Georg Simon Ohm 1789-1854 German Physicist

HIGH SCHOOL TEACHER!!!!

Using equipment he built himself Determined there was a direct

proportionality between voltage applied across a conductor and the resultant current (OHM’s LAW)

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

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)

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)

e.g. After electrons flow through a light bulb, they have less energy than before they went through it (some is used)

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)

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

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