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PHYS 2421 - Fields and Waves
25.1 Current
25.2 Resistivity
25.3 Resistance
25.4 Electromotive force and circuits
25.5 Energy and power in circuits
25.6 Theory of metallic conduction
Batteries and power supplies are examples of
“Electromotive forces”
To have a current, a pump and a circuit are needed
Analogy Electric circuit View of flow
Current flows equally in all points of circuit
It does not get “used up”
EMF
Electromotive forces (EMF)
• are not forces (wrong name!)
• are electric potentials
• are measured in volts
• examples:• batteries
• solar cells
• electric generators
• thermocouples
• fuel cells
• Ideal EMF maintains same Vab
independent of the current flowing
abV = EIdeally, EMFs should not have any resistance:
But in practice they do: abV Ir= E
r is the internal resistance of EMF, the is reduced by Ir
0I I No current flows through A= ?
0 12 Vab abV V Ir r= ? = E E E
How to measure?Use voltmeters and ammeters
Assumptions
• Voltmeters draw zero current• Ammeters have zero resistance
Current?
Voltage between a and b??
122
2 4
V AI
r R
E
V = ?Remember: voltmeters draw zero current
and ammeters have zero resistance
Equivalent
diagram
' ' 2 4 8 A Va bV IR
12 2 8 V
or
A 2 V
abV IrE
I = ?
Homework : Problem 25.33 (11th Ed.) or 25.35 (12th Ed.)
V = ?I = ?
Homework : Problem 25.35 (11th Ed.) or 25.37 (12th Ed.)
Equivalent
diagram
0I
12
2 4
2
V
A
Ir R
E
V = ?I = ?
126
2 0
V AI
r R
E
12 6 2
0
V A
V IrE
Equivalent diagram
Homework : Problem 25.37 (11th Ed.) or 25.39 (12th Ed.)
Voltage rises and drops in a circuit
Voltage
increases
by 12 V
Voltage drops by
Ir=2 A x 2 =4 V
Voltage drops by
IR=2 A x 4 = 8 V
Summary of Section 25.4
Hmwk Sect. 25.4: Probls 25.33, 25.35 and 25.37 (11th Ed.) or
25.35, 25.37 and 25.39 (12th Ed.)
Electromotive forces (EMF)
• are not forces (wrong name!)
• are electric potentials
• are measured in volts
• Have a small internal resistance
abV Ir= E
Assumptions• Voltmeters draw zero current• Ammeters have zero resistance
Power = energy output per unit time
W Fd qEd qP V IV
t t t t
Power consumed by a resistor:
Units? Amp Volt WattC J J
Ws C s
2 2 /P IV I R V R
Power output of a source:
2P IV I Ir I I rE E
Power input to a source:
2P IV I Ir I I rE E
Rate of energy conversion in battery
122
2 4I
r R
V A
E
V = ?
I = ?
' ' 2 4 8 A Va bV IR
2 8 16P I A V WE
Rate of energy dissipation in battery22 2 2 8P I Ir I r A W
Net power output of battery2 16 8P I I r W W=8 WE
Hmwk : Problem 25.45 (11th Ed.) or 25.47 (12th Ed.)
Rate of energy dissipation in resistor22 2 4P I R A =16 W
121.2
2 8I
r R
V A
E
V = ?
I = ?
' ' 1.2 9.6a bV IR A 8 V
Rate of energy dissipation in resistor22 1.2 12P I R A 8 W
Increasing R reduces power consumption by resistor !
Question: which light bulb has a filament with a larger resistor, a 50 W or a 100 W?
Rate of energy dissipation in resistor22 2 4P I R A =16 W
Change R . . .
Hmwk : Problem 25.49 (11th Ed.) or 25.53 (12th Ed.)
Rate of energy conversion in battery
126
2
V A
I
r
EI = ?
6 12 72 A V WP IE
Rate of energy dissipation in battery22 6 2 72 A WP I Ir I r
Net power output of battery2 72 72 W W=0 WP I I rE
Summary of Section 25.5
Hmewk for Sect. 25.5: Probls 25.45, 25.49 (11th Ed.) or Probls 25.47, 25.53 (12th Ed.)
Power = energy output per unit time P IV
Power consumed by a resistor:
Units: W= WattJ
s2 2 /P IV I R V R
Power output of a source:2P IV I Ir I I rE E
Power input to a source: 2P IV I Ir I I rE E
Electrons are pushed by F=qE
They move and collide with atoms.
The less collisions the better the conduction
Volume ddQ q n qnAv dt
Before we obtained:
d
dQI nqv A
dt d
IJ nqv
A Current density
0d
qEF ma qE a v v
m
qEa
m
The drift velocity is:
Where is the average time between collisions
The longer , the better the conduction 2
d
nqJ nqv E
mThe current density becomes:
A microscopic view of resistivity
And since
Where m and q are the mass and charge of the electrons, n is
the charge density, is the average time between collisions
The longer is, the smaller the resistivity
2 2/
V EL IR I L IE J
V IR L L A A
E E m
J nq E m nq
2
m
nq
Hmwk for Section 25.5: Probl 25.52 (11th Ed.) or 25.56 (12th Ed.)
2
d
nqJ nqv E
m
Hmwk for Ch. 25:
Probs. 1, 3, 7, 11, 13, 27, 33, 35, 37, 45, 49 and 52 (11th Ed.) orProbs. 1, 5, 9, 17, 13, 27, 35, 37, 39, 47, 53 and 56 (12th Ed.)
2
m
nq