Ohms Law
Mitsuko J. Osugi
Physics 409D
Winter 2004
UBC Physics Outreach
Ohm’s Law
Current through an ideal conductor is
proportional to the applied voltage
– Conductor is also known as a resistor
– An ideal conductor is a material whose resistance does not change
with temperature
For an ohmic device,
Voltage  Current  Resistance
V  I R
V = Voltage
I = Current
R = Resistance
(Volts = V)
(Amperes = A)
(Ohms = Ω)
Current and Voltage Defined
Electric Current: flow of
electrons from the negative
terminal to the positive one
Conventional Current:
(the current in electrical circuits)
Flow of current from positive
terminal to the negative
terminal.
Current has units of Amperes (A)
is measured using ammeters.
Voltage:
Energy required to move a charge
from one point to another.
Think of voltage as what pushes
the electrons along in the circuit,
and current as a group of
electrons that are constantly
trying to reach a state of
equilibrium.
If there is no voltage, electrons
don’t move, therefore there is no
current.
Difference in electrical charge
between two points creates
difference in potential energy,
which causes electrons to flow
from an area with lots of electrons
(negative terminal) to an area
with few electrons (positive
terminal), producing an electric
current.
Linearity of Voltage and Current for
Resistors which Obey Ohm’s Law
Current (A)
Voltage versus Current
for a 10 ohm Resistor
0.6
0.5
0.4
0.3
0.2
0.1
0
0
1
2
3
4
5
6
Voltage (V)
Voltage and current are linear when resistance is held constant.
Ohmic Resistors
• Metals obey Ohm’s Law so long as their
temperature is held constant
– Their resistance values do not fluctuate with
temperature
• i.e. the resistance for each resistor is a constant
Most ohmic resistors will behave
non-linearly outside of a given range of
temperature, pressure, etc.
Ohm’s Law continued
Ohm’s Law continued
The total resistance of a circuit is dependant on
the number of resistors in the circuit and their
configuration
Series Circuit
Rtotal   Resistors  R1  R2  ...
Parallel Circuit
1
1
1
1

 
 ...
Rtotal
Resistors R1 R2
Kirchhoff’s Current Law
Current into junction = Current leaving junction
 Current  0
The amount of current that enters a junction is
equivalent to the current that leaves the junction
Iin
I1
I1
I2
I2
Iout
I in  I1  I 2  I out
I in  I out  0
Kirchhoff’s Voltage Law
Vin  VoltageAcrossEachResistor
Net Voltage for a circuit = 0
Sum of all voltage drops and voltage rises in a
circuit (a closed loop) equals zero
V1
V2
V  V1  V2
V  V1  V2  0
V
Series Circuit
Current is constant
• Why?
– Only one path for the
current to take
– Kirchhoff’s Current
Law
• Voltages through
circuit equals zero
– Kirchhoff’s Voltage
Law
V  I R
V  V1  V2  V3
R  R1  R2  R3
Series Equivalent Circuit
V1  I  R1 V2  I  R2 V3  I  R3
R  R1  R2  R3
V
V1  V2  V3
V  I  R1  I  R2  I  R3
V  I   R1  R2  R3 
V  IR
Experiment 1
One 10Ω resistor connected to the 6V power
source (batteries). Add another 10Ω resistor to
the circuit in series to the first resistor.
Q:
What is the equivalent resistance, R? What
will happen to the value of the current through
each resistor? What will happen to the value of
the voltage across each resistor?
V  IR
R  R1  R2  ...
V  V1  V2  ...
Experiment 1:
What is happening in theory
Initially,
V =6V, R  10
V  IR  I 
V
6V

 0.6A
R 10
Add the second resistor:
V  6V, R  R1  R2  10  10  20
I 
V
6V

 0.3A  The current through the circuit is halved!
R 20
V1  I  R1  0.3A  10  3V
V2  I  R2  0.3A  10  3V
V  6V  V1  V2  The voltage across each resistor is halved!
Experiment 1:
The actual data
In reality, the data we get is not the same as what
we get in theory.
Why?
Because when we calculate numbers in theory, we
are dealing with an ideal system. In reality there
are sources of error in every aspect, which make
our numbers imperfect.
Parallel Circuit
• Voltage is constant
• Why?
1 1 1
1
 

R R1 R2 R3
I  I1  I 23  I1  I 2  I 3
where I 23  I 2  I 3
– There are 3 closed
loops in the circuit,
which means the
voltage though each
loop is equivalent to the
voltage supplied (like in
a series circuit)
• Branch currents add
to equal total current
Parallel Equivalent Circuits
1 1
1
1
1
1
1
1 1
1 
 

let


so   

R R1 R2 R3
R 23 R2 R3
R  R1 R23 
1
1
1
1
and
 
  R  R123
I  I1  I 2  I 3
R123 R1 R23 R
I1  I 2  I 3
1
V  I  R   I1  I 2  I 3  

1
1
1
1
1
1




R1 R2 R3 R1 R2 R3
Experiment 2
One 10 ohm resistor connected to the battery.
Connect a second 10Ω resistor in parallel to the
first one
Q:
What will the new resistance be? What will
happen to the current through each resistor and
the voltage across each component of the
circuit?
V  IR
 1

