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Electrical Work and Power Electrical Work and Power I + Higher V1 Resistance R I Lower V2 Current I flows through a potential difference DV Follow a charge Q : at positive end, U1 = QV1 at negative end, U2 = QV2 P.E. Decreases: DU QDV 0 The speed of the charges is constant in the wires and resistor. What is electrical potential energy converted to? Electrical resistance converts electrical potential energy to thermal energy (heat), just as friction in mechanical systems converts mechanical energy to heat. This thermal energy means the atoms in the conductor move faster and so the conductor gets hotter. The average kinetic energy of the electrons doesn’t increase once the current reaches a steady state; the electrons lose energy in collisions with the atoms as fast as it is supplied by the field. Power dissipated by a resistor: DU (DQ)V P IV Dt Dt Power dissipated= current x potential difference Units: 1 volt x 1 amp = 1 watt (= 1 J/s) For Resistor: V = IR, so there are 2 other equivalent formulas: 2 V P VI I R R 2 This power dissipated is called “Joule heating” in a resistor Examples a) What is the resistance of a “60 watt” bulb? (for a 120-V supply) b) Find R for a 60-W headlamp (12-V battery). c) What power do you get from a “60-W” household bulb if you connect it to a 12-V car battery? “Electromotive force ε ” (emf) ε external work per unit charge Units: J/C = volts (not actually a force) but it “pushes” the charges through the circuit. + I Eg: Battery (chemical energy electrical energy) Generator (mechanical energy electrical energy) I=2A ε=12 V R When current leaves the battery the battery supplies power equal to: P = Iε = 24W If current were forced to enter the battery, (as in charging it) then it absorbs the same power What does the energy balance look like in this circuit? 12V R 6 so Joule heating=I R 24W 2A 2 Resistor: Electrical energy heat Real Batteries I A r VB + r = “internal resistance” of the battery I I RL (external resistance, “load”) B -Ir=V A VA – VB = V = “terminal voltage” measured Ir (V V ) A ε - Ir = V B “Terminal voltage” Example A battery has an emf of 12V and an internal resistance of 0.05Ω. Its terminals are connected to a load resistance of 3Ω. Find: a) The current in the circuit and the terminal voltage b) The power dissipated in the load, the internal resistance, and the total power delivered by the battery Example Show that the maximum power lost in the load resistance R occurs when R=r, that is, when the load resistance matches the internal resistance of the battery. Example Automobile battery: At terminals Find: 12.8 V (with 20 A current) 9.2 V (with 200 A current) E and rinternal of battery Resistance and Temperature Over a limited temperature range, the resistivity of a metal varies approximately linearly with T according to: ρ=ρo[1+α(T-To)] where T is in oC and ρ is the resisitivity. To is usually taken to be 200C and temperature coefficient is given by: 1 D o DT where D o , DT T To Example A resistance thermometer measures temperature by measuring the chance in resistance of a conductor. Made of platinum with a resistance of 50.0Ω at 20.0oC is immersed in melting indium and its resistance increases to 76.8Ω. Find the melting point of indium. Example What is the fractional change in the resistance of an iron filament when its temperature changes from 25.0°C to 50.0°C?