AS Electricity (DONE)

Question 1

What is electric current?

Answer 1

• electric current is the rate of flow of charged particles.

- The symbol for current is I, and measured in the ampere A.

Question 2

Is electric current always in wires?

Answer 2

• If we consider a piece of wire, current is the flow of electrons within the wire.
• However not all electricity is in wires, if you think about copper sulfate it is not about the electrons moving it is about all of the various ions moving inside.
• Although we consider the charged particles to be electrons they could also be positrons, protons or various ions.

Question 3

What is one Ampere?

Answer 3

• The ampere is defined as the current flowing in 2 parallel wires 1m apart in a vacuum such that there is an attractive force of 2.0 x 10^-7 N per metre length of wire between them.
• A better way to think of an amp is 1 A is the same as 1 coulomb of charge passing a point in 1 second.
• The ampere is an SI base unit.

Question 4

Which direction does current flow?

Answer 4

• The conventional current flows from positive to negative in a circuit, however we know that electrons are negatively charged particles and therefore they will be repelled from the negative end and attracted to the positive end.
• The electrons are going to be moving in the other direction.

Question 5

How do you measure electric current?

Answer 5

• In order to measure current we need an ammeter, we need to connect to a circuit in series with the other components.
• The ammeter always goes in series and it should have a very low resistance

Question 6

What is charge?

Answer 6

• charge is a characteristic of a unit of matter that shows the extent to which it has more or fewer electrons than protons.

Question 7

How do charged objects interact?

Answer 7

• When it comes to objects that are charged, objects with similar charge repel and objects with opposite charges attract.

Question 8

How do we measure charge?

Answer 8

• Charge has the symbol Q, due to the fact charge is quantised.
• In GCSE we say protons have charge +1 and electrons have charge -1, hwoever charge comes in small building blocks.
• The elementary charge which is the smallest charge we can get to is 1.6 x 10^-19
• If it is an electron it will be -1.6 x 10^-19
• The unit for charge is the coulomb C.
• The equation for charge is Q = It

Question 9

What is a coulomb?

Answer 9

• 1 coulomb is the charge that flows past a point in one second when ther eis a current of 1 amp

Question 10

What is Kirchhoffs first law?

Answer 10

• the sum of the currents into and out of a junction are equal.

Question 11

How does current act in a series circuit and what happens when current reaches a junction?

Answer 11

• Current is the flow of electrons and although conventionally they go from positive to negative the electrons are going the other way round.
• If we were to measure the current at any point in the circuit by putting an ammeter in series we would have the same value of current throughout.
• When electrons move along the circuit and reach a junction we find that everything into the junction will also leave the junction.
• Therefore some of the current will move to the left of the junction and some will move to the right.
• Meaning that current I1 turns into I2 and I3.
• Therefore the sum of the currents into the junction I1 = I2 + I3
• Or you could say that I1 – I2 – I3 = 0 therefore the sum of the currents into and out of the junction is 0.

Question 12

Why do we need to know about mean drift velocity?

Answer 12

• When it comes to turning on a light switch although the light seems to turn on instantaneously, the electrons moving in the wire are not moving at the speed of light.
• The electrons in the wire are moving very slowly at mms or cms per second.
• Inside a metal you have a metal lattice which is vibrating slightly and we have free delocalised electrons moving randomly in all directions.
• As soon as a charge is applied to each end the electrons will be repelled form the negative end and move towards the positive end.

Question 13

How can we derive an equation for current using mean drift velocity?

Answer 13

• If you take a thick piece of wire with a current flowing through it.
• Firstly the wire has a cross sectional area A, it also has a length L, and flowing through the wire will be a current I.
• There is going to be a certain amount of charge carriers per unit volume which is symbol n.
• All of these individual charge carriers have an individual charge (if they are electrons they will have charge e), and drift velocity v.
• Volume V = LA, therefore the number of free electrons = volume x charge density
• Meaning that number of free electrons = nLA
• So the total charge free to move = nLAe
• The time it takes for the charge to leave the bit of wire = L/v
• As we know Q = It, we can rearrange to say I = Q/t
• We know total charge free to move and the time it takes for charge to leave the wire is L/v.
• This means that if we substitute it into the equation for current we get
• I = (nLAe)/(l/v)
• We can simplify to say that I = Anev

Question 14

What arrangement do conductors have?

