Transcript Electricity

Electricity
Part 2: Electric Current
A conductor/dielectric is
1. A material in which
charges are free of
move about
2. A material in which
charges are NOT
free to move about
3. A material that
shields a region
from electric fields
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When a charged particle with charge, Q, is
moved through a distance, d, in a uniform
electric field E, the work done is given by
1. QE/d
2. Qd/E
3. QEd
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The Unit for electric fields is
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Volt
Newton/Coulomb
Volt/meter
All of the above
1 and 2
2 and 3
None of the above
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Electric current is the amount of
charge moving past a point
Maxwell Demon counting the
charges passing a given point
Definition of current
I = current
Q = amount of charge that passes point.
t = time for charge to pass by.
Q
I
t
Units of Current
 Q  Coulomb 
I


 t  sec 
1 Ampere (Amp) = 1 Coulomb/second
Electric currents only flow in
wires.
1. True
2. False
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Examples of Currents
Solar wind interacting with
the earths magnetic field
1) Coronal Mass Ejection
2) Aurora from space
3) Aurora from ground
CME animation
Aurora from space
Aurora effects
3 electron beams in a color TV
Tokamak Fusion Experiments
JET discharge
We usually think of currents in wires
Opening switch
Simple circuit
When a charged
particle passes
through the battery, it
gains energy.
When the particle
passes through the
light bulb it gives up
the energy as heat.
Ohm’s Law
V=IR
V= Voltage of the
Battery.
I=current in circuit.
R=Resistance in the
bulb/resistor.
(Depends on
materials and
geometry.)
Units of Resistance
R=V/I (volts/amps)
By definition, 1Ohm = 1 volt/amp,
or
1=1V/A.
Log Ride Analogy
Water circuit analogy
Example problem
How many amps of current would flow in
a light bulb that has a resistance of 60 if
it is connected to a 12 V battery.
I V/R
12V
I
 0.2 A
60
Power in a circuit
When Charge Q
passes through the
battery it gains an
amount of energy
E=(Q)V
(This is the amount of
work the battery does
on the charge.)
If the charge takes an amount of time t to
pass through the battery, the battery supplies
a power of (does work at a rate of)
E Q
P

V
t
t
But
Q
 I
t
Thus
P  IV
The power supplied by the battery must be
dissipated in the resistor.
We also know the V=IR.
Power dissipated in resistor
P  IV  I ( IR)
P I R
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Example Calculation
What is the resistance
and how much
current flows through
a 100 W bulb?
Note: The wattage on a
bulb is its power
output and assumes
that you will use it in
the US where the
voltage in 110 V.
I  P /V
I  100W

0
.
91
A
110V
RV/I
R  110V

121
0.91 A
Redo the calculations for a 60 W bulb
I  P /V
I  60W 110V  0.55 A
RV/I
R  110V 0.55 A  200
If we want to increase the power
output of a light bulb we should
1. Increase the
resistance of the
filament
2. Decrease the
resistance of the
filament
3. None of the
above.
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If we were to use a bulb made for the US
(where V=110V) and use it in Europe
(V=220V) what would happen to the current
in the bulb relative to68%
the US value.
1. The current would
double
2. The current would
be cut it half
28%
3. The current would
remain the same.
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What would happen to the
power dissipated by the bulb?
1. It would quadruple
2. It would double
3. It would not
change
4. It would be cut in
half
5. It would be cut by
a factor of 4.
44%
33%
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