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Topic 4: Oscillations and Waves TEST th 12 Friday March Topic 5: Electric currents Can you look through the contents and definitions? Definition TEST Friday 19th March Stand up! ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ cell ☺ ☺ ☺ energy ☺ electron ☺ ☺ ☺ ☺ lamp ☺ ☺ ☺ Electrons Hi, I’m Eleanor the electron. ☺ Coulomb of charge (electrons) Think of it as a “bag of electrons” (containing 6000000000000000000 electrons!) ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺☺ ☺ ☺ ☺ ☺ ☺☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ Current ☺ ☺ ☺ ☺ ☺ ☺ The rate of flow of electric charge (number of Coulombs flowing past a point in the circuit every second). I = Q/t ☺ ☺ I’m counting how many coulombs of electrons go past me every second ☺ A ☺ ☺ 1 Amp = 1 coulomb per second Let’s build some circuits! ☺ In a series circuit Current is the same at any point in the circuit 2.5 A 2.5 A 2.5 A 2.5 A In a parallel circuit The current splits (total current stays the same) 2.5 A 2.5 A 1.25 A 1.25 A Voltage(emf) ☺ ☺ ☺ V ☺ ☺ ☺ I’m checking the difference in energy (per coulomb) between the 2 red arrows ☺ ☺ ☺ ☺ ☺ 1 Volt = 1 Joule per coulomb Voltage (p.d.) ☺ ☺ I’m checking the difference in energy (per coulomb) before and after the lamp ☺ ☺ ☺ ☺ V ☺ ☺ ☺ ☺ ☺ 1 Volt = 1 Joule per coulomb p.d. and e.m.f Electric potential difference between two points is the work done per unit charge to move a small positive charge between two points. Electromotive force is the total energy difference per unit charge around the circuit (it is the potential difference when no current flows in a circuit). Let’s build some circuits! ☺ In a series circuit The sum of the p.d.s across the lamps equals the emf across the cells 9V 3V 3V 3V In a parallel circuit In a simple parallel circuit, p.d. across each lamp equals the e.m.f. across the cells 5V 5V 5V Stand up! Resistance Measures how difficult it is for current to flow. Measured in Ohms (Ω) V Resistance = voltage/current A R = V/I Ohm’s Law • V = IR V I X R Let’s do a practical! Resistance • R is proportional to the length of wire – WHY? RαL • R is inversely proportional to the cross sectional area of wire – WHY? R α 1/A • R depends on the type of material – WHY? Resistivity R = ρL A where R = resistance in Ohms L = Length of conductor in metres A = cross sectional area of conductor in m2 ρ = resistivity of the material in Ohms.meters Example The resistivity of copper is 1.7 x 10-8 Ωm. What is the resistance of a piece of copper wire 1 m in length with a diameter of 0.1mm? Example The resistivity of copper is 1.7 x 10-8 Ωm. What is the resistance of a piece of copper wire 1 m in length with a diameter of 0.1mm? radius = 0.05mm = 5 x 10-5m cross sectional area = πr2 = 3.14x(5 x 10-5)2 = 7 x 10-9 m2 R = ρL/A = (1.7 x 10-8 x 1)/ 7 x 10-9 = 2.42 Ω Let’s do another practical! Homework • BOTH electricity practicals (resistance of different thicknesses of wire AND voltagecurrent characteristics of filament lamps) to be handed in Wednesday 14th April. Resistance of a lamp Vary the voltage and current using a variable resistor (rheostat). Plot a graph of resistance against current V Resistance = voltage/current A R = V/I Resistance of a lamp • As the current in a lamp increases, its resistance increases. Why? Ohmic behaviour • p.d. is proportional to the current Metal wires at constant temperature Non-Ohmic behaviour • p.d. is not proportional to the current Power The amount of energy used by a device per second, measured in Watts (Joules per second) A V Power = voltage x current P = VI Power dissipated in a resistor/lamp • P = VI • From Ohm’s law, V = IR • So P = VI = I2R • From Ohm’s law also, I = V/R • So P = VI = V2/R Total energy So the total energy transformed by a lamp is the power (J/s) times the time the lamp is on for in seconds, E = VIt E = energy transformed (J) V = Voltage (also called p.d.) I = current (A) t = time (s) Electronvolt • Electronvolt – the energy gained by an electron when it moves through a potential difference of one volt. Questions! • Page 316 and 317 questions 2, 5, 8, 9, 10, 12, 13, 14, 15, 17, 18. Internal resistance • Connecting a voltmeter (VERY high resistance) across the terminals of a cell measures the EMF of the cell (no current flowing) V Internal resistance • We have assumed so far that the power source has no resistance…….not a good assumption! Internal resistance • In actuality the p.d. across a cell is less than the EMF due to energy lost in the INTERNAL RESISTANCE Internal resistance • To help us visualize this, a cell is represented as a “perfect” cell attached in series to the internal resistance, given the symbol r. Internal resistance • The p.d. across a cell (V) is then equal to the EMF (ε)minus the voltage lost across the internal resistance (=Ir) V = ε - Ir Example • A cell of emf 12V and internal resistance 1.5 Ω produces a current of 3A. What is the p.d. across the cell terminals? • V = ε - Ir • V = 12 – 3x1.5 • V = 7.5 V Another practical! Adding resistances In series and parallel Reading and taking notes • Pages 320 to 328 • Read and make your OWN NOTES. • I will collect these in the first lesson back after the holiday. Ideal meters • Voltmeters – infinite resistance! • Ammeters – Zero resistance! Potential divider LDRs, Thermistors in potential divider circuits Simple! Let’s try some IB questions!