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Transcript 
Ohm’s Law
Physics
Dr. Robert MacKay
Voltage (Volts)
Electrical Pressure
-
+
+
-
V
Current (Amps)
Charge Flow
I
+
+
+
V
-
+
+
+
+ +
+
+
+
Resistance, R (Ohms, Ω)
Impedence
I
-
-
-
V
-
-
+
-
- -
-
-
R Depends on
Temperature
Material of wire
Area of wire
Length of wire
-
Resistance, R (Ohms, Ω)
Impedence
Area
-
-
-
-
-
- -
-
Length
R Depends on:
* Temperature
* Material of wire
* Cros sectional Area
* Length
-
-
Resistance and Resistivity
L
R 
A
= resistivity
R=resistance
Material
Silver
Copper
Carbon
Silicon
Glass
Rubber
 (ž- m)
-8
1.59x10
-8
1.70x10
-5
3.6x10
2
2.5x10
12
10
15
10
 (°C-1)
-3
6.1x10
-3
6.8x10
-4
-5.0x10
-2
-7.0x10
1


  0T
R  R0 T
R  R 0 (1 T)   0 (1 T)
Ohm’s Law
V= I R
I
-
-
-
V
-
-
+
-
- -
-
-
-
Exercise
 Find
the electric voltage required to have
10 Amps of current flow through a 2 Ω
resistor.
 Given: I= 10 A and R=2 Ω
 Wanted: V
 Solution: V=I R
 V= (10 A) (2 Ω) = 20 Volts
Exercise
 Find
the electric current flowing in a 10 Ω
resistor (light bulb) when connected to a 20
Volt battery.
 Given: V= 20 Vand R=10 Ω
 Wanted: I
 Solution:
 I=
V=I R or I=V / R
(20 V) / (10 Ω) = 2 Amps
Exercise
 When
a 60 Volt battery is connected to a
circuit 4.0 amps of current flow from the
battery. What is the circuit’s resistance.
 Given: V= 60 V and I=4.0 Amps
 Wanted: R
 Solution:
 I=
V=I R or R=V / I
(60 V) / (4.0 Amps) = 15 Ω
DC
200
OFF 20M
2M
200 k
V
20
AC
20 k
2
Ω
2k
200
200 m
200 µ
2m
.995
20 m
A
2
200 m
A
V-Ω COM
DC
200
OFF 20M
2M
200 k
V
20
AC
20 k
2
Ω
2k
200
200 m
200 µ
2m
1 .
20 m
A
2
200 m
A
V-Ω COM
DC
200
OFF 20M
2M
200 k
V
20
AC
20 k
2
Ω
2k
200
200 m
200 µ
2m
2.41
20 m
A
2
200 m
A
V-Ω COM
DC
200
OFF 20M
2M
200 k
V
20
AC
20 k
2
Ω
2k
200
200 m
200 µ
2m
122.
20 m
A
2
200 m
A
V-Ω COM
DC
200
AC
OFF 20M
2M
200 k
V
20
20 k
2
Ω
2k
200 m
200
200 µ
2m
0.082
20 m
A
2
200 m
A
V-Ω COM
DC
200
AC
OFF 20M
2M
200 k
V
20
20 k
2
Ω
2k
200
200 m
200 µ
2m
65
20 m
A
2
200 m
A
V-Ω COM
DC
200
AC
OFF 20M
2M
200 k
V
20
20 k
2
Ω
2k
200
200 m
200 µ
2m
65
20 m
A
2
200 m
A
V-Ω COM
DC
200
OFF 20M
2M
200 k
V
20
AC
20 k
2
Ω
2k
200
200 m
200 µ
2m
156
20 m
A
2
200 m
A
V-Ω COM
DC
200
OFF 20M
2M
200 k
V
20
AC
20 k
2
Ω
2k
200
200 m
200 µ
2m
156
20 m
A
2
200 m
A
V-Ω COM
Electric Power (Watts)
 Power
= Current x Voltage
 1 Watt = Amp x Volt
Curcuits
R=10 Ω
R
10 Ω
I
P= ?
I =?
A
V
P
V= 20 V
20 V
V= I R
or
I=V/R
Curcuits
R=10 Ω
P= ?
R
10 Ω
I
2.0 A
V
20 V
I =?
A
P
V= 20 V
V= I R
or
I=V/R
Curcuits
R=10 Ω
P= ?
R
10 Ω
I
2.0 Amps
V
20 V
P
40.0 Watts
I =?
A
V= 20 V
P= I V = 2A(20V) = 40W
Curcuits
R=?
R
I
P= 60 W
I =?
A
V= 120 V
V
120 V
P
60 W
Curcuits
V= I R
R=?
R
I
P= 60 W
I =?
A
V
120 V
P
60 W
V= 120 V
P= I V or
I=P/V=60W/120V=0.5A
Curcuits
R=?
R
P= 60 W
I
0.50A
V
120 V
P
60 W
I =?
A
V= 120 V
Curcuits
R=?
R
P= 60 W
I
0.50A
V
120 V
P
60 W
I =?
A
V= 120 V
R=V/I=120V/0.5A=240W
Curcuits
R=?
P= 60 W
R
240W
I
0.50A
V
120 V
P
60 W
I =?
A
V= 120 V
Series Curcuits
R1=10 Ω
A
Req= 20 Ω
V= 20 V
Req= R1+R2
R2=10 Ω
I=?
I1 = I2 = I = V/Req =
V1+V2=20V
Series Curcuits
R=10 Ω
R=10 Ω
Req= 20 Ω
A
V= 20 V
I=?
I1 = I2 = I = V/Req = 1.0 A
Parallel Curcuits
V1=V2=V=20V
R=10 Ω
I1
I2
I1
I1 =V/R1=20V/10W= 2A
I2 = V/R2 = 2A
R=10 Ω
A
I
+
I
I1 + I2 = I= 4A
Req=V/I=20V/4A= 5 Ω
V= 20 V
Series Curcuits
R=8 Ω
R=2 Ω
Req= 10 Ω
A
V= 20 V
I=?
I1 = I2 = I = V/R = 2.0 A
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