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Batteries
Current and Resistance
Chapter 31
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
Circuits
-e
+
e
e e
Chemical
Reaction
that
produces
electrons
Chemical
Reaction
that absorbs
electrons
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
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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 = DQ
Dt
I = dQ
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)
I = eie
electron current
Current Density (J)
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A 1.0 A current passes through a 1.0 mm
diameter aluminum 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 copper wire.
a. Calculate the current density. (6.2 X 105 A/m2)
b. Calculate the drift speed of the electron. (0.046
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 = DQ
Dt
DQ = IDt
DQ = (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 (s)
Resistivitiy (r)
Current density
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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
DV = IR
V = Voltage (V)
I = Current (A)
R = Resistance (Ohms, W)
Resistors
• Color coded to determine resistance
• Devices that heat have high resistance (light
bulbs, electric stoves, toasters)
(only works for metal conductors, not
semiconductors (nonohmic))
A small flashlight bulb draws 300 mA
from a 1.5 V battery.
a. Calculate the resistance of the bulb
(5.0 W)
b. If the voltage dropped to 1.2 V and the
resistance stayed at 5.0 W, what
current would flow. (240 mA)
Resistivity
R = rL
A
R = Resistance
L = Length (longer wire, greater resistance)
A = Area (wider wire, less resistance)
r = Resistivity of the material
http://www.earthsci.unimelb.edu.au/ES304/MODULES/RES/NOTES/resistivity.html
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What is the resistance of a 2.00mm diameter,
10.0 meter copper wire?
R = 0.0535 W of 53.5 m W
A speaker wire must be 20.0 m long and have a
resistance of less than 0.100 W 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)
R = rL
A
A = rL
R
A = [(1.68 X 10-8 Wm)(20.0 m)]/0.100 W
A = 3.36 X 10-6 m2
A = pr2
r = (A/p)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
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 = 4R)
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
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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
The total charge that passes through a wire is
given by the following function: Q = (3t2 +
2t)C.
a. Sketch a quick graph for the interval 0-5
seconds.
b. Determine the expression for current in the wire
at any given time.
c. Calculate the current at 2 seconds.
The current in a wire at a given time is given by
the function I = 5t2.
a. Determine the expression for the charge.
b. Determine th total charge that has passed
through the wire from 0 to 5 seconds
The current in a wire at a given time is given by
the function I = 5e-t/4ms
a. Sketch a graph of I from 0 to 8 ms
b. Determine the expression for the charge.
c. Determine the total charge that has passed
through the wire from 0 to 8 ms
A 3.0 V flashlight has a
resistance of 15.0
ohms. The voltage
decreases according to the
following graph.
a. Determine the formula for
the current that passes
through the wire in that
time interval.
b. Determine the formula for
the total charge moved by
the batteries.
c. Calculate the total charge
that moved in those 5.0
hours.
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2. 3.0 days
4. 2.4 X 1019 s-1
6. 0.023 N/C
8. 4.3 X 10-12 m
10. a) 7.43 X 10-6 m/s
12. a) 1.73 X 107 A/m2
14. 0.141 mA
16. 3.2 mA
18. a) 6.3 X 105 A/m2
22. 1.68 A
24. 5.0 X 10-8 Wm
54.a)
b)
d)
64.a)
c)
b) 2.1 X 10-14 s
b) 5.3 X 1018 s-1
b) 6.5 X 10-5 m/s
26.
30.
32.
34.
36.
38.
40.
48.
50.
a) 1.64 X 10-3 N/C
a) 1.00
5.5 X 10-8 Wm
a) 3.0 W, 30 m
5.5 W
50 W
0.104 V/m
3.2 X 105 C
0.50 mm
b) 1.10 X 10-5 m/s
b) 0.50
b) 1.0 A
0. 7.87, 12.6, 17.3, 19.0, 19.6, 19.9
(10A)e-t/2
c) 10.0 A at t=0 s
10, 6.07, 3.68, 1.35, 0.50, 0.18, 0.07
4.2 X 105 A
b) 1.80 X 105 V/m
Current decreases
d) 1.2 X 10-5 J
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