Physics 05: Electric Currents BY HEI MAN KWOK 12N03S 5.1 ELECTRIC POTENTIAL DIFFERENCE, CURRENT AND RESISTANCE Model of electric conduction in a metal – energy transfer 1. 2. 3. 4. 5. Charge carriers have kinetic energy These collide with lattice ions Increasing amplitude of vibrations This is seen in an increase in temperature Electrons loses it energy and transfers it to thermal energy Speed of electron through a conductor = drift speed Charge (Q) • Measured in Coulombs (C) • Charge of 1 electron = 1.6 π₯ 10−19 πΆ • 1 Coulomb = 1 1.6π₯ 10−19 = 6.2 x 1018 electrons Electric Potential Difference • Energy taken per unit charge πΈ (ππππππ¦) π(π£πππ‘πππ) = π (πΆβππππ) 1 volt = wd per unit charge, 1 joule per coulomb Change in potential energy when a charge moves between two points at different potentials • Potential energy (J) = Voltage (V) x Charge (C) πΈπ = ππ Insert diagram ο Electronvolt (eV) • Amount of energy (work done) one electron would gain by moving through a potential difference of 1 volt • eV = 1.6 π₯ 10−19 π½ • Charge of a electron = −1.6 π₯ 10−19 πΆ Electric Current (in a conductor) The rate flow of electric charge (from positive to negative potential) π (πΆβππππ) πΌ ππ’πππππ‘ = π‘ (π‘πππ π‘ππππ ππ π ππππππ ) Ampere = force per unit length between parallel current-carrying conductors why? 2x 10^7 N Resistance • Ratio of Voltage to Current • How easily does current flow through π (πππ‘πππ‘πππ ππππππππππ) π π ππ ππ π‘ππππ = πΌ (ππ’πππππ‘) Resistor = a component with a known resistance Resistivity πΏ (πππππ‘β) π (πππ ππ π‘ππππ) = π π΄ (ππππ π − π πππ‘πππππ ππππ) π = πππ ππ π‘ππ£ππ‘π¦ ππ π‘βπ πππ‘πππππ More current flows through a short fat conductor than a long thin one Ohm’s Law • The ratio of potential difference to current are proportional at constant temperature • V = IR • Eg. Wires and fixed resistors Non-Ohmic – Filament lamp • As current increases, temperature increases, atoms vibrate more, collisions between electrons and metal atoms are more frequent so resistance increases, and graph flattens • The current is not directly proportional to the voltage – disobeys ohm’s law • Graph = symmetrical Diode • One way value for electrons • No current when negative V • Positive = current flows easily as diode has low resistance above about 0.7V • LEDs also one way :D Component’s Potential Difference Cannot be equal to the p.d. of the battery: • Some voltage will be lost to internal resistance and/ or the resistance in wires Cannot equal to zero • Low voltage requires very high resistance, max. resistance of the variable resistor cannot be infinite, there will always be some resistance from the component 5.2 ELECTRIC CIRCUITS Electromotive Force (emf) • Work done per unit charge made available by the energy source (cell or battery) • Power supplied by the cell per unit current from the cell Law of Conservation of Energy • Energy cannot be created or destroyed – the energy converted from chemical to electrical in the cell must be equal to the amount converted from electrical to heat in the resistor πΉπππ πβπ′ π πΏππ€: π = πΌπ Internal Resistance • The resistance of the cell is ^ π = πΌπ + πΌπ • As current flows through the internal resistance – some energy is converted from electrical to heat inside the cell (so the cell gets hot) • This means that there is less energy to be converted to heat in the resistor – p.d. across the resistor is therefore less than the emf of the cell Finding Internal Resistance Experimentally • V = E – Ir • By recording values of current and terminal pd as the external resistance changes you can plot the graph and find the internal resistance and the emf of the cell. • If there is more than one cell in series the internal resistances of the cells must be added. Power Dissipated π = πΌπ π = πΌ2 π π2 π= π P = Power; I = Current; V = Voltage; R = Resistance Power Delivered π = ππΌ In a real battery – actual power delivered will be a bit less, since there will be some power dissipated in the internal resistance V, I, R in Series ππππ‘πππ = π1 + π2 + π3 πΆπ’πππππ‘ ππ π πππ π ππ ππ π‘ππππ = π 1 + π 2 + π 3 V, I, R in Parallel πΆπ’πππππ‘ = πΌ1 + πΌ2 + πΌ3 ππππ‘πππ ππ π πππ 1 1 1 1 = + + π ππ ππ π‘ππππ π 1 π 2 π 3 Ideal Ammeter • Zero resistance so it does not change the current in the circuit • Connected in series so the current will flow through the ammeter as it flows through a component Ideal Voltmeter • Infinite resistance so it does not take any current from the circuit • Connected in parallel to see the difference in potential energy between two points Thermistor and LDR • Made of semi-conducting material • Heat and light frees more charge carriers: as the temperature/ light intensity increases, the resistance decreases • The current is not directly proportional to the voltage LDR Thermistor Strain Gauge • Think metal wire • If stretch – length increases and crosssectional area decreases = increase in resistance Potential Divider πππ’π‘ π 2 = πππ π 1 + π 2