Answer 14

• Inside a conductor we have charged particle which are free to move around, they can move around and flow as an electric current transferring energy around a circuit.
• You can give a number to the amount of charged particles which are free to move n, this value of n will be large.

Question 15

What arrangement do insulators have?

Answer 15

• insulators are materials where the charged particles are locked into the atomic structure, the density of the charge carriers that are free to move is very low meaning it has a low value of n and it is therefore an insulator.

Question 16

What arrangement do semi-conductors have?

Answer 16

• There is a middle ground where we have a few charge carriers free to move but the other charge carriers are locked into the structure, we can liberate the charge carriers by applying external force.
• the value n will be between the conductor and insulator.

Question 17

What is an example of a semi-conductor and how does it work?

Answer 17

• A good example of this is a thermistor, when you apply heat it allows electrons to escape where they are locked meaning the number of charge carriers increases and so it becomes a better conductor as heat increases.
• Instead of giving energy in the form of heat we could apply in the form of light, the photons of light can cause electrons to escape and increase charge carriers such as in an LDR.

Question 18

what symbol does a variable resistor have?

Answer 18

• variable resistors are able to change their value and because of this we put an arrow diagonally across the resistor protruding at both ends.

Question 19

what symbol does an LDR have?

Answer 19

• light dependant resistors (LDR) change resistance depending on light, this has a rectangle with a circle around it along with 2 arrows of light pointing diagonally towards the rectangle.

Question 20

what symbol does a thermistor have?

Answer 20

• a thermistor changes resistance depending on temperature, this symbol is a rectangle with what looks like a hockey stick through it.

Question 21

What symbol do diodes have?

Answer 21

• diodes are like a one way valve which lets the current through in one direction only.
• the symbol is an arrow against a horizontal T, there is also sometimes a circle around it and diodes which give out light (light emitting diodes) will have 2 arrows of light heading out of the symbol.

Question 22

What will happen if you apply opposite charges to each side of a conductive wire?

Answer 22

• If you have a wire with a charge carrier inside, it will have no flow in any direction as there is no difference in the electric potential at each end.
• however if you make one side +ve and the other –ve and the charge carrier represents the conventional current, it will flow from +ve to –ve.

Question 23

How can a battery gain a current?

Answer 23

• Inside a battery we have various chemicals which can create a difference in electric potential between one end and the other.
• If we connect this into a circuit we have a different electric potential at each end and therefore we have a flow of charged particles and therefore a current.

Question 24

How does energy flow through a circuit with a cell and a filament lamp?

Answer 24

• By convention we have current going +ve to –ve, the cell is our source of energy.
• Inside the cell we have chemical energy, if you add a filament lamp into a circuit you can look at the flow of energy.
• Provided there is a current flowing, the chemical energy will be transferred to electrical energy, move around the circuit and be emitted by the filament lamp.
• In order for the energy to be emitted by the lamp the charge carrier will transfer electric energy into other forms and the energy will be emitted .
• As the charge carriers pass through the filament lamp they then go past the cell again, picking up some more energy and the process will continue.

Question 25

How is emf and potential difference induced in a circuit with a cell and a filament lamp?

Answer 25

• Across the cell we have an emf, which is the amount of energy transferred from chemical to electrical per unit charge.
• as the charge carriers move through the cell, chemical energy is transferred to electrical energy which flows through the circuit and is emitted by the time it again reaches the cell.
• The pd is the energy transferred per unit charge from electrical energy into other forms of energy.
• the charge carriers will reach the filament and transfer electrical energy into other forms which are then emitted by the lamp.
• There is going to be a difference in pd on each side of the filament lamp.
• We can use a voltmeter in parallel to measure the pd across a component.

Question 26

Why must the voltmeter be in parallel and how doesn’t it impact the rest of the circuit?

