the circuit.

http://www.mpoweruk.com/chemistries.htm

Extension

http://www.mpoweruk.com/chemistries.htm


A battery is connected across a wire. The current in the wire is 40 mA.

Starter Question….

iSlice Basic

(i) Calculate the total charge that flows past a point in the conductor in 3 minutes. (ii) Using data from the Data Sheet calculate the number of electron charge carriers passing the same point in the conductor in this time.

(iii) If 8.6 J of energy are transferred to the conductor in this time, calculate the potential difference across the conductor. (iv) Calculate the resistance of the conductor.


4.2 Potential difference and power

Specification link-up 3.1.3: Charge, current and potential difference; Current/voltage characteristics What do we mean by potential difference? How can we calculate electrical power? How do energy transfers take place in electrical devices?

P = VI

P=E/t

V=IR

P = I2R

P=V2/R

V=W/Q

Q=I∆t E=IV∆t

E=VIt

Potential Difference (Theory)

If we think of potential difference in terms of energy transfer;

1) To bring two like charges near each other work must be done. 2) To separate two opposite charges, work must be done. 3) Whenever work gets done, energy changes form.

Monkey example; Imagine two positive spheres in space and a monkey. Imagine that a small monkey does some work on one of the positive charges. He pushes the small charge towards the big charge;

+ +

+ +

Work

This is all about a separation of charge between two points. The closer he brings it, the more potential energy it has since the two charges want to repel

When he releases the charge, work gets done on the charge. This is a change in energy;

Electrical potential energy to Kinetic energy.

If the monkey brought the charge closer to the other object, it would have more potential energy.

If he brought 2 or 3 charges instead of one, then he would have had to do more work.

Since the potential energy can change when the amount of charge you are moving changes, it is helpful to describe the potential energy per unit of charge. This is known as electrical potential or potential difference.

Potential Difference (Theory)

V = potential difference in volts, V, JC -1 W = Energy or work done in Joules, J Q = charge in coulombs, C

This is all about a separation of charge between two points.

To understand this we should remember that the charge on one electron is;

1e = 1.6 x 10-19C

Inversely to this the number of electrons that makes up 1 Coulomb of charge is

1C = 6.25 x 1018 electrons

The volt (V) defines the work done per coulomb of charge transferred between two points where;

1V = 1 JC-1

We can also derive a smaller unit of energy (J) for an electron which is the electron volt. Sub 3 into 1 i.e. x each side directly and multiply out the units

1eV = 1.6 x 10-19J

Potential Difference (Definition)

NB: Remember this definition for Unit 1 as well!

1

2

3

4


Potential Difference: * is defined as the work done (or energy transfer) per unit charge

V (Volt) = W (work done, J) Q (charge, C)

B

A +

+

If 1J of work is done in moving 1 C of positive charge from A to B then the Pd is 1V

1V = 1 J / C

Summary

1.5 V 1.5 V

1.5 V

1.5 V 1.5 V 1.5 V

1.5 V V

Voltage or pd:

* is electrical pressure

* causes current to flow

* is measured in Volts (V)

with a voltmeter in parallel


Series & Parallel PD’s


A

In a series circuit

the voltage is shared

across components

V V

3.0 V V

V V

3 V 4

3 V 4

3 V 4

3 V 4

0.1 A

Each bulb adds more resistance to the

series circuit. The current flowing drops.

The voltage across each bulb drops

Series PD


For a component with


HSW - PC Energy Use Question

Energy = Power x Time Energy = 154W x 1 hour Energy = 154W x 3600s E = 154J/s x 3600s E = 554400J E = 554.4kJ

Here is an example of the energy usage of a home computer. Each internal component transfers a variable amount of energy. If you left this PC on for an hour it would transfer;

Current flow to the PC would then be… P = VI or P/V = 0.67A


Power Delivery

When we talk about Power what we mean is “the amount of energy delivered per second”

1 Joule / 1 Second = 1 Watt

It then makes sense that the Power used by a component can be found from the product of current through and voltage across the component;

Power = Voltage x Current

P = V x I


Example

In this circuit the Voltmeter = 4V and the Current = 1A

Power = Voltage x Current

P = V x I

P = 4V x 1A

P = 4J/C x 1C/s

P = 4J/s

P = 4W

4V

1A


Analogy

Another way of thinking about it is saying that the current carries the energy;

4V

1A

C C C C

C C

C C C

1J

1J

1J

1J

C 1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

As the Coulombs of Charge move they release their energy as heat and light (through the bulb)

