Prof. Dr. Eser OLĞAR

April 7,20201

EP106 General Physics II

Current and Resistance

Content

Electric Current

Current Density

Resistivity and Ohm’s Law

Resistivity vs. Resistance

Temperature Dependence of Resistivity.

Electrical Power

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To define the current, we consider positive charges moving perpendicularly onto a surface area A as shown in Fig. The

current is the rate at which

charge flows through this

surface. If Q is the amount of

charge that passes through this

area in a time interval t, the

average current Iav is equal to

Δ

Δav

QI

t3

Electric Current

If the rate at which charge flows varies in time, then the current varies in time; we define the instantaneous current Ias the differential limit of average current:

Electric Current

dQI =

dt

The SI unit of current is the ampere (A):1C

1sI =

4

positive charge flows right or negative charge flows left

Typical currents:

• 100 W light bulb: roughly 1A

• car starter motor: roughly 200A

• TV, computer, phone: nA to mA “m” for milli = 10-3

Current is a scalar (not a vector)

• has a sign associated with it

• conventional current is flow of positive charge

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+ -

current electrons

An electron flowing from – to + gives rise to the same “conventional current” as a proton flowing from + to -

If your calculation produces a negative value for the current, that means the conventional current actually flows opposite to the direction indicated by the arrow.

In most conductors, charge carriers are negative electrons

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Example 1: 3.8x1021 electrons pass through an area in a wire in 4 minutes. What was the average current?

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av

Q NeI

t t

21 19

av

3.8 10 electrons 1.6 10 C / electronsI

4 60s

avI 2.53A

Example 1: 3.8x1021 electrons pass through an area in a wire in 4 minutes. What was the average current?

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The current across an area can

be expressed in terms of the

motion of the charge carriers.

To achieve this, we consider a

portion of a cylindrical rod that

has a cross-sectional area A,

length x, and carries a

constant current I.

Q = (n Ax) q

For convenience we consider positive charge carriers each having a charge q, and the number of carriers per unit volume in the rod is n. Therefore, in this portion, the number of carriers is nA x and the total charge Q is:

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Electric Current, Microscopic View

Suppose that all the carriers move with an average speed vd

(called the drift speed). Therefore, during a time interval t, all

carriers must achieve a displacement

I= nqAvd

Therefore, the current I = Q/ t in the rod will be given

by:

Q = (n Avd t) q

x =vdt

Rewriting, the total charge Q as

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The charge carriers in a solid conductor are all free electrons.

If the conductor is isolated, these electrons move with speeds

of the order of 106 m/s, and because of their collisions with

the scatterers (atoms or molecules in the conductor), they

move randomly in all directions.

dd

nqv ACurrentJ nqv

Area A

The current density J is defined as the current per unit area,

i.e.:

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Electric Current Density

Example 2: the 12-gauge copper wire in a home has a cross-sectional area of 3.31x10-6 m2 and carries a current of 10 A. The conduction electron density in copper is 8.49x1028

electrons/m3. Calculate the drift speed of the electrons.

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d

Iv

nqA

d

Iv

neA

d 28 -3 19 6 2

10C/sv

(8.49 10 m )(1.60 10 C)(3.31 10 m )

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dv 2.22 10 m/s

Example 2: The 12-gauge copper wire in a home has a cross-sectional area of 3.31x10-6 m2 and carries a current of 10 A. The conduction electron density in copper is 8.49x1028

electrons/m3. Calculate the drift speed of the electrons.

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Example 3: Estimate the drift speed of the conduction electrons in a copper wire that is 2mm in diameter and carries a current of 1A. Comment on your result. The density ofcopper is 8.92×103 kg/m3.

[Hint: Assume that each copper atom contributes one free conduction electron to the current.]

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The molar mass of copper is M(Cu)=63.546 kg/kmol. Recall that the

mass of one kmol of Cu contains Avogadro’s number of atoms

(NA =6.022×1026 atoms/kmol). Thus:

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1J E E

Caution! is not volume density! is not surface density!

As a result of maintaining a potential difference V across a conductor, an electric field E and a current density J are established in the conductor. For materials with electrical properties that are the same in all directions (isotropic conductors), the electric field is found to be proportional to the current density.

• is electrical conductivity• is electrical resistivity• and are material properties

• unit of :

Ohm’s law

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V / m Vm m

A / m A

Ohm

Ohm’s Law and Electric Resistance

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Materials that follow Ohm’s Law are called “ohmic” materials

Resistivity is constant

Graph of J vs. E is linear

J

E

slope=1/

J

E

materials that do not follow Ohm’s Law are called “non-Ohmic” materials

Graph of J vs. E is nonlinear

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Ohmic vs non-Ohmic materials

Example 4: The 12-gauge copper wire in a home has a cross-sectional area of 3.31x10-6 m2 and carries a current of 10 A. Calculate the magnitude of the electric field in the wire.

