Electrical Work and Power. II+- Higher V 1 Lower V 2 Resistance R Current I flows through a potential difference  V Follow a charge Q : at positive end,

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Presentation transcript:

Electrical Work and Power

II+- Higher V 1 Lower V 2 Resistance R Current I flows through a potential difference  V Follow a charge Q : at positive end, U 1 = QV 1 at negative end, U 2 = QV 2 P.E. Decreases: 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: 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: 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) Units: J/C = volts (not actually a force) but it “pushes” the charges through the circuit. Eg: Battery (chemical energy  electrical energy) Generator (mechanical energy  electrical energy) ε  external work per unit charge -+ I

ε=12 V R I = 2 A What does the energy balance look like in this circuit? When current leaves the battery the battery supplies power equal to: If current were forced to enter the battery, (as in charging it) then it absorbs the same power P = I ε = 24W Resistor: Electrical energy  heat

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

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: 12.8 V (with 20 A current) 9.2 V (with 200 A current) Find: E and r internal of battery At terminals

Resistance and Temperature Over a limited temperature range, the resistivity of a metal varies approximately linearly with T according to: ρ=ρ o [1+α(T-T o )] where T is in o C and ρ is the resisitivity. T o is usually taken to be 20 0 C and temperature coefficient is given by:

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 20.0 o C 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?