C

1J

= 1 Coulomb of charge

= 1 Joule of energy

= 1 Second of time


Analogy 2

If the voltage increases, more energy is delivered so the power increases;

5V

1A

C C C C

C C

C C C

1J

1J

1J

1J

C 1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

Power = 5V x 1A = 5J/s = 5W

1J 1J 1J 1J

C

1J


= 1 Joule of energy

= 1 Second of time


Analogy 3

If the current increases, more energy is delivered so the power increases;

4V

2A

C C C C

C C

C C C

C 1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

Power = 4V x 2A = 8J/s = 8W

C C C C

C C

C C C

C

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J 1J 1J

1J

C

1J


= 1 Joule of energy

= 1 Second of time


Starter/Plenary Question……

Energy = Power x Time Energy = 154W x 1 hour Energy = 154W x 1800s E = 154J/s x 1800s E = 277200J E = 277.2kJ v= 230V P = VI or P/V = 0.67A

1. If you left this PC on for 30 minutes how much energy it would transfer in that time (2 marks)

2. Current flow to the PC would then be… (2 marks)


4.3 Resistance & Resistivity

Specification link-up 3.1.3:

Current/voltage characteristics; Resistivity

1. What causes electrical

resistance?

2. When can we use Ohm’s law?

3. What is a superconductor?

L

RA

321

21

1111

......

RRRR

RRR

T

T


GCSE Background

From GCSE physics we already know that;

V = IR So simply in this circuit the resistance of the bulb is 12. This is a measure of how much the bulb resists the flow of electrons.

But where does R and rho come from and how are they linked?

To understand this topic clearly we need to delve into the structure of materials and their electronic configurations.

We must think about the chemistry behind things and the physical dimensions of materials…


http://phet.colorado.edu/simulations/index.php?cat=Physics http://www.batesville.k12.in.us/physics/PhyNet/e&m/current/ECurrent_Notes.htm

Ohm’s law: The pd across a metallic conductor is proportional to the current through it, provided the physical conditions do not change

Q. For the resistor opposite calculate: a) the resistance at this current b) the new pd when the current is 50 A

12V

2mA R ohm

a) R = V = 12 = 6000 I 2.0 x 10-3

b) V = I R = 50x10-6 x 6000 = 0.3 V

Resistance 1

http://phet.colorado.edu/simulations/index.php?cat=Physics

http://phet.colorado.edu/simulations/index.php?cat=Physics

http://www.batesville.k12.in.us/physics/PhyNet/e&m/current/ECurrent_Notes.htm

http://www.batesville.k12.in.us/physics/PhyNet/e&m/current/ECurrent_Notes.htm


Measurement of resistance:

A

V

Record the pd across R for increasing values of current. ( change r to change the circuit current ()

r

The ammeter has a very low resistance ( 0.2 ohm) The voltmeter has a very high resistance ( 20,000 ohm) Why?

Pd / V

I / A

Gradient = V

I

Gradient = resistance

R

Resistance 2


Conductors

• Metals are generally known as conductors.

• Copper electricity cables is an example and what we mean is that they conduct electrons very well.

• Metals are conductors rather than insulators as they have a unique "property".

• This "property " is that the outermost electron on the atom is relatively loosely held and can hop from atom to atom when pushed!


Ionisation Energies

Energy in J/mol

Metal n=1 n=2 n=3 Free Electrons

Li 520 7298 11815 1

K 419 3051 4412 1

Cu 746 1958 3554 1 or 2

Zn 906 1733 3833 2

The table below shows how much energy it takes to remove a whole mole of electrons from their respective atoms.

The blue shaded regions show which are the easy ones.

It is a similar type of thing that is happening in a conductor….

NB: Don’t need to learn these energies just be aware of the idea of the stripping away!


Modelling

So what we are talking about is an electron being ripped from its atom (home) and then moving through the structure of the other adjacent atoms in a form of drift or current.

This is an interesting motion where each electron gains some KE from the e.m.f.

Jumps to another adjacent atom losing the KE and then repeats the process over and over….

Here is a simple example of how the process might work with boron

NB: Boron was picked due to simple structure – it is not a good conductor!

e- e- e- e- e-


Modelling II

Lets examine this rod of an elemental material

You can manipulate the rod and see how many atoms might look.

Then imagine how the electrons would move as on the previous slide

The harder it is to strip away that “free electron” the higher the resistance!