Answer 26

• The voltmeter must go in parallel with the component being measured, so effectively we are looking for the difference in electric potential before and after the lamp.
• In order for the voltmeter to take no effect on the rest of the circuit it must have a high resistance so that no current flows through the voltmeter in parallel.

Question 27

What is the difference between emf and pd?

Answer 27

• Although both emf and pd are measured in volts, the emf is going to be the energy per unit charge that is transferred from chemical to electrical whereas the pd is the energy per unit charge transferred form electrical to other forms.

Question 28

Why is work needed to move a charged particle in a uniform electric field?

Answer 28

• If you have a unit of positive charge and an electric field with neutral and positive plates.
• The unit charge will repel the positive plate and move towards the neutral plate.
• ## In order to move the particle against this natural motion you need to do work, therefore we can look at the amount of work per unit charge to move the charge in the electric field.

Question 29

How can we derive an equation for the work needed to move a charged particle in an electric field?

Answer 29

• You will do a certain amount of work W to move the positive charge which has a charge Q, and if you move the particle through a certain distance it will increase its electric potential so V is the potential difference.
• The greater the pd you need to move the particle through, the greater the energy you need to apply.
• Therefore the equation work done W = QV
• You can also say that V = W/Q

Question 30

How can you define the volt?

Answer 30

• Using the equation V = W/Q you can define the volt.

- 1 volt is the potential difference between 2 points when 1 joule of work is done to move a charge of 1 coulomb.

Question 31

How do particle accelerators work?

Answer 31

• When we have a unit charge and move it through a potential difference towards a plate of the same charge, when we let go it will be similar to how a stone would move in a gravitational field.
• The particle will accelerate towards the opposite plate and it hits the plate with a velocity V, showing how particles are accelerated.

Question 32

What energy changes occur in particle accelerators?

Answer 32

• When the particle is moved through the pd towards there plate of same charge, it will gain in electric potential energy.
• When the particle accelerates towards the plate with opposite charge it will transfer this electric potential into kinetic energy.
• If the energy transfer is 100% efficient then ask electric potential will be converted to kinetic energy when the particle hits the opposite plate.

Question 33

What equation can be used to calculate work done to move a particle through a pd (particle accelerator) and how can it be equated to kinetic energy?

Answer 33

• The work done to move the particle through the pd will be W = QV
• This energy will turn into kinetic energy by the time the particle reaches the bottom if it is 100% efficient.
• The amount of work we do on the particle is the amount of electric potential it has at the start and if it is 100% efficient we can say work done = kinetic energy and therefore QV = 1/2mv^2.

Question 34

How would we calculate the work done to move an electron through a pd in a particle accelerator?

Answer 34

• We can do work to an electron as we let it go it will accelerate towards the positive plate, the charge is e and the energy we need to move it away will be eV.
• This then turns to kinetic energy so eV = 1/2mv^2
• There are similarities between this equation and the definition for the electron volt.
• Equally, if we apply a voltage to the electron we can accelerate it by using potential difference which turns to kinetic energy, this is used in particle accelerators.

Question 35

Why do components have resistance and how can we find a definition for resistance?

Answer 35

• All electric components have resistance and this is due to the fact that it isn’t easy for the electrons to move through it.
• You can find the value of resistance for a component by measuring the current and pd across it.
• This leads to the definition of resistance = pd across component/current in the component, R = V/I

Question 36

What is the definition of 1 ohm?

Answer 36

• 1 ohm is the resistance of a component that has a pd across it of 1v per ampere of current in it.

Question 37

What does ohms law state and what components does it apply for?

Answer 37

• ohms law states that for a conductor at a constant temperature the current is directly proportional to the potential difference across it.
• It is important we have the right kind of conductor which is an ohmic conductor/metal.
• It doesn’t have to be a resistor it could be the piece of wire connecting the resistor.
• provided we have a constant temperature inside the wire we can say the pd is proportional to the current.

Question 38

Why does resistance to current increase as temperature of a conductor increases?

Answer 38

• If you have a metal lattice at a fairly low temperature, although the metal ions are vibrating, the charge carriers can get through with fairly little resistance stopping them.
• If you increase the temperature everything will have more kinetic energy and so the metal ions oscillate at a greater amplitude which means they knock into electrons causing a greater resistance.
• The hotter a metal gets the greater the resistance to the flow of current.

