Download presentation

Presentation is loading. Please wait.

Published byNathan Emery Modified over 2 years ago

1
Ohms Law Physics Dr. Robert MacKay

2
Voltage (Volts) Electrical Pressure V

3
Current (Amps) Charge Flow I V

4
Resistance, R (Ohms, ) Impedence I V R Depends on Temperature Material of wire Area of wire Length of wire

5
Resistance, R (Ohms, ) Impedence R Depends on: * Temperature * Material of wire * Cros sectional Area * Length Area Length

6
Resistance and Resistivity = resistivity R=resistance

7

8
As a wire heats it up its resistance increases. Some thermometers work on this principle.

9
Ohms Law V= I R I V

10
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

11
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

12
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

13
Electric Power (Watts) Power = Current x Voltage 1 Watt = Amp x Volt

14
Curcuits R=10 A V= 20 V I =? P= ? R I V P V

15
Curcuits R=10 A V= 20 V I =? P= ? R I V P V 2.0 A V= I R or I = V / R

16
Curcuits R=10 A V= 20 V I =? P= ? R I V P V 2.0 Amps 40.0 Watts P= I V = 2A(20V) = 40W V= I R or I = V / R

17
Curcuits R=? A V= 120 V I =? P= 60 W R I V P 120 V 60 W

18
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

19
Curcuits R=? A V= 120 V I =? P= 60 W R I V P 120 V 60 W 0.50A

20
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

21
Curcuits R=? A V= 120 V I =? P= 60 W R I V P 120 V 60 W 0.50A 240

22

23
Curcuits R=10 A V= 20 V I =2 A P= 40 Watts R I V P V 2 A 40 Watts

24

25

26
Two light bulbs are connected in series to a battery the resistance of each is 20. What is the total effective resistance of this series combination?

27
Req=R1+R2=

28
Two light bulbs are connected in series to a 40 Volt battery. The resistance of each is 20. What is the total Current leaving the battery?

29
V=I R or I=V/R (R=Req=40 I=40V/40 =1.0 Amp

30

31

32

33

34
Curcuits R=10 A V= 20 V I =2 A P= 40 Watts R I V P V 2 A 40 Watts

35

36

37

38

39

40
A bit of trivia When 2 equal resistors are connected in parallel the effective resistance is half the value of each. (two 10 resistor Req=5 ) When 3 equal resistors are connected in parallel the effective resistance is one-third the value of each. (three10 resistor Req=3.33 ) When 4 equal resistors are connected in parallel the effective resistance is one-forth the value of each. (four10 resistor Req=2.5 ) When N equal resistors are connected in parallel the effective resistance is 1/N the value of each. (N 10 resistor Req=10/N )

41
What is the effective total resistance when 10 equal 10 resistors are connected in parallel?

42
Req=10 /10=1.0

43
Two 20 Resistors are connected in parallel to a 20 V battery. What is the voltage difference across each resistor?

44
V=V1=V2=20 V

45
Two 20 Resistors are connected in parallel to a 20 V battery. What is the current flowing through each resistor?

46
Two 20 Resistors are connected in parallel to a 20 V battery. What is the voltage difference across each resistor? I=V/R I 1 =V 1 /R 1 and I 2 =V 2 /R 2 I 1 =20V/ 20 =1.0 A I 2 =20V/ 20 =1.0 A

47
Two 20 Resistors are connected in parallel to a 20 V battery. What is the total effective resistance of this circuit?

48
Two 20 Resistors are connected in parallel to a 20 V battery. What is the current flowing through each resistor? R=V/I=20V/2.0A=10

49

Similar presentations

© 2016 SlidePlayer.com Inc.

All rights reserved.

Ads by Google