1. As Unit 1 - Electricity What you need to know.

2.  Current is the rate of flow of charged particles.  In metals these are conduction electrons, most electrons are attached to atoms but some are not. When a voltage is applied these electrons are attracted to the +ve terminal.  When an electric current is passed through a salt solution the charge is carried by ions.

3.  Unit for Current = Ampere (A) Symbol I = Coulomb per second = Coulomb per second  Unit for Charge = Coulomb (C) Symbol Q  (The charge flow in 1 sec when the current is 1A) Charge on an electron = 1.6 x 10 -19 C

4. Conventional Current +ve to -ve This convention is kept even though we now know that the charge carriers (electrons) are actually flowing in the opposite direction.

5.  Insulators - each electron is attached to an atom and cannot move even when a voltage is applied.  Semi Conductor - number of charge carriers increases with temperature. Resistance therefore decreases as temperature increases.

6. Potential Difference and Power  Electrons are supplied with Electrical Potential Energy (EPE) by a battery.  Electrons do work to pass through a component, using some or all of its EPE.  The work done per unit charge is defined as the POTENTIAL DIFFERENCE or p.d.

7. EMF  EMF ( ε) of a source is defined as the electrical energy produced per unit charge passing through the source. The unit is also the volt. Electrical energy produced when charge Q passes through = W = Qε(Joules) This energy can be transferred in components with resistance, as heat energy ( electric heater), due to charge carriers colliding with atoms, causing them to vibrate and heat up.

8. Electric Power  Current I, time Δt, p.d across component V  Charge flowing through a device = Q = IΔt  Work done (Energy) = QV = IΔt V  Power = Energy = IV so P=IV time time  Unit = Watts or Joules/Second (W or Js -1 )  Energy Used = Power x Time

9. Resistance  Ohms law states that the p.d. across a metallic conductor is proportional to the current through it provided the physical conditions do not change.  R = V/I = p.d across the component the current through it  A resistor is a component that is designed to have a certain resistance regardless of the current through it.

10. Measuring Resistance Plot a graph of p.d. y axis vs current x axis. Line of best fit. Gradient = R

11. Resistivity

12. Super Conductors: used for power cables and electromagnets  A material that has zero resistivity, therefore resistance, at and below a critical temperature, that depends on the material.  So when a current passes through there is no p.d and no heating effect. Highest Tc 150K (-123 o C) Tc above 77K (-196 o C) are called high temperature superconductors.

13.  Resistance and Temperature  Metals have a positive temperature coefficient as its resistance increases with temperature, due to the vibration of the metal lattice (+ve ions).  Semiconductors have a negative temperature coefficient as its resistance decreases with temperature, due to the increase of charge carriers.

14. I is constant (same everywhere). Component voltages adds up to supply V 0 = V 1 + V 2 + V 3 so IR T = IR 1 + IR 2 + IR 3 R T = R 1 + R 2 + R 3 Total resistance

15. Voltage is constant V 0 = V 1 = V 2 Current Splits so I 0 = I 1 + I 2 (as I=V/R) then V 0 /R T = V 1 /R 1 + V 2 /R 2 divide through by V Gives 1/R T = 1/R 1 + 1/R 2

16. Using Resistance to Heat Power supplied = IV and pd V= IR Gives P = I 2 R = V 2 /R Energy = Power x Time If the component heats up its temperature rise depends on the power supplied and heat transfer to the surroundings.

17. Emf and Internal Resistance

18.  The pd across the terminals, V, is the electrical energy produced per unit charge by the source when it is in a circuit with resistance R.  This will be less than the Emf of the source (unless otherwise stated) due to the internal resistance (r) of the source. R T = R + r and V = IR

19. Power = IV

20. Measuring Internal resistance Measure I and V Plot V vs I When I = 0

21. y = mx + c V vs I means that gradient = -r Y intercept =

22. More Complicated Circuits 1. Sketch the circuit if not drawn 2. Calculate cell current. Cell current = cell emf total circuit resistance (inc internal) total circuit resistance (inc internal) 3. Work out current and so pd across resistors in series with cell first (I is constant) using V = IR 4. Work out current through any parallel resistors Current through each resistor = pd across parallel resistance resistance

23. Cells in series  Cells in same direction: total emf is sum of individual emfs.  Cells connected oppositely net emf is the difference between the individual emfs.  Total internal resistances are the sum of the individual internal resistances regardless.

24. Cells in parallel  Current through each = I/n  pd lost in each = (I/n) x r = Ir/n

25. The Potential Divider Uses:  To supply a variable pd from 0 to V 0  To supply a pd that varies with physical condition; light, temperature, pressure etc.

26. The ratio of the pds across each resistor is equal to the the ratio of the resistances  Volume or light control:  An audio signal is supplied to the potential divider, the variable output from the potential divider is supplied to the loudspeaker.

27. Temperature Sensor  At constant temperature the pd is shared between the variable resisitor and thermistor.  By adjusting the variable resistor the pd across the thermistor can be set at any desired value.  When the temperature changes the reistance will change so then the pd across it will change.  If the temperature rises the resistance of the thermistor will drop, and the pd will drop. Light Sensor  When light intensity increases the resistance of the LDR falls and so the pd across it falls.

29. Alternating Current and Power An alternating current is one in which the current repeatedly reverses its direction over time. (Sinusoidal)  Frequency (f) is the number of cycles per second. (Hz or S -1 )  Mains = 50Hz (each cycle takes 0.020s)

30.  Peak value is the maximum pd V 0 or current I 0, and depends on the peak pd or current of the source and the components within the circuit.  In mains circuits this peak value is 325V

31. Alternating Current and Power P = I 2 R when I is negative, I 2 will still be positive

32. At Peak current I 0 maximum power is I 0 2 R At zero current, zero power is supplied. Power varies from 0 to I 0 2 R. So the mean Power is 1/2 x I 0 2 R The direct current that would give the same power as the mean power is called the ROOT MEAN SQUARE value of the alternating current = I rms

34. Using an Oscilloscope  An oscilloscope consists of an electron tube and control circuits.  An electron gun fires electrons towards a screen that fluoresces when electrons hit it.  A pair of electric plates can be used to deflect the beam horizontally and vertically.  The displacement is proportional to the applied p.d. http://www.virtual-oscilloscope.com/simulation.html#

35. X-Scale (time base)  Vertical plates move the beam in the x direction.  These are connected to the TIME BASE CIRCUIT moving the spot at a constant speed left to right.  The x-scale can be calibrated usually in milli or micro seconds per cm. Y-Scale (y-sensitivity/gain)  Horizontal plates move the beam in the y direction.  The p.d. to be displaced is connected to these plates, so the spot moves up and down, and if the beam is moving left to right, a wave form will be created.  The Y-input is calibrated in volts per cm Vcm-1

36. To measure the peak p.d. measure height from the bottom to the top of the wave. The amplitude (peak value is half of this). If the y-gain is set to 0.5Vcm-1, then what is the peak value from this trace? = 1.25V

37. To measure the frequency of the alternating p.d. measure the time period (T) for one cycle. The frequency f = 1/T If the time base is set to 2mscm -1. What is the time period? = 10ms or 0.01s What is the frequency? = 100Hz To measure the time period accurately, measure over a number of cycles and then divide by the number of cycles.

38. Measuring D.C. Voltage  A D.C. input voltage will make the spot move at a constant displacement from the zero p.d. line.  Knowing the y-gain the D.C. voltage can be calculated and can be either positive or negative. 3.3cm p.d = 3.3 x 0.1 = +0.33V 2.4cm p.d = 2.4 x 0.1 = - 0.24V