Question 39

How can you setup a circuit to measure the pd across a resistor and current flowing through it?

Answer 39

• You need a circuit with a power supply connected various components, first of all the resistor needs to be connected.
• You can then use a voltmeter in parallel with the resistor to measure the pd across it.
• An ammeter needs to be connected in series with the resistor to measure the current passing through it.

Question 40

What can variable resistors do within a simple circuit?

Answer 40

• The resistance value of the variable resistor can be changed to alter the value of the current moving through the component and also the share of pd in the circuit.

Question 41

What does it mean to switch the polarity of a cell and what impact would this have on readings for pd and current?

Answer 41

• Changing the polarity of the cell is to switch the +ve and –ve sides of the cell.
• This causes the conventional current to flow in the opposite direction, meaning that we will have a negative current and pd across the component.

Question 42

What are the IV characteristics of a fixed resistor?

Answer 42

• The resistor is designed to have a constant resistance therefore it acts as an ohmic conductor meaning current is directly proportional to the pd.
• If we were to plot the IV graph for a resistor the line would be linear and would show a directly proportional relationship between I and V meaning resistance is constant.
• A resistor with a higher resistance will cause a lower current for the same pd.

Question 43

What are the IV characteristics of a filament bulb?

Answer 43

• The IV graph of a filament bulb is shaped a bit like an S.
• It passes through the origin and at a low voltage the gradient is steep but the gradient is shallower at higher voltages.
• It behaves the same for negative and positive pd.
• The resistance increases as the current increases however in this graph the resistance is not equal to the gradient of the line it is equal to the pd/current at any point.

Question 44

How does a filament bulb emit light?

Answer 44

• The filament bulb has a small filament of tungsten inside which has a high melting point.
• This means if a current is passed through it, it will illuminate and the hotter something is the higher wavelength of light will be given out.

Question 45

Why does the filament bulb not follow ohms law?

Answer 45

• The filament lamp does not follow ohms law, as there is not a constant of proportionality between current and pd.
• The resistance increases as current increases as the greater the flow of charge carriers, the more they knock into the metal lattice and therefore the higher the resistance.

Question 46

What is a diode?

Answer 46

• A diode acts as a one-way valve which lets conventional current through but any current flowing in the other direction it will stop.
• The most common diode is the LED light emitting diode, which will only light up when current is conventional.

Question 47

What are the IV characteristics for an LED?

Answer 47

• If you look at the IV for an LED you will have no current for all values of negative pd.
• This continues slightly into the +ve values of pd and then the gradient becomes very steep.
• For all –ve values of pd, the resistance will be infinite.
• On the +ve side we have a point called the threshold voltage where below this values there is no current flowing at all which is important when you investigate plancks constant.
• After the threshold voltage we have a very steep line where the resistance gets very small.

Question 48

What does an LDR consist of and how does it work?

Answer 48

• Inside an LDR is cadmium sulphite with metallic connecters within it.
• This is underneath a plastic case which acts as a lense to focus the light onto the surface.
• The LDR works through the photoconductive effect which means the material becomes more conductive when photons of light land onto it.
• This causes the LDR to change resistance as light varies.

Question 49

What will a resistance against light intensity graph show about LDRs?

Answer 49

• We can look at the value of resistance against the light intensity (Wm^-2).
• When it is dark and there is low light intensity the resistance is high.
• As the light intensity increases we liberate more charge carriers and the resistance decreases.
• This creates a curved line on the graph of R against light intensity.

Question 50

How do heat sensors work?

Answer 50

• Inside a heat sensor is a thermistor which acts as a temperature switch.
• It changes its resistance as the temperature changes.
• This is useful in kitchens as if you had a smoke detector which detects smoke particles it would go off all the time.

Question 51

What are NTC thermistors?

Answer 51

• NTC thermistors are resistors with a negative temperature coefficient, which means that the resistance decreases with increasing temperature.

Question 52

What does the graph of resistance against temperature for an NTC thermistor show?