4.3 Resistance Resistivity/resistivity.swf


Other physical factors…

e-

e-

e-

e-

e-

At any temperature above 0K atoms will jiggle around and impede any flow of electrons

Only the electrons move as they are about 2000 times lighter than the atoms they are attached to and pick up the e.m.f

In a Copper wire with 1 x 1028 electron carriers per m3 you would have to accelerate 1 x10-26kg's of electrons.

This is quite a mass of electrons and the more you have the more push you have to use overall to get them moving!

Area and length must also effect such a problem


Resistance or Resistivity

The formulae used to take into account how the physical factors of a wire effect resistance is chicken or egg as you can either consider using it from the R or the perspective.

A

lR

l

RA

The way I remember is resistivity = RAL and of course remember to put the L underneath to make the units correct!

NB: when A = 1m2 and l = 1m = R


Resistivity Formula

There are two main principles at play here;

The resistance is proportional to length i.e. the longer the wire the more resistance there is

The resistance is inversely proportional to the area of the wire i.e. the bigger the area the smaller the resistance.

where R = Resistance in ohms

= Resistivity in ohm metres m A = Cross sectional area in metres squared m2 l = length in metres m

A

lR

lR

AR

1

NB: is the taken as the value of the Resistivity at room temperature 20C


Facts and Figures

Type Material Resistivity in m

Metal

Copper 1.7 x 10-8

Gold 2.4 x 10-8

Aluminium 2.6 x 10-8

Semiconductors Germanium (pure) 0.6

Silicon (pure) 1.7 x 103

Insulators

Glass 1.7 x 1012

Perspex 1.7 x 1013

Polyethylene 1.7 x 1014

Sulphur 1.7 x 1015

This table shows some examples you should be familiar with.

Quite simply they mean that for each substance the resistivity will be …… m (at 20C) as its constant


Simple Example…

Question Calculate the resistance of a one metre length of 24 SWG Nichrome wire.

Answer The key to getting the correct value is to make sure you use consistent units for the values of resistivity, area and length. In this example SI units are used throughout.

There is one other formula that is needed. To calculate the cross-sectional area of the wire we will assume that the wire is circular. So the cross-sectional area, A, is the area of a circle with a radius r. A = π × r²

Resistivity of Nichrome = 110 × 10-8 Ω m Length = 1.00 m

Diameter of 24 SWG wire = 0.559 mm Radius = 0.559/2 = .2795 mm = 2.795 × 10-4 m

Cross-sectional area = π × (2.795 × 10-4)2 = 2.45 × 10-7 m² R = [(110 × 10-8) × 1.00] / *π × (2.795 × 10-4)2]

R = 4.48 Ω


Simple Application….

Material Resistivity in

m Thickness

cm Length

cm

Copper 1.7 x 10-8 0.1 20

Gold 2.4 x 10-8 0.2 10

Aluminium 2.6 x 10-8 0.4 12

Robert Oppenheimer was making a complex atomic bomb called “Fat Man”. He needed to use a certain type of wire in his detonator circuits to connect to the primary so that the overall resistance in that part of the microcircuit was exactly 0.0008 . Then his timing would be just right to produce the maximum number of neutrons possible and thus kill as many Japanese civilians in Hiroshima as possible. He had a choice of copper, gold or aluminium wire to use in the circuit. However, each wire was of a different thickness and length.

Without cutting the wires work out which one he could use…you need to do three separate calculations using the following data;

Hint: tabulate your data or write it out for each question with conversions in full


Finding the Resistivity of a Wire

Arad

g

This new formulae has a simple characteristic which again would fit the;

y = mx + c principle

One way of working out a resistivity of any material is to set up a circuit with a sample of the substance. Measure the area of the wire several times and take the average. Then measure the current flow through and potential difference across the wire. This enables you to work out R. Then repeat the experiment for several different lengths.

NB: is the taken as the value of the Resistivity at room temperature 20C

l

R

lA

R


Resitivity?

Arad

g

Length in m Resistance in m

0.10 0.18

0.20 3.70

0.30 5.50

0.40 7.10

0.50 9.20

0.60 11.00

0.70 13.00

0.80 14.50

0.90 14.60

1.00 18.00

Use the graphing technique to find out the resistivity of a sample wire which is 0.26cm thick. What metal is the wire made from?

NB: calculated at 20C

l

R


Looks like Gold to me!

