Transcript PowerPoint

Today’s agenda:
Potential Changes Around a Circuit.
You must be able to calculate potential changes around a closed loop.
Emf, Terminal Voltage, and Internal Resistance.
You must be able to incorporate all of the above quantities in your circuit calculations.
Electric Power.
You must be able to calculate the electric power dissipated in circuit components, and
incorporate electric power in work-energy problems.
Examples.
Electric Power
Last semester you defined power in terms of the work done by
a force.
dWF
PF 
dt
We’d better use the same definition this semester! So we will.
We focus here on the interpretation that power is energy
transformed per time, instead of work by a force per time.
energy transformed
P
time
The above equation doesn’t appear on your equation sheet, but
it should appear in your brain.
However, we begin with the work aspect. We know the work
done by the electric force in moving a charge q through a
potential difference:
Wif  Uif  qVif .
Using the above, the work done by the electric force in moving
an infinitesimal charge dq through a potential difference is:
dWif  dq Vif .
The instantaneous power, which is the work per time done by
the electric force, is
dWif
dq Vi f
P

.
dt
dt
Let’s get lazy and drop the  in front of the V, but keep in the
back of our heads the understanding that we are talking about
potential difference. Then
dW
dq
P
  V.
dt
dt
But wait! We defined I = dQ/dt. So
P  IV.
And one more thing… the negative sign means energy is being
“lost.” So everybody writes
P  IV
and understands that P<0 means energy out, and P>0 means
energy in.
Also, using Ohm’s “law” V=IR, we can write P = I2R = V2/R.
I can’t believe it, but I got soft and put P = I2R = V2/R on
your starting equation sheet.
Truth in Advertising I. The V in P=IV is a potential
difference, or voltage drop. It is really a V.
Truth in Advertising II. Your power
company doesn’t sell you power. It sells
energy. Energy is power times time, so a
kilowatt-hour (what you buy from your
energy company) is an amount of energy.
Example: an electric heater draws 15.0 A on a 120 V line. How
much power does it use and how much does it cost per 30 day
month if it operates 3.0 h per day and the electric company
charges 10.5 cents per kWh. For simplicity assume the current
flows steadily in one direction.
What’s the meaning of this assumption about
the current direction?
The current in your household wiring doesn’t flow in one
direction, but because we haven’t talked about current other
than a steady flow of charge, we’ll make the assumption. Our
calculation will be a reasonable approximation to reality.
An electric heater draws 15.0 A on a 120 V line. How much
power does it use.
P  IV
P  15 A 120 V   1800 W = 1.8 kW
How much does it cost per 30 day month if it operates 3.0 h
per day and the electric company charges 10.5 cents per kWh.
 3 h   $0.105 
cos t  1.8 kW  30 days  


 day   kWh 
cos t  $17.00
How much energy is a kilowatt hour (kWh)?
1 kW 1 h   1000 W 3600 s 
J

 1000   3600 s 
s

= 3.6 106 J
So a kWh is a “funny” unit of energy. K (kilo) and h (hours) are
lowercase, and W (James Watt) is uppercase.
How much energy did the electric heater use?
Paverage 
Wdone by force
time
Energy Transformed

time
Energy Transformed   Paverage   time 
 3 h used   3600 s 
J

Energy Transformed  1800   30 days  


s
day
h





Energy Transformed  583, 200, 000 Joules used
Energy Transformed  583, 200, 000 Joules used
That’s a ton of joules! Good bargain for $17. That’s about
34,000,000 joules per dollar (or 0.0000029¢/joule).
OK, “used” is not an SI unit, but I stuck it in there to help me
understand. And joules don’t come by the ton.
One last quibble. You know from energy conservation that you
don’t “use up” energy. You just transform it from one form to
another.