Answer 52

• The graph of resistance against temperature shows that as temperature increases, resistance decreases.
• The first part of the graph shows a small change in temperature (x axis) and large change in resistance (y axis), this large change in resistance makes the thermistor useful.
• It has the large change in resistance because It is a semiconductor.
• This means that charge carriers are locked in the material but when the temperature reaches a certain temperature there is enough energy for charge carriers to escape causing resistance to increase very quickly.

Question 53

When are NTC thermistors often used?

Answer 53

• NTC thermistors are used to monitor temperatures in for example laptops, car engines, heat alarms and kettles.

Question 54

What are the IV characteristics of an NTC thermistor?

Answer 54

• The IV characteristics of the NTC thermistor is the opposite way round to the filament lamp.
• This is because resistance gets lower at higher temperatures and so the line gets steeper rather than shallower.

Question 55

What is the relationship between resistance and length of a material?

Answer 55

• Perhaps a conductor is twice as long as another of the same material, if the same charge carrier had to move through twice the length there’s going to be twice the resistance.
• We can then say the resistance is proportional to the length of the object.

Question 56

Why do conductive materials have a value of resistance?

Answer 56

• A metal lattice will vibrating slightly and when an electron has to move from one side to the other, as it moves through it will feel resistance from the vibrating lattice.

Question 57

What is the relationship between the area of a conductive material and resistance?

Answer 57

• If we had an object which was twice as wide as another object of the same material, the charge carrier will be able to find a path of shortest resistance to reach the other side therefore the resistance is proportional to 1/Area.

Question 58

How can we derive an equation for resistance using resistivity?

Answer 58

• We know that resistance is directly proportional to the length and inversely proportional to the area.
• Therefore if we bring these 2 relationships together we can say that the resistance is proportional to L/A.
• Provided we have the same metal at the same temperature we can then give this a constant of proportionality so we can say R = (L)/A where  = resistivity.
• The resistivity is specific to the material.

Question 59

What units does resistivity have?

Answer 59

• If we think about the units of resistivity we can say  = (RA)/L where resistance is measured in (ohms x m^2)/m we get the units of resistivity to be ohm metres.

Question 60

What is 1 ohm metre of resistivity?

Answer 60

• 1 Ohm metre is defined as a resistance of a conductor, which is 1 m long, has a cross-sectional area of 1 m² and resistance of 1 Ohm.
• This will vary from metal to metal and will increase as temperature increases

Question 61

How can we derive all 3 equations for power dissipated in a component?

Answer 61

• If you have a component and there is energy being transferred to it per unit time, we can calculate the power dissipated in the component.
• We know that power P = W/t
• We also know that work done W = VQ and we can substitute this into the previous equation giving us P = VQ/t
• There is a link between the charge transferred and the time.
• The equation Q = It can also be written as I = Q/t
• We can replace the Q/t term in the power equation to get P = IV which means if you have something with a greater current or pd there is going to be a greater amount of energy transferred or dissipated into that component.
• Electric power P = IV but another equation says that V = IR or I = V/R
• We can use these equations to create more forms of the equation for electrical power.
• We can say the power P = IIR = I^2R
• We can also say that P = VV/R = V^2/R
• Power is still the rate of energy transferred but we can calculate it using the 3 equations above.

Question 62

How can we use power and time to calculate the energy transferred?

Answer 62

• Electrical power is the rate at which work is done P = W/T and if you rearrange to make energy the subject you can say work done W = Pt.
• Normally we measure power in Watts and time in seconds.
• Ws^-1 means the energy has units Joules.

Question 63

Why do we use the kilowatt hour to measure energy rather than the joule in some cases?

Answer 63

• Joules are only useful if we think about everyday small scale objects.
• A kilowatt hour is a much larger unit and so if we think of something such as a house where a large amount energy is transferred, a kilowatt hour is a lot more useful.

Question 64

How can we get work done/energy in the units kilowatt hours?

Answer 64

• If we measure power in Kw and time in hours we will get work done in units kilowatt hour.
• One kilowatt is 1000 watts and 1 hour is 3600 seconds so this means 1 kilowatt hour = 1000 x 3600 which = 3.6 x 10^6 J.