Graph to show resistance against length

y = 18.253x

0

2

4

6

8

10

12

14

16

18

20

0 0.2 0.4 0.6 0.8 1 1.2

Length of wire in meters m

Re

sis

tan

ce

in

milli o

hm

s

1.32739x10-6m2 x 18.253x10-3 m-1 =2.42 x 10-8

m


Electrical Resistance Data SWG - Standard Wire Gauge


Material Resistivity at 20°C

Ω·m µΩ·cm

silver 1.6 × 10-8 1.6

copper 1.7 × 10-8 1.7

gold 2.2 × 10-8 2.2

aluminium 2.7 × 10-8 2.7

magnesium 4.2 × 10-8 4.2

nickel 6.9 × 10-8 6.9

iron 10.1 × 10-8 10.1

chromium 13.2 × 10-8 13.2

manganese 160 × 10-8 160

carbon (graphite) 3 000 × 10-8 3 000

Manganin 44 × 10-8 44

Constantan (Eureka) 49 × 10-8 49

Nichrome 110 × 10-8 110


This is the property of a material which is at or below a critical temperature Tc where it has zero resistivity .

Implications: * Zero resistance * no pd exists across a superconductor with a current flowing * the current has no heating effects

temp Tc

Superconductivity


temp Tc

Properties of a superconductor: * material losses the effect above the critical temperature Tc.

* If Tc is above 77K ( -196 C) it’s a high temperature superconductor current Tc max = 150 K - 123C

Superconductivity


4.4 Components and their characteristics

Specification link-up 3.1.3: Circuits How does the current through a filament lamp vary with pd? What are the characteristics of a diode? What can we use a thermistor for?

ammeter

voltmeter

cell

Indicator / light source

diode

Light emitting diode

resistor

Variable resistor

thermistor

Light dependent resistor

heater

Electric motor

Many cells = a battery

No more semi circles!

of a specific value

Resistance falls as temp rises

Low resistance, connect in series

High resistance, connect in parallel

Emits light when forward biased

Resistance falls as light level rises

Conducts when forward biased

TASK: Link these symbols to names & functions….

ammeter

voltmeter

cell

Indicator / light source

diode

Light emitting diode

resistor

Variable resistor

thermistor

Light dependent resistor

heater

Electric motor

Many cells = a battery

No more semi circles!

of a specific value

Resistance falls as temp rises

Low resistance, connect in series

High resistance, connect in parallel

Emits light when forward biased

Resistance falls as light level rises

Conducts when forward biased

V

I

+ -

-

+

wire

V

I

+ -

-

+

lamp

V

I

+ -

-

+

V

I

+ -

-

+

0.6 V

thermistor diode

TASK: Draw out the axes and sketch these graphs….

V

I

+ -

-

+

wire

V

I

+ -

-

+

lamp

V

I

+ -

-

+ High temp

low temp

V

I

+ -

-

+

0.6 V

thermistor diode

Amps

Milli-amps


Resistance and temperature

For metals: * resistance increases with temperature * positive ions vibrate more * conduction electrons movement is impeded * positive temperature coefficient

For intrinsic semiconductors: * resistance decreases with temperature * more charge carriers become available * negative temperature coefficient * much larger change in resistance /C than with metals * thermistors used as temperature sensitive component in a sensor

V

I

+ -

-

+

High temp

low temp



Resistance and temperature

For metals: * resistance increases with temperature * positive ions vibrate more * conduction electrons movement is impeded * positive temperature coefficient

For intrinsic semiconductors: * resistance decreases with temperature * more charge carriers become available * negative temperature coefficient * much larger change in resistance /C than with metals * thermistors used as temperature sensitive component in a sensor

V

I

+ -

-

+

High temp

low temp

V

I

+ -

-

+ High temp

low temp



Resistor

As the voltage increases the current also increases at the same rate.

This is what is called “ohms law”

True for resistor at a constant temperature

It is directly proportional


Filament Lamp / Bulb

The resistance of a filament lamp increases as the temperature of the filament increases.

i.e. the resistance changes as the temperature of the wire changes.

The gradient of the graph represents the resistance


Thermistor

The resistance of a thermistor decreases as the temperature increases so if we look at it from the VI perspective it is the opposite of a bulb!


How do they work?

The exact conduction mechanisms are not fully understood but metal oxide NTC thermistors behave like semiconductors, as shown in the decrease in resistance as temperature increases. The physical models of electrical conduction in the major NTC thermistor materials are generally based on this theory;

A model of conduction called "hopping" is relevant for some materials. It is a form of ionic conductivity where ions (oxygen ions) "hop" between point defect sites in the crystal structure.

