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Text sections 27.4, 27.6, 28.1 • Drude Model • Electrical work and power • “emf” Practice problems: Chapter 27, problems 17, 24, 27, 47, 49 Chapter 28, problems 1, 3 From before: I =nqAvd • We assumed that the drift velocity is proportional to the electric field, thus leading to Ohm’s Law, • vd is the average velocity of an electron in a metal (net displacement per second) and is small • Typical speeds are in km/s for electrons in metals 1 • Random velocities of electrons are large (several km/s) With no field, the electrons scatter randomly (due to impurities and ions in the metal), so on average they go nowhere. An electric field will cause the electron’s path to curve slightly between collisions, but always in the same direction, leading to a slow but steady net “drift”. end start E - end start end net displacement 2 Assumptions: 1) Electrons have large random velocities (no field). 2) Electrons collide with atoms (time ); is not affected by the field. 3) After a collision, velocity is random again. q Averages: 3 So… drift velocity: and (if |q| = e) So… Therefore… I + V1 Resistance R V2 I Current I flows through a potential difference ΔV Follow a charge Q : at positive end, U1 = QV1 at negative end, U2 = QV2 P.E. Decreases: Electrical energy is converted to other forms of energy. 4 Quiz The derivation assumed positive charges, and found that the potential energy decreases with time. If the current is actually carried by negtive charges, then: A) the potential energy will increase with time B) the potential energy will still decrease C) the potential energy will not change Quiz If potential energy is steadily lost, what happens with the kinetic energy of the moving electrons ? A) the average kinetic energy increases steadily B) the average kinetic energy decreases steadily C) none of the above 5 Electrical resistance converts electrical potential energy to thermal energy (heat), just as friction in mechanical systems converts mechanical energy to heat. The average kinetic energy of the electrons doesn’t increase once the current reaches a steady state; the electrons lose energy in collisions as fast as it is supplied by the field. Negative charge carriers would move in the opposite direction (from low V to higher V); but this still means they lose potential energy. P = work per unit time Power = current x (potential difference) Units: 1 volt x 1 amp = 1 watt (= 1 J/s) Eg. Resistor: ΔV = IR P = I2 R (rate of “Joule heating” in resistor) 6 1) What is the resistance of a “60 watt” bulb? (for a 120-V supply) 2) Find R for a 60-W headlamp (12-V battery). 3) What power do you get from a “60-W” household bulb if you connect it to a 12-V car battery? 4) How many 200-A, 750 kV lines are needed to carry the power from a 1,500,000 kW hydro dam? E ≡ external work per unit charge Units: J/C = volts (not actually a force) e.g.: Battery (chemical energy electrical energy) Generator (mechanical energy electrical energy) Solar cell, etc. 7 - Example: battery. I=2A Eg. What does the energy balance look like in this circuit? R 12 V What is the resistance R? - Example: battery Eg. 12 V + + I=2A R Battery: supplies power, P = IE = 24W (Chemical energy electrical energy) Resistor: Electrical energy heat, P = 24W=I2R ⇒ R = 6Ω 8 I A r C • I I r = “internal resistance” RL (external “load”) B VA – VB = “terminal voltage” measured VC – VB = E (but “C” is not accessible for measurement) And VA – VC = - I r “Terminal voltage” At terminals Find: Automobile battery: 12.8 V (with 20 A current) 9.2 V (with 200 A current) E and rinternal of battery What is the maximum current the battery can provide? 9 Notes: i) As I 0, Vload E ii) Imax = E /r (when RL = 0, a short circuit) Question: How would you select a load (e.g., lightbulb) to get the maximum power output, for a given battery? A famous theorem: Maximum power transfer is achieved when the load resistance matches the source resistance. Maximum load power when RL= r Proof: Set and solve for RL (exercise) 10 VL = E – rIL Max. power point VLoad ILoad Pmax PLoad r RLoad 11