Tuesday, March 2, 1999
Announcements: Midterm 2 scheduled for March 18, 1999

Lecture notes:

•  R= r L/A
• For metals: r =r_o [ 1+ a(T- T_o)],  a > 0
• 
• For Semiconductors such as germanium and silica:
• 

• b to c and d to a the circuit is ideal: There is no resistance
• c to d is a resistor
• Let's follow a quantity of positive charge DQ that flows around the circuit.
• As DQ moves from a to b (through the battery), its electrical potential energy increases by an amount [VDQ ]
• The chemical potential energy in the battery decreases by the same amount.
• As DQ  moves from c to d, through the resistor, the charge carried suffers "collisions" with the (misplaced) atoms in the the resistor and loses energy, by the same amount [VDQ ]
• The energy is disipated as heat or thermal energy
• By the time DQ  reaches point a, it is at the same electric potential again.
• DU/ Dt = (DQ /Dt) V = IV = P
• P is the power lost in the resistor; conversely the power supplied by the battery
• Let's check the units for P...
• Units of I = coul/sec
• Units of V= Joul/ Coul
• Therefore the units of IV= the units of power = Joul/ sec = Watt
• P =IV if the resistor is ohmic
• then, if V= IR.... P= I²R  or P = (V/R)V = V²/ R
• Question: What is the resistance of a light bulb rated at 50 Watts? The effective voltage in a typical household circuit is 110V.
• P = V²/ R = 50 watts so  V²/ P = R = 242W
• What is the resistance of a bulb of 100 Watts?
• P = V²/ R = 100 watts so  V²/ P = R = 121W
• Therefore, the higher the wattage, the lower the resistance

• Let us consider the simple circuit again....
• 
• The battery acts as a pump that moves positive charges from lower to higher potential.
• By convention, any device that can increase the potentail energy of positive charges is called an "emf" device
• There are many different kinds of emf devices; i.e. batteries, solar cells, etc
• 
• Va + x - iR = Va  so that x - iR = 0 or x = iR
• x is emf.