The probability of point defects in the crystal lattice increases as temperature increases, hence the "hopping" is more likely to occur and so material resistivity decreases as temperature increases.


Temperature Sensors?

They are inexpensive, rugged and reliable. They respond quickly to changes and are easy to manufacture in different shapes.

An example could be made from a combination of

Fe3O4 + MgCr2O4 (metallic oxides) A NTC thermistor is one in which the resistance

decreases with an increase in temperature. The circuit shows how you can use the thermistor as

a potential divider. As the temperature changes the division of voltage or energy will change. You need the 5k resistor or the voltage would be that of the cell a constant 3V.

A common use is the glass heat sensor in a car or

the temperature sensor in a conventional oven.


LDR

The resistance of a light-dependent resistor (LDR) decreases as light intensity increases. This is a similar process to a thermistor


Semi Conducting Diode

The current through a diode flows in one direction only. The diode has a very high resistance in the reverse direction.

Often used in transformers to change A.C to D.C. currents or computers!

Why does this happen?


Diode – example...


Real Data logging Information LED + 220 resistor in unit


Diode on its own


Semiconductors?

Carbon, silicon and germanium (germanium, like silicon, is also a semiconductor) have a unique property in their electron structure - each has four electrons in its outer orbital.

This allows them to form nice crystals. The four electrons form perfect covalent bonds with four neighbouring atoms, creating a lattice.

In carbon, we know the crystalline form as diamond.

In silicon, the crystalline form is a silvery, metallic-looking substance. Which is a near perfect insulator.

Well on its own that is not amazing but a semi-conductor is made from a silicon lattice with an impurity which will enable it to conduct. We often refer to these types of materials as intrinsic semiconductors.


Doping p or n?

In N-type doping, phosphorus or arsenic is added to the silicon in small quantities. Phosphorus and arsenic each have five outer electrons, so they're out of place when they get into the silicon lattice. The fifth electron has nothing to bond to, so it's free to move around. It takes only a very small quantity of the impurity to create enough free electrons to allow an electric current to flow through the silicon. N-type silicon is a good conductor. Electrons have a negative charge, hence the name N-type. In P-type doping, boron or gallium is the doping agent. Boron and gallium each have only three outer electrons. When mixed into the silicon lattice, they form "holes" in the lattice where a silicon electron has nothing to bond to. The absence of an electron creates the effect of a positive charge, hence the name P-type. Holes can conduct current. A hole happily accepts an electron from a neighbour, moving the hole over a space. P-type silicon is a good conductor.

p-type

n-type


PN junctions?

N-type and P-type silicon are not that amazing by themselves; but when you put them together, you get some very interesting behaviour at the junction. That's what happens in a diode.

The place where they meet in the diagram forms what is called a depletion zone.

Where free electrons have filled positive “holes” to form an area where there are no free charge carriers.

P N P N P N


PN junctions - behaviour to applied pd

V< 0.7V V> 0.7V

This typically takes 0.7V to overcome and bridge the gap and make it conduct!

If we reverse the flow then the gap gets larger and does not conduct (up to 30V)

V< 0V V<< 0V

P N P N P N

P N P N P N


Summary

• Ohms law states that V I when T = const

• If we plot a graph of a component of v on y-axis and I on the x-axis the gradient is the resistance of the component.

• Light bulb and diode are examples of non-ohmic components

Flow of Charge in Circuits

P = VI

P=Et

V=IR

P = I2R

P=V2/R

Q=It

E = VIt

E = VQ

Ideas for Circuits….

Flow of Charge in Circuits

Previously we have looked at equations of Power, Resistance & Flow of Charge;

P = VI P = Et V = IR P = I2R P = V2/R Q = It E = VIt E = VQ We can conduct some simple experiments and mathematical substitutions to prove and verify these

formulae. Form a group of 3-4 and work it out for yourselves. Be ready to present to the rest of the class.

12V D.C. Supply

Voltmeter(s)

Ammeter(s)

3x 12V 21W bulbs

Stopwatch

Hint: Try connecting the bulbs in different configurations whilst monitoring the voltage and current.

Then map out the connections on a sheet of blank A3 paper & graph paper as you wish with results, images and calculations. You may also want to try out the formulae for resistors in series or parallel from further on in the course!

AS Physics Unit 1 4 Electric Current - Animated Science

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