Transcript Electricity

Electricity
Part 2: Electric Current
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
50%
1
50%
2
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
2
Which type of light bulb has a
larger resistance, a 100W bulb or a
60W bulb?
25%
25%
25%
2
3
25%
1. The 100 W bulb
2. The 60 W bulb
3. They will have the
same resistance
4. None of the above
1
4
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