510 likes | 608 Views
Topic 4: Oscillations and Waves. TEST Friday 12 th March. Topic 5: Electric currents. Can you look through the contents and definitions? Definition TEST Friday 19 th March. Stand up!. ☺. ☺. ☺. ☺. ☺. ☺. ☺. ☺. ☺. ☺. ☺. ☺. ☺. ☺. ☺. ☺. ☺. ☺. ☺. ☺. ☺. ☺. cell. energy.
E N D
Topic 4: Oscillations and Waves TEST Friday 12th March
Topic 5: Electric currents Can you look through the contents and definitions? Definition TEST Friday 19th March
☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺
☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ 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!) ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺
A ☺ ☺ ☺ ☺ ☺ I’m counting how many coulombs of electrons go past me every second Current ☺ ☺ The rate of flow of electric charge (number of Coulombs flowing past a point in the circuit every second). I = Q/t ☺ ☺ ☺ ☺ 1 Amp = 1 coulomb per second
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
☺ ☺ ☺ V ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ Voltage(emf) I’m checking the difference in energy (per coulomb) between the 2 red arrows 1 Volt = 1 Joule per coulomb
☺ ☺ ☺ V ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ Voltage (p.d.) I’m checking the difference in energy (per coulomb) before and after the lamp 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).
In a series circuit The sum of thep.d.s across the lamps equals the emf across the cells 9 V 3 V 3 V 3 V
In a parallel circuit In a simple parallel circuit, p.d. across each lamp equals the e.m.f. across the cells 5 V 5 V 5 V
V A Resistance Measures how difficult it is for current to flow. Measured in Ohms (Ω) Resistance = voltage/current R = V/I
V I R X Ohm’s Law • V = IR
Resistance • R is proportional to the length of wire – WHY? R α L • 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 10-8Ωm. 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 10-8Ωm. 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
Homework • BOTH electricity practicals (resistance of different thicknesses of wire AND voltage-current characteristics of filament lamps) to be handed in Wednesday 14th April.
A V Resistance of a lamp Vary the voltage and current using a variable resistor (rheostat). Plot a graph of resistance against current Resistance = voltage/current R = V/I
Resistance of a lamp • As the current in a lamp increases, its resistanceincreases. 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
V A Power The amount of energy used by a device per second, measured in Watts (Joules per second) 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
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!