Current and Resistance

Current and Resistance

Chapter 31

Batteries

Battery

• Batteries provide Chemical Electricity

• Electrons “bunch up” or have the potential to flow from the negative end

• Electrons can’t flow in an isolated battery

+-

Chemical Reaction that produces electrons

Chemical Reaction that absorbs electrons

e

e

e

e

Circuits

Drift Speed

• Electrons do not flow through wires like pipes

• Electric field gives direction to the random motion of electrons. (vD = drift speed)

• 0.05 mm/s

• About 5 ½ hours to travel one meter (coin waterfall at Chuck-E-Cheese)

• About a year to go 1 mile

Electron Current (ie)

ie = neAvD

ne = electron density

Calculate the electron current in a 2.0 mm diamter copper wire if the electron drift speed is 1.0 X 10-4 m/s. (2.7 X 1019 s-1)

• Conventional Current – Flows positive to negative– Opposite of electron flow (electron current)

Current (I)

• Current – Net amount of charge per unit time

• 1 coulomb/second = 1Ampere

I = Q

t

I = dQ I = eie electron current

dt

The electron current through a wire is 1.2 X 1019

electrons/s.

a. Calculate the current, I (1.9 A)

b.Calculate the amount of charge that flows each hour (6800 C)

Current Density (J)

A 1.0 A current passes through a 1.0 mm diameter wire.

a. Calculate the current density. (1.3 X 106 A/m2)

b.Calculate the drift speed of the electron. (0.13 mm/s)

A 5.0 A current passes through a 3.2 mm diameter wire.

a. Calculate the current density. (6.2 X 105 A/m2)

b.Calculate the drift speed of the electron. (0.05 mm/s)

Current: Ex. 1

A steady current of 2.5 A flows through a wire for 4.0 min. How much charge passed through any point in the circuit?

I = Q

t

Q = It

Q = (2.5 A)(240 s) = 600 C

How many electrons would this be?

1 electron = 1.60 X 10-19 C

600 C 1 electron = 3.8 X 1021electrons

1.60 X 10-19 C

Current Density (J)

Conductivity ()

Resistivitiy ()

Current density

A 2.0 mm diameter aluminum wire carries a current of 800 mA.

a. Calculate the current density using J = I/A (2.55 X 105 A/m2)

b.Calculate the electric field inside the wire (0.0072 V/m)

A copper wire has a diameter of 3.2 mm. The current is 5.0 A.

a. Calculate the current density of the wire (6.2 X 105 A/m2)

b.Calculate the electric field inside the wire (0.01 V/m)

Ohm’s Law

V = IR V = IR

V = Voltage (V)

I = Current (A)

R = Resistance (Ohms, )

(only works for metal conductors, not semiconductors (nonohmic))

Resistors

• Color coded to determine resistance

• Devices that heat have high resistance (light bulbs, electric stoves, toasters)

A small flashlight bulb draws 300 mA from a 1.5 V battery.

a. Calculate the resistance of the bulb (5.0 )

b.If the voltage dropped to 1.2 V and the resistance stayed at 5.0 , what current would flow. (0.24 A)

Resistivity

R = L

A

R = Resistance

L = Length (longer wire, greater resistance)

A = Area (wider wire, less resistance)

= Resistivity of the material

http://www.earthsci.unimelb.edu.au/ES304/MODULES/RES/NOTES/resistivity.html

What is the resistance of a 2.00mm diameter, 10.0 meter copper wire?

A = r2 = (3.14)(0.001 m)2 = 3.14 X 10-6 m2

R = L = (1.68 X 10-8 m)(10.0 m)

A (3.14 X 10-6 m2)

R = 0.0535 of 53.5 m

A speaker wire must be 20.0 m long and have a resistance of less than 0.100 per wire.

a. What diameter copper wire should be used? (2.06 mm)

b.What is the voltage drop across each wire at a current of 4.00 A? (0.40 V)

A wire of length L is stretched to twice its normal length.

a. Calculate the new cross sectional area (assume the volume does not charge (Anew =1/2A)

b.Calculate the new resistance (Rnew = R)

R = L

A

A = L

R

A = [(1.68 X 10-8 m)(20.0 m)]/0.100 A = 3.36 X 10-6 m2

A = r2

r = (A/)1/2

r = (3.36 X 10-6 m2 /3.14)1/2 = 1.03 X 10-3 m

D = 2r = 2.06 X 10-3 m or 2.06 mm

Resistance and Temperature

• Metals– Resistance increases with temp.– Atoms more disorderly– Interferes with flow of electrons

• Semiconductors– Resistance sometimes decreases with temperature– Some electrons become excited and able to flow

Superconductivity

• Superconductivity – resistance of a material becomes zero

• No loss of current over a wire

• Generally near absolute zero

• Record as of 2007 is 138 K

• Maglev trains