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Electrical Work and Power

Electrical Work and Power. Electrical Work and Power. Resistance R. I. +. -. I. Lower V 2. Higher V 1. Current I flows through a potential difference D V . Follow a charge Q : at positive end, U 1 = QV 1 at negative end, U 2 = QV 2 P.E. Decreases:.

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Electrical Work and Power

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  1. Electrical Work and Power

  2. Electrical Work and Power Resistance R I + - I Lower V2 Higher V1 Current I flows through a potential differenceDV Follow a charge Q : at positive end, U1 = QV1 at negative end, U2 = QV2 P.E. Decreases: The speed of the charges is constant in the wires and resistor. What is electrical potential energy converted to?

  3. 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.

  4. Power dissipated by a resistor: Power dissipated= current x potential difference Units: 1 volt (=J/C) x 1 amp (=C/s) = 1 watt (= 1 J/s) For Resistor: V = IR, so there are 2 other equivalent formulas: This power dissipated is called “Joule heating” in a resistor

  5. Example 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?

  6. Example The tungsten filament of a light bulb dissipates 60.0W of power from a 120-volt source. If the filament is a wire 50.0 mm long and 0.250 mm in diameter, calculate the resistivity of tungsten at the operating temperature of the filament.

  7. Quiz Car batteries are usually rated in ampere-hours. This information designate the amount of:a) currentb) powerc) energyd) chargee) potential that the battery can supply

  8. “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)

  9. When current leaves the battery the battery supplies power equal to: I = 2 A P = Iε=24W ε=12 V R 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? Resistor: Electrical energy  heat

  10. Real Batteries I r = “internal resistance” of the battery A r I RL (external resistance, “load”) I B VB + -Ir=VA VA – VB = V = “terminal voltage”  measured “Terminal voltage” ε - Ir = V

  11. 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: • The current in the circuit and the terminal voltage • The power dissipated in the load, the internal resistance, and the total power delivered by the battery

  12. 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

  13. Automobile battery: 12.8 V (with 20 A current) 9.2 V (with 200 A current) Find: E and rinternal of battery Example At terminals

  14. Shocking questions • Why is it safer to touch wires with the back of your hand? • If you fall from a building and on your way down grab a high-voltage line, will you be electrocuted?

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