Ohm’s Law


Experiments show that for many
materials, including most metals, the
resistance remains constant over a wide
range of applied voltages or currents
This statement has become known as
Ohm’s Law
•V=IR

Ohm’s Law is an empirical relationship
that is valid only for certain materials
• Materials that obey Ohm’s Law are said to be
ohmic
V=RI
Resistance, R = V/I
[R] = V/A = W (Ohm)
For a fixed potential difference across a resistor,
the larger R, the smaller current passing through it.
Req
Parallel connection
Series connection
R1
R2
R1
R2
R3
R3
Req = R1 + R2 + R3
1/Req =
1/R1+1/R2+1/R3
Q2. What is the ratio of the current flowing
through each resistor (I1:I2) in the circuit?
R1 = 10
R2 = 30
6V
1.
2.
3.
4.
1:1
3:1
1:4
Need more info.
Q3. What is the ratio of the current flowing
through each resistor (I1:I2)?
R1 = 10
R2 = 30
6V
1.
2.
3.
4.
1:1
3:1
1:4
None of above
• Electrical wires can be bent and/or stretched.
• A Node point (branching point) can be moved arbitrarily
along the wire.
There are n identical resistors connected in parallel.
Req?
1/Req = 1/R + 1/R + 1/R + … + 1/R
= n/R
Req = R/n
Ra
(1) 1/Req = 1/Ra + 1/Rb
(2) Req is smaller than Ra and Rb
Rb
20
25
Req ≈ 10
1000 = 1k
2
Req < 2
Practically all the current flows
Though the bottom one!!
Ohm’s law:
e = R·I
I = e/R
= (6 V)/(6 Ohm)
= 1.0 A
R1 = 6
6V
What is the electric potential at ?
We cannot tell the absolute potential at this point.
If e at
is +6 V, then 0 V at
If e at
is +3 V, then -3 V at
For both, the potential diff. is 6 V.
To be able to specify absolute potential at a given point,
we need to specify a reference point “0” potential.
GROUND
R1 = 6
6V
Then, e at
is +6 V.
e = “0”
De = 4 + 2 = 6 (V)
De = R2I = 4 (V)
De = R3I = 2 (V)
R1 = 6
R2=4
R3=2
6V
e=2V
Resistivity

The resistance of an ohmic conductor
is proportional to its length, L, and
inversely proportional to its crosssectional area, A
L
R
A
• ρ is the constant of proportionality and
is called the resistivity of the material
A
L
R =  L /A
Resistivity: material parameter
same for any shape in a given material.
[] = W.m
e.g. for copper
 = 1.7 x 10-8
gold
 = 2.44 x 10-8
tungsten  = 5.6 x 10-8
nickel-chrome
 = 150 x 10-8
Example
A silver wire has a resistance of 2W.
What would be the resistance of a
silver wire twice its length and half
its diameter?
Electrical Energy and Power

In a circuit, as a charge moves through
the battery, the electrical potential energy
of the system is increased by QV
• The chemical potential energy of the battery
decreases by the same amount

As the charge moves through a resistor, it
loses this potential energy during
collisions with atoms in the resistor
• The temperature of the resistor will increase
Electrical Energy and Power,
cont

The rate at which the energy is lost
is the power
Q
P  V  IV
t

From Ohm’s Law, alternate forms of
power are
2
V
PI R
R
2
Electrical Energy and Power,
final

The SI unit of power is Watt (W)
• I must be in Amperes, R in ohms and V
in Volts

The unit of energy used by electric
companies is the kilowatt-hour
• This is defined in terms of the unit of
power and the amount of time it is
supplied
• 1 kWh = 3.60 x 106 J
A light bulb with 60 W/120 V
When 120 V applied, consume 60 W of power
POWER = Energy produced or dissipated per unit time
= dq.e/dt = (dq/dt)e  I.e
P = I2R = e2/R
[P] = J/s = W (watt)
+dq
dt
High potential
Low potential
e
Example



Light bulb 60 W, 120 V. Find
resistance of the light bulb.
Bulbs in series
Bulbs in parallel
Spec: 60 W/120 V
P = e2/R = 60 (W)
R = e2/60 = 240 (W)
Rmeasured ≈ 26 W
WHY are they so different?
 depends on temperature!!!
Example



Light bulb 60 W, 120 V. Find
resistance of the light bulb.
Bulbs in series
Bulbs in parallel
Temperature Variation of
Resistivity

For most metals, resistivity increases
with increasing temperature
• With a higher temperature, the metal’s
constituent atoms vibrate with
increasing amplitude
• The electrons find it more difficult to
pass the atoms
Temperature Variation of
Resistivity, cont

For most metals, resistivity increases
approximately linearly with
temperature over a limited
temperature range
  o [1 + (T  To )]
• ρo is the resistivity at some reference
temperature To


To is usually taken to be 20° C
 is the temperature coefficient of resistivity
Temperature Variation of
Resistance

Since the resistance of a conductor
with uniform cross sectional area is
proportional to the resistivity, you
can find the effect of temperature on
resistance
R  Ro [1+ (T  To )]
Ag: 3.8 x 10-3 /C
Cu: 3.9 x 10-3 /C
Fe:5.0 x 10-3 /C
Tungsten: 4.5 x 10-3 /C
Example
Light bulb (60 W; 120 V; 240 W)
operates at 1800 C. What is the
resistance of the filament (tungsten)
at 20 C?

= o + AT
for metals
Temperature
However, insulators have the opposite tendency.
The higher temperature is, the lower .
Q. A 40 W/120 V light bulb produces 40 W of power when
connected to 120 V. How much of power would be produced
if the same bulb is connected to 60 V?
You can solve this problem by calculating R of the filament and use
the formula for power, P = V2/R. But we know the resistance of
the filament is fixed. Therefore, power is just proportional to V2.
P (for 120 V) = 40 (W),
p (for 60 V) = 40*(60/120)2 = 10 (W)
A certain 1400 W space heater is designed to operate on 120 V.
How much current flows through it when it is operating? What is
Its resistance when operating?
Power = IV = I2R = V2/R
Since we know P and V, use P = IV to calculate I.
I = P/V = (1400 W)/(120 V) = 11.7 A
Now, we can use Ohm’s law to calculate R.
When it is operating on 120 V, 11.7 A current flows.
R = V/I = (120 V)/(11.7 A) = 10.25 W
Statement: P = IV If I hook up this heater on 60 V, the power
Produced by this heater would drop to ½.
So far, we have considered circuits which can be reduced
to a single loop circuit and applied Ohm’s law to extract
I and/or V in the circuit.
5
6V
3