Week 26 (Feb. 17 - 21)

Reading: Current and resistance (Chap. 27)

Key Topics:
electric current, resistance, resistivity, ohm's law, joule heating
Week 26 Homework Problems:
  1. Electron accelerator: Suppose that two large parallel plates are connected to a 1 kilo-volt power supply so that the left plate is maintained at an electric potential of +500 volts and the right plate is maintained at an electric potential of -500 volts. The plates are separated by distance of 1 mm. (a) First, sketch the electric field in the region between the plates. (b) What is the strength of the electric field between the plates? (Answer: 1000 volts / 1 mm = 1,000,000 volts/meter) (c) If an electron is placed very near the +500 volt plate, what is the force exerted on the electron by the electric field? (Answer: the force is just the electric field times the electron charge. This gives 1.6 e -13 Newtons) (d) How much work does it take to move the electron from the +500 volt plate over to the -500 volt plate? (Answer: this is the force times the distance. It is 1.6 e-16 joules) (e) If the electron, placed near the -500 volt plate is released, what will be its acceleration toward the +500 volt plate? (Answer: the acceleration is the force divided by the electron mass. it is a =1.76e17 meters/second^2) (f) what will be the electron's speed just before striking the +500 volt plate (Answer: using conservation of energy, we can set the work equal to the change in kinetic energy. This gives v =1.88 e 7 m/s )?
  2. Power lines: High-voltage power lines deliver electrical current to your neighborhood. The aluminum wire used for these lines has a cross sectional area of about 5 square centimeters. What is the resistance of ten kilometers of this wire? (Answer: resistivity of aluminum * length / area = 0.52 ohms ) Hint: you will need to look up the resistivity of aluminum.
  3. Car battery: A car battery has a rating of 220 ampere-hours. This rating is an indication of the total charge that the battery can provide to a circuit before failing. What is the total charge (in coulombs) that this battery stores? (Answer: an ampere is a coulomb per second. So 220 ampere-hours * 60 minutes/hour * 60 seconds/minute = 792,000 coulombs). What is the maximum current that the battery can provide for 38 minutes (Answer: 792,000 coulombs / 38 minutes = 347 amperes).
  4. Mystery resistor: The current in a series circuit is 15.0 amperes. When an additional 8.00 ohm resistor is inserted in series, the current drops to 12.0 amperes. What is the resistance of the original circuit? (Answer: Since the voltage didn't change, the product of the current and the resistance remains constant. Thus, 15 amps * R = 12 ams * (R + 8). Solving for R gives R = 32 ohms)
General College Physics