Question 65

How are Kwh referred to on house bills etc?

Answer 65

• Kwh are referred to as ‘the unit’ in order to prevent confusion on house bills etc, it then gives you cost per unit .
• If you know the total amount of kilowatt hours used and the cost of per unit you can work out the total costs.

Question 66

How do charge carriers operate in a simple circuit with a source of emf and 2 components?

Answer 66

• In a circuit with a cell which is the source of emf and 2 components which are resistors, we can use this analogy to eventually explain Kirchhoffs second law.
• The source of emf is the energy source, and moving around the circuit are charge carriers carrying energy from the cell to the various components.
• The charge carriers move from positive to negative in the circuit.
• If you think of the circuit as an assault course you have the cell as an energy station and every time charge carriers move through the cell they pick up energy.
• Around the assault course you have various obstacles represented by resistors.
• Effectively there is no resistance in the wire but when the charge carriers get to a resistor they have to give off energy to get past it, this creates a pd across the resistor where electrical energy is transferred to other forms.
• The charge carrier will give off energy at both resistors then return to the cell where it gains more energy.

Question 67

How can you use the way that charge carriers behave to explain Kirchhoffs second law?

Answer 67

• If you had a circuit with a source of emf and 2 resistors in parallel we can use the same assault course analogy to show the behaviour of the charge carriers.
• Everyone will start at the same place but there is a junction and perhaps half the people go route 1 and the other half route 2.
• The charge carriers will distribute in the same way, this also relates to kirchoffs first law.
• If you split the circuit into 2 loops and take the first loop, this is one possibility for the charge carriers to travel round the circuit and the other way is the other loop, we can say we have a closed circuit.
• Whatever energy is given to the charge carrier at the emf must also be given out by the charge carrier throughout the circuit, if the charge carrier travels around loop 1 it will see there is only one place it can release its energy at the resistor, therefore the emf is going to be equal to the pd across the resistor.
• Kirchhoffs second law states that the sum of the emfs in any closed loop in a circuit is equal to the pd around that closed loop.

Question 68

How can you derive an equation for the total resistance of resistors in a series circuit?

Answer 68

• In a circuit with 2 resistors in series the total pd across them is the sum of their individual pds.
• So Vt = V1 + V2 (this can be found from kirchhoffs second law).
• However V = IR so therefore IRt = IR1 + IR2…
• Because of kirchhoffs first law, the current everywhere in the circuit is going to be the same therefore we can cancel the I values in the equation above to give Rt = R1 + R2…

Question 69

How can you derive an equation for the resistance of resistors in a parallel circuit?

Answer 69

• If we have 2 resistors in parallel in a circuit with many other components as well, we can look at their overall resistance considering kirchhoffs first and second law.
• If we have a current I flowing into a junction it leads to a current I1 in first resistor and I2 in second resistor;
• Using kirchhoffs first law we can say current into junction I = I1 + I2
• The potential difference V = IR therefore we can rearrange this to say I = V/R.
• If you use the equation to replace the I terms we can say V/Rt = V/R1 + V/R2…
• Using kirchhoffs 2nd law we know that the pd across both components will be equal therefore V = V1 = V2.
• This equation therefore means the Vs in the previous equation can cancel as they are all equal leaving the equation 1/Rt = 1/R1 + 1/R2 …

Question 70

How can you rearrange the equation for resistance of 2 parallel resistors to make Rt the subject?

Answer 70

• If you have 2 components in parallel you can rearrange the equation to make Rt the subject.
• Firstly you need to multiply everything by Rt, when we do that we get 1 = Rt/R1 + Rt/R2
• If we then multiply by R1 we get R1 = Rt + (RtR1)/R2
• We can then multiply everything by R2 to get R1R2 = RtR2 +RtR1
• We can say R1R2 = Rt(R1 + R2)
• Meaning Rt = (R1R2)/(R1 +R2)
• Therefore Rt = product/sum

Question 71

What is a battery and what does it consist of?

