6/1/2016
Batteries
Current and Resistance
Chapter 18
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
1
6/1/2016
• 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
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
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
The current through a wire is 1.90 A.
a. Calculate how long it would take 2280 C of
charge to flow (20 minutes)
b. Calculate the amount of charge that flows each
hour (6840 C)
c. How many electrons flowed in one hour? The
charge on an electron is 1.60 X 10-19 C. (4.28 X
1022 electrons)
2
6/1/2016
In one minute, 720.0 C of charge flows through a
wire in a car engine.
a. Calculate the current (12.0 A)
b. Calculate how many electrons flowed in one
minute. The charge on an electron is 1.60 X 1019 C. (4.50 X 1021 electrons)
In two minutes, 6.75 X 1021 electrons flowed
through a flashlight circuit.
The total amount of charge that enters a wire is
given by the expression Q = 2t2 – t, where t is in
seconds.
a. Determine the expression for the current in the
wire.
b. Calculate the current at 2 seconds and at 4
seconds.
c. Determine the expression for the current using
Q = 1.5sint as the expression.
Ohm’s Law
a. Calculate how many coulombs flowed through
the circuit (1080 C)
b. Calculate the current (9.00 A)
V = IR
DV = IR
V = Voltage (V)
I = Current (A)
R = Resistance (Ohms, W)
(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 W)
b. If the voltage dropped to 1.2 V and the
resistance stayed at 5.0 W, what
current would flow. (240 mA)
3
6/1/2016
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
What is the resistance of a 2.00mm diameter,
10.0 meter copper wire?
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 = 0.0535 W or 53.5 m W
A wire of length L is stretched to twice its
normal length.
a. Calculate the new cross sectional area (assume
the volume does not change (Anew =1/2A)
b. Calculate the new resistance (Rnew = 4R)
The resistance of nichrome wire in a toaster is 9.8
Ω. The wire is 220 cm long and the resistivity is
110x10-8 Ω•m. Calculate the diameter of the
wire.
(0.56 mm)
4
6/1/2016
A wire of uniform cross-section has a resistance of
R . What would be the resistance of a similar
wire, made of the same material, but twice as
long and of twice the diameter?
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
The resistance of a platinum resistor at 20oC is
164.2 W. Calculate the resistance at 56.0 oC.
• Superconductivity – resistance of a material
becomes zero
• No loss of current over a wire
• Generally near absolute zero
• Record as of 2015 is 203 K
• Maglev trains
Electric Power
• Watt
• 1 Watt = 1 Joule
1 second
P = I2R
P = IV
P = V2
R
5
6/1/2016
Calculate the resistance of a 40-W auto headlight
that operates at 12 V.
An electric heater draws 15.0 A on a 120-V line.
How much power does it use?
P = I2R
P = I2R
V = IR so R = V/I
P = I2V
I
P = IV = (15 A)(120V) = 1800 W or 1.8 kW
(3.6 W)
A lightning bolt can transfer 109 J of energy at a
potential difference of 5 X 107V over 0.20 s.
What is the charge transferred?
V = PE/Q
Q = E/V = 109 J/ 5 X 107V = 20 C
Household Electricity
• Kilowatt-hour
• You do not pay for power, you pay for energy
1 kWh = (1000 J)(3600 s) = 3.60 X 106 J
1s
What is the current?
I = DQ/Dt
I = 20 C/0.2 s = 100 A
What is the power?
P = I2R
V = IR so R = V/I
P = I2V
I
P = IV = (100 A)(5 X 107V ) = 5 X 109 W
How much does it cost to run an 1800 W dryer
for 30 days if it operates 3.0 h per day and the
electric company charges 10.5 cents per kWh?
Hours = 30 days X 3.0 h/day = 90 h
Cost = (1.80 kW)(90 h)($0.105/kWh) = $17
6
6/1/2016
Household Electricity
• Circuit breakers – prevent “overloading” (too
much current per wire)
• Metal melts or bimetallic strip expands
Household Electricity:
Ex 1
Determine the total current
drawn by all of the
appliances shown.
P = IV
I = P/V
Ilight = 100W/120 V = 0.8 A
Ilight = 100W/120 V = 0.8 A
Iheater = 1800W/120 V = 15 A
Istereo = 350W/120 V = 2.9 A
Ihair = 1200W/120 V = 10.0 A
You leave a 60 W lightbulb on outside your
house for 12 hours a day for an entire month
(30 days). How many kiloWatt hours will you
be charged for?
Itotal = 0.8A + 15.0A + 2.9A + 10.0A = 28.7 A
This would blow the 20 A fuse
You decide to turn your computer (75 W) off
every night. Assume that it is off for 9 hours
every night for 30 days. How many kilowatthours did you save?
Suppose you are charged for 3000 kWh for a
given month. How many Joules of energy did
you use that month? (Hint: Start from the
power formula)
7
6/1/2016
Jump Starting a Car
DC vs AC
POSITIVE TO POSITIVE
(or your battery could explode)
DC
AC
•Electrons flow constantly
•Electrons flow in short burst
•Electrons flow in only one
direction
•Electrons switch directions
(120 times a second)
•Batteries
•House current
DC vs AC
DC vs AC
• DC = Direct current
– Electrons flow constantly
– Electrons only flow in one direction (negative to
positive)
– Batteries provide DC current
http://www.ibiblio.org/obp/electricCircuits/AC/AC_1.html
DC vs AC
• AC = Alternating current
–
–
–
–
Electrons switch directions
“Pulsed current”
Home electricity
More efficient for power transmission over large
distances
– USA uses 60 Hertz (60 cycles per second), many
other countries use 50 Hz
AC Current
I = Iosin2pft
Io = Peak current (highest possible)
Vrms = root-mean square voltage
Irms = root-mean square current
8
6/1/2016
In the United States, the rms voltage is 120 V.
Calculate the peak voltage. (170 V)
In Europe, the rms voltage is 240 V. Calculate the
peak voltage. (340 V)
A hair dryer has a power of 1000-W and is
connected to a 120 V line (rms).
a. Calculate the rms current that runs through the
dryer. (8.33 A)
b. Calculate the resistance of the dryer.
c. Calculate the power through the dryer if you
plug it into a 240 V line in Europe.
A stereo reciever has an average power of 100.0
W and a resistance of 8.00 W
a. Calculate the rms voltage. [28.3 V]
b. Calculate the rms current. [3.534 A]
c. Calculate peak voltage and peak current. [40.0
V, 5 A]
d. Calculate the rms voltage if you turn down the
speaker so it is only 1.00 W. [2.82 V]
An AC source has an rms voltage of 200.0 V.
The resistance of the circuit 20.0 W
a. Calculate rms current [10.0 A]
b. Calculate peak current [14.1 A]
c. Calculate the average power [2000 W]
d. Calculate the maximum power [4000 W]
An AC source has an rms Voltage of 17.0 V and is
connected to a circuit with 265 W of resistance.
a. Calculate peak voltage [24.0 V]
b. Calculate the rms current [64.1 mA]
c. Calculate the average power [1.09 W]
d. Calculate the maximum power. [2.18 W]
9
6/1/2016
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 the 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.
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
26.
30.
32.
34.
36.
38.
40.
48.
50.
b) 2.1 X 10-14 s
b) 5.3 X 1018 s-1
b) 6.5 X 10-5 m/s
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
10
6/1/2016
54.a)
b)
d)
64.a)
c)
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
11
6/1/2016
12
6/1/2016
13