Transcript Powerpoint Slides

Lab 2: Ohm’s Law
Only 10 more labs to go!!
When trying to understanding electricity it is helpful to draw analogies to the classical
mechanics.
Gravitational
Field
In this example we have flow of water
(current) from the storage tank because
the gravitational field (force) is pushing
the water downwards (in order to
decrease the water’s potential.
PE
flow, current
Electric current flow
Analogous to the water flow due to the
gravitational field, charge will flow from
it’s storage device (a battery) when
in the presences of an electric field.
Electric Field
+
+
+
+
+
PE
When charges flow we can do work, in fact batteries are characterized by the
amount of work they can do.
PE
V 
q
Electric potential- the amount of work each Coulomb of
charge delivers, measured in VOLTS, V
So 110 V means that 110 Joules of work can be delivered
per 1 Coulomb of charge
The current, I in a circuit is defined as the amount of charge that flows per
second.
Q
I
t
Current, I, is measured in Amperes or Amps, A.
1 A of current means that 1 Coulomb of charge has flowed past
a point in a second.
We electrical current is not free to flow we call this Electrical Resistance
As charge flows through something that has resistance, some of the charge’s
energy is dissipated. We call this type of device a resistor. Resistance is measured
in ohms, .
a bad, circuit diagram of a resistor
Electric potential, current and resistance are related through Ohm’s law:
V I R
As current flows through a resistor, the charges dissipate energy. Each Coulomb of
charge will deposit one joule of energy per volt. If one Coulomb of charge flows per
second, then one Watt of power is dissipated. Remember:
E
P
t
Also remember
from before:
E
qV
V 
 E  qV  P 
q
t
qV
P
t
q
recall that I  so P  IV
t
We have several variations of the power equation:
P  IV  I IR   I R also
2
V2
V 
P  IV   V 
R
R
What if I gave you two light bulbs, one rated at 60 W and the other at 100 W at
120 Volts. Which bulbl will have the largest resistance? From the equation above:
V2
R
P
V 2 (120V ) 2 14400V 2
R60 


 240
P60
60W
60W
V 2 (120V ) 2 14400V 2
R100 


 144
P100
100W
100W
The first thing you are going to do today:
V
A
Construct a graph of current vs. voltage. Increment the voltage by 5 to 7 Volts ,
when the voltage exceeds 40 Volts, increment by 10 to 15 Volts.
Current vs. Voltage for an Ideal Resistor
What’s the slope of this line?
From Ohm’s law:
14
V I R
12
I Current (A)
10
1
I V
R
8
6
4
2
0
0
2
4
6
8
V Voltage (Volts)
10
12
1
1
slope   R 
R
slope
14
The next procedure you’re going to
examine the lamp circuit:
V
A
Construct a graph of current vs. voltage. Increment the voltage by 5 to 7 Volts ,
when the voltage exceeds 40 Volts, increment by 10 to 15 Volts.
I Current (Amps)
Current vs. Voltage for a Lamp
Notice how the slope of the tangent
line is changing.
(For smaller voltages the slope is higher
therefore the resistance is lower)
In this case the temperature of the bulb
is increasing with increasing power.
As the temperature increases the
resistance also increases
V Voltage (Volts)
The next procedure you’re going to examine a light emitting diode (LED) circuit.
An LED is a special semiconductor device.
According to device physics:
If I take the natural log
of both sides:
I ce
V
A
V
ln( I )  ln( c) 
a
Construct a graph of the natural log of current vs. voltage. Increment the voltage
by 5 to 7 Volts for all data points.
On a log graph
On a regular graph
I
I
R
Voltage
1
slope
Voltage