Transcript DC Circuits (Chapter 28) - McMaster Physics and Astronomy

Electrical Work and Power
Electrical Work and Power
I
+
Higher V1
Resistance R
I
Lower V2
Current I flows through a potential difference DV
Follow a charge Q : at positive end, U1 = QV1
at negative end, U2 = QV2
P.E. Decreases:
DU  QDV  0
The speed of the charges is constant in the wires and
resistor.
What is electrical potential energy converted to?
Electrical resistance converts electrical potential
energy to thermal energy (heat), just as friction in
mechanical systems converts mechanical energy to
heat.
This thermal energy means the atoms in the
conductor move faster and so the conductor gets
hotter.
The average kinetic energy of the electrons
doesn’t increase once the current reaches a
steady state; the electrons lose energy in
collisions with the atoms as fast as it is supplied
by the field.
Power dissipated by a resistor:
 P
DU (DQ)V

 IV
Dt
Dt
Power dissipated= current x potential difference
Units: 1 volt x 1 amp = 1 watt (= 1 J/s)
For Resistor: V = IR, so there are 2 other equivalent
formulas:
2
V
P  VI  I R 
R
2
This power dissipated is called “Joule heating” in a resistor
Examples
a) What is the resistance of a “60 watt” bulb? (for
a 120-V supply)
b) Find R for a 60-W headlamp (12-V battery).
c) What power do you get from a “60-W” household
bulb if you connect it to a 12-V car battery?
“Electromotive force
ε ” (emf)
ε  external work per unit charge
Units: J/C = volts (not actually a force)
but it “pushes” the charges
through the circuit.
+

I
Eg: Battery
(chemical energy  electrical energy)
Generator
(mechanical energy  electrical energy)
I=2A
ε=12 V
R
When current leaves the battery
the battery supplies power
equal to: P = Iε = 24W
If current were forced to enter
the battery, (as in charging it)
then it absorbs the same power
What does the energy balance
look like in this circuit?
12V
R
 6 so Joule heating=I R  24W
2A
2
Resistor: Electrical energy  heat
Real Batteries
I
A
r

VB +
r = “internal resistance” of the battery
I
I
RL (external resistance, “load”)
B
 -Ir=V
A
VA – VB = V = “terminal voltage”  measured
   Ir  (VA  VB )
ε - Ir = V
“Terminal voltage”
Example
A battery has an emf of 12V and an internal resistance of 0.05Ω. Its
terminals are connected to a load resistance of 3Ω. Find:
a) The current in the circuit and the terminal voltage
b) The power dissipated in the load, the internal resistance, and the
total power delivered by the battery
Example
Show that the maximum power lost in the load resistance R occurs
when R=r, that is, when the load resistance matches the internal
resistance of the battery.
Example
Automobile battery:
At terminals
Find:
12.8 V (with 20 A current)
9.2 V (with 200 A current)
E and rinternal of battery
Resistance and Temperature
Over a limited temperature range, the resistivity of a
metal varies approximately linearly with T according to:
ρ=ρo[1+α(T-To)]
where T is in oC and ρ is the resisitivity.
To is usually taken to be 200C and temperature coefficient
is given by:
1 D

o DT
where D    o , DT  T  To
Example
A resistance thermometer measures temperature by measuring the
chance in resistance of a conductor. Made of platinum with a
resistance of 50.0Ω at 20.0oC is immersed in melting indium and
its resistance increases to 76.8Ω.
Find the melting point of indium.
Example
What is the fractional change in the resistance of an iron
filament when its temperature changes from 25.0°C to
50.0°C?