1 1
1
1
 
 ...  R   
 ... 
R R1 R2
 R1 R2

I  I1  I 2  ...
1
Experiment 2:
What
is
happening
in
theory
Initially,
V  6V, R  10
V
6V
V  IR  I 

 0.6A
R 10
Add the second resistor:
1
1
1
1 
1 
 1
V  6V, R  



 5  Resistance of circuit halves!



 10 10 
 R1 R2 
V
6V
V  IR  I 

 1.2A  Current through circuit doubles!
R 5
I1 
V
6V

 0.6A
R1 10
I2 
V
6V

 0.6A
R2 10
I  I1  I2  1.2A  0.6A+0.6A
Experiment 2:
The actual data
We’ve now looked at how basic electrical
circuits work with resistors that obey
Ohm’s Law linearly.
We understand quantitatively how these
resistors work, but lets see qualitatively
using light bulbs.
Let us also determine if light bulbs also obey
Ohm’s Law linearly.
The Light Bulb and its Components
• Has two metal contacts at
the base which connect
to the ends of an
electrical circuit
• The metal contacts are
attached to two stiff wires,
which are attached to a
thin metal filament.
• The filament is in the
middle of the bulb, held
up by a glass mount.
• The wires and the
filament are housed in a
glass bulb, which is filled
with an inert gas, such as
argon.
Light bulbs and Power
Power dissipated by a bulb relates to the
brightness of the bulb. The higher the power, the
brighter the bulb.
Power is measured in Watts [W]
2
V
P  I2  R  V  I 
R
For example, think of the bulbs you use at home.
The 100W bulbs are brighter than the 50W
bulbs.
Experiment 3
One bulb connected to the 6V power source. Add another
bulb to the circuit in series.
Q: What is the initial current through the circuit? When
the second bulb is added, will the bulbs become
brighter, dimmer, or not change?
• We can use Ohm’s Law to approximate what will
happen in the circuit:
V  IR
P V I
Experiment 3:
The qualitative results
(with some theory)
Recall:V  I  R  I 
V
R
When we add the lightbulbs:
V constant for the circuit, but R increases
 I decreases
P  V  I  decreases
 The bulbs get dimmer because the power dissipated decreases
Experiment 3:
Some numbers
Let’s see what kind of values we find for the
voltage across the light bulbs, and the current
through the bulbs.
Experiment 4
Have one bulb connected to the 6V power source. Add
a second bulb to the circuit in parallel.
Q: What is the initial current through the circuit? What
happens when the second bulb is added?
 We can use Ohm’s Law to approximate what will
happen in the circuit:
V  IR
P V I
1
1 
R


R
R
2 
 1
1
Experiment 4:
The qualitative results
(with some theory)
V  IR I 
V
R
P V I
1
1
1 
R


R
R
2 
 1
 R1 and R2 > 1,  R < R1 and R2
V constant for the circuit, R decreases  I increases
 P increases as R decreases
The bulbs do not change in brightness,
but the total power of the circuit is increased
Experiment 3:
Some numbers
Let’s see what kind of values we find for the
voltage across the light bulbs, and the current
through the bulbs.
Light bulbs do not obey Ohm’s Law
• Bulbs are non-linear conductors
(R increases with temperature)
R  Ro  1   T  To 


R  Conductor resistance at temperature T []
Ro  Conductor resistance at reference To []
  Temperature coefficient of resistance [C 1]
T  Conductor temperature [C ]
To  Reference temperature  specified for [C ]
The filaments of light bulbs are made of Tungsten,
which is a very good conductor. It heats up easily.
 Tungsten  0.004403 / C at 20C (i.e. To  20C )
As light bulbs warm up, their resistance increases.
If the current through them remains constant:
2
P  I R
They glow slightly dimmer when first plugged in.
Why?
R increases but I remains constant  P increases
Most ohmic resistors will behave non-linearly outside of
a given range of temperature, pressure, etc.
Voltage versus Current for
Constant Resistance
The light bulb does not have a linear relationship. The resistance
of the bulb increases as the temperature of the bulb increases.
“Memory Bulbs” Experiment
• Touch each bulb in succession with the
wire, completing the series circuit each
time
Q:
What is going to happen?
Pay close attention to what happens to each
of the bulbs as I close each circuit.
“Memory Bulbs” Continued…
How did THAT happen??
Temperature of bulbs increases
 resistance increases
 power dissipation (brightness) of bulbs
increases
• Filaments stay hot after having
been turned off
• In series, current through each
resistor is constant
– smallest resistor (coolest bulb)
has least power dissipation,
therefore it is the dimmest bulb
RHot  RCold
P  I2  R  R 
PHot
PCold
 2  2
I
I
 PHot  PCold
P
I2
Conclusion
• Ohmic resistors obey Ohm’s Law
V  IR
• Resistance is affected by temperature. The
resistance of a conductor increases as its
temperature increases.
• Light bulbs do not obey Ohm’s Law
– Tungsten is such a good conductor that their
resistance depends on their temperature
– As their temperature increases, the power dissipated
by the bulb increases
• i.e. They are brighter when they are hotter