Ohm’s LawOhm’s Law

Experiments show that for many Experiments show that for many materials, including most metals, the materials, including most metals, the resistance remains constant over a wide resistance remains constant over a wide range of applied voltages or currentsrange of applied voltages or currents

This statement has become known as This statement has become known as Ohm’s LawOhm’s Law• V = I RV = I R

Ohm’s Law is an empirical relationship Ohm’s Law is an empirical relationship that is valid only for certain materialsthat is valid only for certain materials• Materials that obey Ohm’s Law are said to be Materials that obey Ohm’s Law are said to be

ohmicohmic

V = R IResistance, R = V/I[R] = V/A = 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

R3

R1R2 R3

Req = R1 + R2 + R3

1/Req = 1/R1+1/R2+1/R3

Q2. What is the ratio of the current Q2. What is the ratio of the current flowing through each resistor (Iflowing through each resistor (I11:I:I22) in ) in the circuit?the circuit?

1.1. 1:11:1

2.2. 3:13:1

3.3. 1:41:4

4.4. Need more info.Need more info.

6 V

R1 = 10

R2 = 30

Q3. What is the ratio of the current Q3. What is the ratio of the current flowing through each resistor (Iflowing through each resistor (I11:I:I22)?)?

R1 = 10

R2 = 30

6 V

1.1. 1:11:1

2.2. 3:13:1

3.3. 1:41:4

4.4. None of aboveNone 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

Rb

(1) 1/Req = 1/Ra + 1/Rb

(2) Req is smaller than Ra and Rb

20

25

Req ≈ 10

1000 = 1k

2

Req < 2Practically all the current flowsThough the bottom one!!

Ohm’s law:Ohm’s law: = R = R·I·I

R1 = 6

6 V

I = /R = (6 V)/(6 Ohm) = 1.0 A

What is the electric potential at ? We cannot tell the absolute potential at this point. If at is +6 V, then 0 V at If 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

6 V = “0”

Then, at is +6 V.

R1 = 6

6 V

R2=4 R3=2

= R3I = 2 (V) = R2I = 4 (V)

= 4 + 2 = 6 (V)

= 2 V

ResistivityResistivity

The resistance of an ohmic conductor The resistance of an ohmic conductor is proportional to its length, L, and is proportional to its length, L, and inversely proportional to its cross-inversely proportional to its cross-sectional area, Asectional area, A

• ρ is the constant of proportionality and ρ is the constant of proportionality and is called the is called the resistivityresistivity of the material of the material

A

LR

L

A

R =

Resistivity: material parameter same for any shape in a given material.

[] = .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

L /A

ExampleExample

A silver wire has a resistance of 2A silver wire has a resistance of 2. . What would be the resistance of a What would be the resistance of a silver wire twice its length and half silver wire twice its length and half its diameter?its diameter?

Electrical Energy and PowerElectrical Energy and Power

In a circuit, as a charge moves through the In a circuit, as a charge moves through the battery, the electrical potential energy of battery, the electrical potential energy of the system is increased by the system is increased by QVQV• The chemical potential energy of the battery The chemical potential energy of the battery

decreases by the same amountdecreases by the same amount As the charge moves through a resistor, it As the charge moves through a resistor, it

loses this potential energy during loses this potential energy during collisions with atoms in the resistorcollisions with atoms in the resistor• The temperature of the resistor will increaseThe temperature of the resistor will increase

Electrical Energy and Power, Electrical Energy and Power, contcont

The rate at which the energy is lost is The rate at which the energy is lost is the powerthe power

From Ohm’s Law, alternate forms of From Ohm’s Law, alternate forms of power arepower are

IVVt

QP

R

VRIP

22

Electrical Energy and Power, Electrical Energy and Power, finalfinal

The SI unit of power is Watt (W)The SI unit of power is Watt (W)• I must be in Amperes, R in ohmsI must be in Amperes, R in ohms and V and V

in Voltsin Volts The unit of energy used by electric The unit of energy used by electric

companies is the companies is the kilowatt-hourkilowatt-hour• This is defined in terms of the unit of This is defined in terms of the unit of

power and the amount of time it is power and the amount of time it is suppliedsupplied

• 1 kWh = 3.60 x 101 kWh = 3.60 x 1066 J 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

=

q

High potential Low potential

t

q./t = (q/t)

P = I2R = 2/R [P] = J/s = W (watt)

ExampleExample

Light bulb 60 W, 120 V. Find Light bulb 60 W, 120 V. Find resistance of the light bulb.resistance of the light bulb.

Bulbs in seriesBulbs in series Bulbs in parallelBulbs in parallel

Spec: 60 W/120 V

P = 2/R = 60 (W)

R = 2/60 = 240 ()

Rmeasured ≈ 26

WHY are they so different?

depends on temperature!!!

ExampleExample

Light bulb 60 W, 120 V. Find Light bulb 60 W, 120 V. Find resistance of the light bulb.resistance of the light bulb.

Bulbs in seriesBulbs in series Bulbs in parallelBulbs in parallel

Temperature Variation of Temperature Variation of ResistivityResistivity

For most metals, resistivity increases For most metals, resistivity increases with increasing temperaturewith increasing temperature• With a higher temperature, the metal’s With a higher temperature, the metal’s

constituent atoms vibrate with constituent atoms vibrate with increasing amplitudeincreasing amplitude

• The electrons find it more difficult to The electrons find it more difficult to pass the atomspass the atoms

Temperature Variation of Temperature Variation of Resistivity, contResistivity, cont

For most metals, resistivity increases For most metals, resistivity increases approximately linearly with approximately linearly with temperature over a limited temperature over a limited temperature rangetemperature range

• ρρoo is the resistivity at some reference is the resistivity at some reference temperature Ttemperature Too

TToo is usually taken to be 20° C is usually taken to be 20° C is the is the temperature coefficient of resistivitytemperature coefficient of resistivity

)]TT(1[ oo

Temperature Variation of Temperature Variation of ResistanceResistance

Since the resistance of a conductor Since the resistance of a conductor with uniform cross sectional area is with uniform cross sectional area is proportional to the resistivity, you proportional to the resistivity, you can find the effect of temperature on can find the effect of temperature on resistanceresistance

)]TT(1[RR oo Ag: 3.8 x 103.8 x 10-3-3 /C /C

Cu: 3.9 x 103.9 x 10-3-3 /C /C Fe:5.0 5.0 x 10x 10-3-3 /C /C

Tungsten: 4.5 x 10Tungsten: 4.5 x 10-3-3 /C /C

ExampleExample

Light bulb (60 W; 120 V; 240 Light bulb (60 W; 120 V; 240 ) ) operates at 1800 C. What is the operates at 1800 C. What is the resistance of the filament (tungsten) resistance of the filament (tungsten) at 20 C?at 20 C?

Temperature

= o + AT

for metals

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 whenconnected 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 usethe 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/RSince 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

Statement: P = IV If I hook up this heater on 60 V, the powerProduced by this heater would drop to ½.

No! No!

So far, we have considered circuits which can be reducedto a single loop circuit and applied Ohm’s law to extractI and/or V in the circuit.

6 V5 3