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EMF

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E.m.f and p.d - Learning Outcomes You should all be able to: define potential difference (p.d.); select and use the equation W = VQ; define the volt; describe how a voltmeter may be used to determine the p.d. across a component; define electromotive force (e.m.f.) of a source such as a cell or a power supply; describe the difference between e.m.f. and p.d. in terms of energy transfer

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Electric Potential (a bit like Gravitational Potential) To give the ball more gravitational ‘potential’ energy work must be done on it. Now it is ‘higher’ it has more ‘potential’. This is a bit like electrical ‘potential’. The position of a charge (rather than in a ball) in an electric field (rather than a gravitational field) determines its potential.

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Electric Potential (a bit like Gravitational Potential) The direction of current flow is conventionally taken to be from points of higher potential (i.e. the top of a hill) to a point of lower potential. Note: Conventional current flows from +ve to –ve, but the electrons (that are –ve) flow to the positive terminal.

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If the work done in causing one coulomb of electric charge to flow between two points is one joule, then the PD between the points is one volt (i.e. 1V = 1 JC -1 ) It follows that W = QV Where W = the work done (J) Q = the charge (C) V = the potential difference (V)

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e.m.f is the energy transferred per unit charge when one type of energy is converted into electrical energy. Potential difference is the energy transferred per unit charge when electrical energy is transferred into another form of energy. Key Definitions

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Potential difference 12V What is the voltage across each resistor? 6V What happens to the charge as it flows through the resistor? Electrical energy is transformed into another energy (heat in this case)

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Resistance

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Resistance - Learning Outcomes You should all be able to: define resistance; select and use the equation for resistance (V = I.R) define the ohm; state and use Ohm’s law; describe the I–V characteristics of a resistor at constant temperature, filament lamp and light-emitting diode (LED); describe an experiment to obtain the I–V characteristics of a resistor at constant temperature, filament lamp and light-emitting diode (LED); describe the uses and benefits of using light emitting diodes (LEDs).

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AV

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AV

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AV bit of wire http://phet.colorado.edu/en/simulation/circuit-construction-kit-dc How to measure resistance

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Ohm’s Law Resistance = Potential Difference (volts) (ohms) Current (amperes) V is Potential Difference measured in voltage, V I is current measured in Amps, A R is resistance measured in Ohms, Ω R = V I V = I R

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Ohm’s Law The current through a resistor at a constant temperature is directly proportional to the potential difference across the resistor. This means if you double the current you double the voltage over a component. It also means that the resistance of the component does not change when you put more current through it.

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Candidates should be able to: define resistivity of a material; select and use the resistivity equation ; describe how the resistivities of metals and semiconductors are affected by temperature; describe how the resistance of a pure metal wire and of a negative temperature coefficient (NTC) thermistor is affected by temperature.

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Resistivity The resistivity ρ of a wire of length l, resistance R and cross sectional area A is given by: ρ = RA l

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Power

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Candidates should be able to: describe power as the rate of energy transfer; select and use power equations P = VI, P = I 2 R and P = V 2 / R explain how a fuse works as a safety device determine the correct fuse for an electrical device; select and use the equation W = IVt; define the kilowatt-hour (kW h) as a unit of energy; calculate energy in kW h and the cost of this energy when solving problems.

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Power Power (watts) = Energy Transformed (joules) Time (seconds) This means the more powerful something is, the more energy is transferred every second.

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Power Power (watts) = Energy Transformed (joules) Time (seconds) For example: If a bulb transforms 300 J of electrical energy into light in 3 second, the power is: P = Energy Transformed ÷Time P = 300 (J) ÷ 3 (s) P = 100 W

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Power in a circuit Power = Current x Potential Difference (watts, W) = (ampere, A) x (volts, V) P = I x V For example: If a bulb has a p.d. across it of 3.0V, and a current flowing through it of 2.0A then the power is: P = I x V P = 2.0 (A) x 3.0 (V) P = 6.0 W

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Charge in a circuit Charge = Current x Time (coulomb, C) = (ampere, A) x (seconds, s) For example: How much charge flows if a current of 2.0 A flows for 60 seconds? Charge = Current x Time Charge = 2.0 (A) x 60 (s) Charge = 120 C

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Energy in a circuit E nergy Transformed = Potential Difference × Charge (joules, J) = (volts, V) x (coulomb, C) For example: How much energy is transformed when the p.d. is 30V and the charge is 2.0 C? Energy = Potential Difference × Charge Energy = 30 (V) x 2.0 (C) Energy = 60 J

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SOUND WAVES Why do car headlights dim when you turn on the ignition? What has this got to do with internal resistance?

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Lesson Objectives To understand a method by which internal resistance may be calculated. To analyse and evaluate quantitative data. To develop your scientific approach to carrying out practical work.

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Task – To calculate the internal resistance of a power source Physics by Robert Hutchings (the white book) pages 250 – 251 may help. Use the sheet to help you get started. Think about your practical skills.

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Candidates should be able to: (a) state Kirchhoff’s second law and appreciate that this is a consequence of conservation of energy; (b) apply Kirchhoff’s first and second laws to circuits; (c) select and use the equation for the total resistance of two or more resistors in series; (d) select and use the equation for the total resistance of two or more resistors in parallel; (e) solve circuit problems involving series and parallel circuits with one or more sources of e.m.f.; (f) explain that all sources of e.m.f. have an internal resistance; (g) explain the meaning of the term terminal p.d.; (h) select and use the equations e.m.f. = I (R + r) = V + Ir.

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Kichhoff’s Second Law In any closed loop in a circuit the sum of the e.m.f.s is equal to the sum of the p.d.s.

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