2.  Georg Simon Ohm  1789-1854  German  Physicist  HIGH SCHOOL TEACHER!!!!

3.  Using equipment he built himself  Determined there was a direct proportionality between voltage applied across a conductor and the resultant current (OHM’s LAW)

4.  Good conductors easily allow electrons to move through their atoms  In a good conductor little energy is lost as the electrons pass along the conductor’s atoms  In other materials such as a light bulb the electrons do not move as easily.  This ability of a substance to impede the flow of electrons is called electrical resistance

5.  In other materials such as a light bulb the electrons do not move as easily.  This ability of a substance to impede the flow of electrons is called electrical resistance  Different loads have different resistances to achieve the desired conversion of energy  e.g. : light bulb filament = high resistance (produces light)  toaster = lower resistance (produces heat)  The symbol for electrical resistance is R  The standard unit is (SI) is the Ohm (Ω)

6.  The symbol for electrical resistance is R  The standard unit (SI) is the Ohm (Ω)  When electrons flow through different materials, the electrical resistance causes a loss of voltage  There is a difference in the amount of voltage (electric potential) that an electron has after it has flowed through the material  This is known as potential difference or voltage drop  e.g. After electrons flow through a light bulb, they have less energy than before they went through it (some is used)

8.  Ohm’s Law  Ohm’s Law describes the relationship between potential difference and current in a conductor  Potential difference = electric current x Electrical Resistance  (voltage drop)

9. V = Voltage (Potential Difference) I = electric current (measured in Amps) R = electrical resistance (measured in Ω)

10.  V = IxR (units for V are Volts)  V/I = R or R = V/I (Units for R are Ω)  V/R = I or I = V/R (Units for I are Amps)  Sample problem:  What is the voltage drop across the tungsten filament in a 100W light bulb? The resistance of the filament is 144 Ω and the current is 0.833A.  V = ? I = 0.833A R = 144 Ω  V = I x R V = 0.833A x 144 Ω = 120V  Therefore the voltage drop at the light bulb is 120V