Electromotive force

Learning Objectives

• (a) recall and use appropriate circuit symbols as set out in SI Units, Signs, Symbols and Abbreviations (ASE, 1981) and Signs, Symbols and Systematics (ASE, 1995);

• (a) define potential difference (p.d.);• (b) select and use the equation W = VQ;• (c) define the volt;• (d) describe how a voltmeter may be used to• determine the p.d. across a component;• (e) define electromotive force (e.m.f.) of a source• such as a cell or a power supply;• (f) describe the difference between e.m.f. and• p.d. in terms of energy transfer.

What is the point of a circuit?

• Circuit are there to deliver electrical energy to a device.

• At GCSE you learnt that Power = Energy / time• (you may not remember it but you did!)

Definition time

• Electromotive force is the energy transferred per unit charge when one type of energy is converted into electrical energy

• This makes more sense with examples: a standard cell works by turning chemical energy into electrical energy so is producing an emf

• A dynamo turns kinetic energy into electrical energy so is producing an emf

Calculation time

• Electromotive force is the energy transferred per unit charge when one type of energy is converted into electrical energy

• Energy per unit charge is the same as energy divided by charge so

• Electomotive force = Electrical energy transferred (e.m.f.) charge

Units time

• Electomotive force = Electrical energy transferred (e.m.f.) charge

• Energy is measured in Joules (J)• Charge is measured in Coulombs (C)

• So e.m.f is measured in JC-1

• So e.m.f is measured in JC-1

• But this is given it’s own unit called the Volt

• Definition• 1 volt is 1 Joule per Coulomb

What does this mean?

• So a 1.5V cell will provide each Coulomb of charge with 1.5J

• If you needed 6V to make a device work . Each 1.5V cell would provide 1.5J to every Coulomb so 4 of the 1.5V cells would provide 1.5J each to every Coulomb meaning that every Coulomb was carrying a total of 6J of electrical energy.

How about UK mains

• Well the e.m.f for UK mains is 230V so this means that every Coulomb is carrying 230J of energy with it (which is why it is much more dangerous)

This will annoy you!

• In equations to do with electricity we give e.m.f the letter E

• Therefore we need to give energy a letter so we give it W

Potential Difference

• For this part we are going to assume that all of the electical energy made is used. For most devices this is true but in A2 Capacitors are introduced which store some of the electrical energy. Let’s not worry about them for now!

Potential difference

• Definition time• Potential difference is the electrical energy per

unit charge when electrical energy is converted to another form of energy

• (Like our definition for e.m.f. but in reverse)

So imagine……

• A battery of cells has an e.m.f of 6V. It is connected to one bulb. Therefore each coulomb is supplied with 6J of energy. All of that energy is used in the bulb to light up and make heat etc so it has a potential difference of 6V

The Voltmeter

• P.d is what you have referred to as voltage up to now.

• A voltmeter measures p.d. by measuring the difference in electrical potentials

• (think of this as comparing the energy per coulomb before entering a device and then after leaving a device)

• Voltmeters are connected in parallel (i.e. across the component)

Equations can be mixed and matched

• 1 ampere= 1 coulomb per second• That means I= Q/t so Q = It• 1 volt = 1 joule per coulomb• That means V=W/Q so W=QV• 1 watt = 1 joule per second• That means P=W/t so W=Pt

We can combine these

• Q=It; W= QV; W=Pt

• So W = VIt• P=W/t = QV/t = VIt/t = VI

• a) define resistance;• (b) select and use the equation for resistance R= V/I• (c) define the ohm;• (d) state and use Ohm’s law;• (e) describe the I–V characteristics of a resistor at constant

temperature, filament lamp and light-emitting diode (LED);• (f) describe an experiment to obtain the I–V characteristics

of a resistor at constant temperature, filament lamp and light-emitting diode (LED);

• (g) describe the uses and benefits of using light emitting diodes (LEDs).

Resistance and Ohm’s Law

• Definition• The current through a conductor is

proportional to the potential difference across it provided physical conditions, such as temperature remain constant

Resistance

• Resistance = Potential difference/current

• Therefore it is measured in volts per ampere which is known as the ohm

Resistance

• A resistor of resistance 4.6Ω needs a p.d of 4.6V to allow the current to be 1 ampere

The filament lamp

• As the p.d. increases the lamp becomes hot.• This in turn increases the lamp’s resistance

The diode and therefore the LED

• Diodes allow current to pass only in one direction• LEDs do the same but emit visible light when a

current goes through them• Advantages of an LED are;• they switch on instantly• Are very robust• Are very versatile• Operate on low p.ds• Have a long working life