Answer 71

• The battery is the source of emf which provides energy for the circuit.
• You can look inside a battery and what you will find is a number of cells.
• One end is negative and the other is positive and if we look at the individual cells we can look at the materials they are made out of.
• Because there is a difference in electrical potential between one end and another we therefore have a source of emf.
• It is this difference in electrical potential between different layers of chemicals which drive electrons around a circuit.

Question 72

Why do cells/batteries have an internal resistance and how does this impact the max energy given to each unit charge?

Answer 72

• Every time you have a current moving through a cell the electrons have to pass through the various materials meaning that a cell has its own internal resistance.
• In a circuit the emf provides the maximum amount of energy given to each unit of charge.
• In reality, although we may have a high emf not all of the energy is given into the circuit as some of the energy is used up as charge carriers move through the power source.

Question 73

What is the terminal pd and how do you measure it?

Answer 73

• The terminal pd is the output you get from the cell.

- Measured using a voltmeter in parallel across the terminals of the cell.

Question 74

What is the relationship between internal resistance, current and pd in a circuit?

Answer 74

• Internal resistance is directly proportional to the current.
• The more current flowing around the circuit, the greater the effect of the internal resistance r.
• However internal resistance and current is inversely proportional to the pd.
• The greater the internal resistance or the greater the current, the smaller the terminal pd.

Question 75

Why is the terminal pd always smaller than the emf?

Answer 75

• The terminal pd V is always going to be smaller than the emf because when we have a current flowing and an internal resistance, some of the energy available to the circuit is used up by the charge carriers moving through the battery.

Question 76

What are the possible equations for emf in a circuit?

Answer 76

• Using kirchhoffs second law we can say the pd across the resistor and cell will be equal to the emf.
• Therefore emf E = V + Ir
• The Ir part of the equation is called the lost volts as the volts are lost within the internal power source.
• We know that V = IR and so we can rewrite the emf equation as E = IR + Ir.
• This can be simplified to E = I(R + r)

Question 77

How can the emf of a cell be investigate using a circuit with a voltmeter, ammeter and variable resistor?

Answer 77

• We can set up a circuit with a cell of internal resistance r, and we can put a voltmeter across it to measure the terminal pd, and put a variable resistor and an ammeter into the circuit.
• We can then investigate the internal resistance and original emf of the cell through altering the resistance of the variable resistor.
• We will find that the greater the current, the smaller the terminal pd.
• We can change the equation E = V + Ir, into the form y = mx + c and through using the graph of V against I we can rearrange the equation to V = -rI + E.
• This means the y intercept is equal to the value of the emf.
• It also means the gradient of the line is equal to –r.

Question 78

How will pd be shared across 2 resistors in a series circuit?

Answer 78

• If we had a simple circuit with a source of emf and a couple of resistors in series we would find that the pd is shared between the 2 components.
• we could make one resistor into a thermistor and see how the pd is shared between R1 and R2.
• This is an example of a potential divider circuit.
• The point of a circuit like this is that we can use it to look at the pd (Vout) that we get from the circuit.

Question 79

How can we calculate Vout of a circuit consisting of a thermistor and fixed resistor in a series circuit?

Answer 79

• The total resistance Rt = R1 + R2
• R1 is going to be a fixed value but R2 is going to change depending on the temperature.
• We can look at the current I, that flows in the circuit and because of kirchhoffs first law we know the current is going to be equal across the circuit.
• If we wanted to work out the value of the current we could use I = Vin/Rt where Vin is the emf.
• If we consider the current that flows through R2 we could also say the current I = Vout/R2.
• If we know the current is the same everywhere in the circuit we can say Vin/Rt = Vout/R2
• If we substitute Rt = R1 + R2 into the equation we get Vin/(R1 + R2) = Vout/R2 which if we know the values of R1 and R2 and the supplying voltage we can write this as Vout = Vin(R2/(R1 + R2)) which is the ratio of R2 to R1.

Question 80

Why is work done in a circuit equal to itV?

Answer 80

• Power P = W/t and if we rearrange this W = Pt
• If we think about electric circuits we know the power P = IV and therefore we can substitute this making the equation W = IVt