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IE J

A

8

6 2

(1.72 10 m) 10AE

(3.31 10 m )

2E 5.20 10 V/m

of copper

Example 4: The 12-gauge copper wire in a home has a cross-sectional area of 3.31x10-6 m2 and carries a current of 10 A. Calculate the magnitude of the electric field in the wire.

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current in a wire:

L

A

Where;L=length L,A=cross section area: material of resistivity

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Resistance

The longer a wire, the harder it is to push electrons through it

The greater the cross-sectional area, the “easier” it is to push electrons through it

The greater the resistivity, the “harder” it is for the electrons to move in the material

Resistance of wire (or other device) measures how easily charge flows through it

LR

A

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Resistance

Caution!!! Resistivity = material’s propertyResistance = device property

R = L / A

A = L / R

A = (d/2)2 geometry!

(d/2)2 = L / R

(d/2)2 = L / R

d/2= ( L / R )½ don’t skip steps!

d = 2 ( L / R )½22

Example 5: Suppose you want to connect your stereo to remote speakers.(a) If each wire must be 20 m long, what diameter copper wire should you use to make the resistance 0.10 per wire.

V = I R

d = 2 [ (1.68x10-8) (20) / (0.1) ]½ m

d = 0.0021 m = 2.1 mm

V = (4.0) (0.10) V

V = 0.4 V

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(b) If the current to each speaker is 4.0 A, what is the voltage drop across each wire?

Most electric circuits use elements callsed resistors to control the current flowing through the circuit. Symbol we use for a “resistor:”

In principle, every circuit component has some resistance

All wires have resistance

For efficiency, we want wires to have low resistance

In idealized problems, consider wire resistance to be zero

Lamps, batteries, and other devices in circuits also have resistance

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Values of the resistance are normally indicated by color-coding as

shown in Tables 24.2 and 24.3.

Resistors in circuits

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How to Read the Color-codingFirst find the tolerance band; it will typically be gold (5%) or silver (10%), and sometimes colorless (20%).

Resistors are often intentionally used in circuits. The picture shows a strip of five resistors (you tear off the paper and solder the resistors into circuits).

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Many materials have resistivities that depend on temperature. We can model this temperature dependence by an equation of the form

0 01 T T ,

where 0 is the resistivity at temperature T0, and is the temperature coefficient of resistivity.

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0 0R R 1 T T

Temperature Dependence of Resistivity

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Resistance thermometers made of carbon (inexpensive) and platinum (expensive) are widely used to measure very low temperatures.

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Example 6: At 20 0C, a copper wire has a diameter of 4 mm, a length of 10m, a resistivity of 1.7×10−8 .m, a temperature coefficient of

resistivity of 3.9×10−3 (0C)-1, and carries a current of 1A.(a) What is the current density in the wire?(b) What is the magnitude of the electric field applied to the wire?(c) What is the potential difference between the two ends of the wire?(d) What is the resistance of the wire?(e) When the wire is used in a thermometer for measuring the melting

point of indium, the resistance calculated in part (d) increases to 0.0207 . Find the melting point temperature of indium.

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(b) The electric field is given by:

(a) The current density in a copper wire of radius 2mm is:

(c) The potential difference will be given by:

(d) The resistance of the wire is:

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e) Solving for T, we get

When a battery is used to establish an electric current in a light bulb, the battery transforms its stored chemical energy to kinetic energy of the electrons.

These electrons flow through the filament of the light bulb, and result in an increase in the temperature of the filament.

It is important to calculate the rate of this energy transfer.

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Electric Power

Imagine, a positive charge dQ flowing clockwise from point a

through the battery and the resistor, and back to the same

point a.

In a time interval dt a quantity of charge dQ enters point a,

and an equal quantity leaves point b.

Thus, the electric potential energy of the system increases

by the amount

dU =dQ V

The rate at which the system loses energy as the charges

pass through the resistor is:

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Electric Power

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This rate is equal to the rate at which the resistor gains

internal energy, and is defined as the power P:

Using the relation V =I R for a resistor of resistance R, the

electric power P delivered in the resistor can be written in the

following form:

Units: 1 Watt = W = V A = (J/C) (C/s) = J/s

Example 7: A 220V potential difference is maintained

across an electric heater that is made from a nichrome

wire of resistance 20 .

(a) Find the current in the wire and the power rating of the

heater.

(b) At an estimated price of 0.35LE (Egyptian pound) per

kilowatt-hour of electricity, what is the cost of

operating the heater for 2 h?

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Thank You.

TAKE CARE YOUR SELVES.