Download Ohm`s Law - Dr. Robert MacKay

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)
As a wire heats it up its resistance increases.
Some thermometers work on this principle.
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
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
Curcuits
R=10 Ω
I =2 A
P= 40 Watts
A
V= 20 V
R
10 Ω
I
2A
V
20 V
P
40 Watts
 Two
light bulbs are connected in series to a
battery the resistance of each is 20 W. What
is the total effective resistance of this series
combination?
 Two
light bulbs are connected in series to a
battery the resistance of each is 20 W. What
is the total effective resistance of this series
combination?
 Req=R1+R2= 20 W20 W40 W
 Two
light bulbs are connected in series to a
40 Volt battery. The resistance of each is
20 W. What is the total Current leaving the
battery?
 Two
light bulbs are connected in series to a
40 Volt battery. The resistance of each is
20 W. What is the total Current leaving the
battery?
V=I R
or I=V/R (R=Req=40 W)
I=40V/40 W =1.0 Amp
Curcuits
R=10 Ω
I =2 A
P= 40 Watts
A
V= 20 V
R
10 Ω
I
2A
V
20 V
P
40 Watts
A bit of trivia




When 2 equal resistors are connected in parallel the
effective resistance is half the value of each. (two 10 W
resistor Req=5 W)
When 3 equal resistors are connected in parallel the
effective resistance is one-third the value of each. (three10
W resistor Req=3.33 W)
When 4 equal resistors are connected in parallel the
effective resistance is one-forth the value of each. (four10
W resistor Req=2.5 W)
When N equal resistors are connected in parallel the
effective resistance is 1/N the value of each. (N 10 W
resistor Req=10/N W)
 What
is the effective total resistance when
10 equal 10 W resistors are connected in
parallel?
 What
is the effective total resistance when
10 equal 10 W resistors are connected in
parallel?
 Req=10
W/10=1.0 W
20 WResistors are connected in
parallel to a 20 V battery. What is the
voltage difference across each resistor?
 Two
20 WResistors are connected in
parallel to a 20 V battery. What is the
voltage difference across each resistor?
 Two
 V=V1=V2=20
V
20 WResistors are connected in
parallel to a 20 V battery. What is the
current flowing through each resistor?
 Two
20 WResistors are connected in
parallel to a 20 V battery. What is the
voltage difference across each resistor?
 I=V/R I1=V1/R1 and I2=V2/R2
 I1=20V/ 20 W=1.0 A
 I2=20V/ 20 W=1.0 A
 Two
20 WResistors are connected in
parallel to a 20 V battery. What is the total
effective resistance of this circuit?
 Two
20 WResistors are connected in
parallel to a 20 V battery. What is the
current flowing through each resistor?
 Two
 R=V/I=20V/2.0A=10
W