# Ohms Law Physics Dr. Robert MacKay. Voltage (Volts) Electrical Pressure + - + - V.

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Ohms Law Physics Dr. Robert MacKay

Voltage (Volts) Electrical Pressure + - + - V

Current (Amps) Charge Flow + + + - + + ++ ++ ++ I V

Resistance, R (Ohms, ) Impedence - - + - - - -- -- -- I V R Depends on Temperature Material of wire Area of wire Length of wire

Resistance, R (Ohms, ) Impedence - - - - -- -- -- R Depends on: * Temperature * Material of wire * Cros sectional Area * Length Area Length

Resistance and Resistivity = resistivity R=resistance

Ohms Law V= I R - - + - - - -- -- -- I V

ExerciseExercise Find the electric voltage required to have 10 Amps of current flow through a 2 resistor. Given: I= 10 A and R=2 Wanted: V Solution: V=I R V= (10 A) (2 ) = 20 Volts

ExerciseExercise Find the electric current flowing in a 10 resistor (light bulb) when connected to a 20 Volt battery. Given: V= 20 Vand R=10 Wanted: I Solution: V=I R or I=V / R I= (20 V) / (10 ) = 2 Amps

ExerciseExercise When a 60 Volt battery is connected to a circuit 4.0 amps of current flow from the battery. What is the circuits resistance. Given: V= 60 V and I=4.0 Amps Wanted: R Solution: V=I R or R=V / I I= (60 V) / (4.0 Amps) = 15

OFF DC AC V A 20M 200 k 200 2 k 2 200 m 20 m 200 µ 200 m 2 20 200 A V- COM 20 k 2 M 2m.995

OFF DC AC V A 20M 200 k 200 2 k 2 200 m 20 m 200 µ 200 m 2 20 200 A V- COM 20 k 2 M 2m 1.

OFF DC AC V A 20M 200 k 200 2 k 2 200 m 20 m 200 µ 200 m 2 20 200 A V- COM 20 k 2 M 2m 2.41

OFF DC AC V A 20M 200 k 200 2 k 2 200 m 20 m 200 µ 200 m 2 20 200 A V- COM 20 k 2 M 2m 122.

OFF DC AC V A 20M 200 k 200 2 k 2 200 m 20 m 200 µ 200 m 2 20 200 A V- COM 20 k 2 M 2m 0.082

OFF DC AC V A 20M 200 k 200 2 k 2 200 m 20 m 200 µ 200 m 2 20 200 A V- COM 20 k 2 M 2m 65

OFF DC AC V A 20M 200 k 200 2 k 2 200 m 20 m 200 µ 200 m 2 20 200 A V- COM 20 k 2 M 2m 65

OFF DC AC V A 20M 200 k 200 2 k 2 200 m 20 m 200 µ 200 m 2 20 200 A V- COM 20 k 2 M 2m 156

OFF DC AC V A 20M 200 k 200 2 k 2 200 m 20 m 200 µ 200 m 2 20 200 A V- COM 20 k 2 M 2m 156

Electric Power (Watts) Power = Current x Voltage 1 Watt = Amp x Volt

Curcuits R=10 A V= 20 V I =? P= ? R I V P 10 20 V

Curcuits R=10 A V= 20 V I =? P= ? R I V P 10 20 V 2.0 A V= I R or I = V / R

Curcuits R=10 A V= 20 V I =? P= ? R I V P 10 20 V 2.0 Amps 40.0 Watts P= I V = 2A(20V) = 40W V= I R or I = V / R

Curcuits R=? A V= 120 V I =? P= 60 W R I V P 120 V 60 W

Curcuits R=? A V= 120 V I =? P= 60 W R I V P 120 V 60 W V= I R P= I V or I=P/V=60W/120V=0.5A

Curcuits R=? A V= 120 V I =? P= 60 W R I V P 120 V 60 W 0.50A

Curcuits R=? A V= 120 V I =? P= 60 W R I V P 120 V 60 W 0.50A R=V/I=120V/0.5A=240

Curcuits R=? A V= 120 V I =? P= 60 W R I V P 120 V 60 W 0.50A 240

Series Curcuits R 1 =10 A V= 20 V R 2 =10 I = ? Req= 20 I 1 = I 2 = I = V/R eq = V 1 +V 2 =20V R eq = R 1 +R 2

Series Curcuits R=10 A V= 20 V R=10 I = ? Req= 20 I 1 = I 2 = I = V/R eq = 1.0 A

Parallel Curcuits R=10 A V= 20 V R=10 Req=V/I=20V/4A= 5 I 1 + I 2 = I= 4A I1 I2 I I I1 V 1 =V 2 =V=20V + I 1 =V/R 1 =20V/10 = 2A I 2 = V/R 2 = 2A

Series Curcuits R=8 A V= 20 V R=2 I = ? Req= 10 I 1 = I 2 = I = V/R = 2.0 A

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