Concept 16.1: A battery changes chemical
energy into electrical energy. It creates an emf.
Two reactions occur, which release energy
16. D.C. Circuits
initial chemical
potential energy
final chemical
potential energy
1
VA – VC = - I r (internal loss)
VA − VB = ε − Ir
Terminal voltage
emf
I
A
r = “internal r
resistance”
C•
ε
I
+
-
RL
I
2
Quiz of concept 16.2
A battery with emf ε and internal
resistance r provides a current
through the resistance R of a
lightbulb. What power is delivered to
the lightbulb?
Battery
-----
emf in V
∆U chem
≡ε
q
Physics 1E03 Lecture 16
Concept 16.2: A real battery can be modeled
as an ideal battery plus an internal resistance.
ε = emf
B
i
f
− U chem
= ∆U chem = ∆E + q∆V
U chem
Physics 1E03 Lecture 16
VC – VB = ε (but point “C”
is not accessible for
measurement)
B-
energy released
Reaction stops at ∆E = 0, ∆V =
VA – VB = “terminal voltage”
A+
ion solution
Charge builds up on the battery terminals, creating a
potential difference ∆V. Energy balance gives
Serway and Beichner
Sections 28.1-.3
To provide charge, the reaction must occur.
Thus ∆E = 0, and the full emf does not
appear at the terminals. This can be
modeled by an internal resistance r.
A
A0 + ∆EA
eA- + A+
BB0 + eB- + ∆EB
__________________________
eA- + A+ + BA0 + B0 + eB- + ∆E
• electromotive “force”
• series and parallel resistors
• Kirchhoff’s Laws
+++++
a) ε2/R
r
R
•
ε
+
-
b) ε2/(R+r)
c) ε2R/(R+r)2 d) ε2r/(R+r)2
B
Internal loss
modeled by
Ohm’s Law
Physics 1E03 Lecture 16
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Physics 1E03 Lecture 16
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Concept 16.3: Resistors in series add simply,
but resistors in parallel add by reciprocals.
R2
R3
ε
ε
I1
Reff
=
I
I
I
R1
I
1
2
1
ε
ε = V =V =V
1
2
3
3
ε
Reff
= I(R2+R2+R3) = IReff
Reff = R1 + R2 + R3
Physics 1E03 Lecture 16
a) 7V/11R
c) 4V/11R
R1
V
b) 11V/7R
d) 11V/4R
R4
R5
Reff
=
ε
R1
+
ε
R2
+
ε
R3
Physics 1E03 Lecture 16
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Concept 16.4: Conservation of charge and
mechanical energy allow the detailed analysis
of circuits.
R3
R2
=ε
R3
1
1
1
1
= +
+
Reff R1 R2 R3
5
Quiz of concept 16.3
What current is
provided by the
voltage source?
I3
and I = I1+ I2 + I3
3
2
R2
The voltage across each resistor in parallel is the same.
The current through each resistor in series is the same.
ε =V +V +V
ε = IR + IR + IR
I2
I
Series:
R1
Parallel:
Junction Rule: conservation of charge.
R6
I1
All resistors are
equivalent
I2
I3
I1 = I2 + I3
Loop Rule: conservation of energy.
∑ q × ( ∆Vi ) = 0
Physics 1E03 Lecture 16
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Sum of voltage changes
around any closed loop is
zero.
Physics 1E03 Lecture 16
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2
Loop analysis of a circuit:
ε1
c
R2
R3
R
A
-
ε2
a
B
R
+
ε
d
I
b
VB-VA= ∆VBA
I
R1
-Q
+Q
∆VBA = -IR
∆VBA = +IR
∆VBA =ε
∆VBA = Q/C
C
Physics 1E03 Lecture 16
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Quiz of concept 16.4
When the switch in the
diagram is closed, the
current through the
switch
a)
b)
c)
d)
moves left to right
moves right to left
is zero
approaches infinity
(there is no resistance)
c
100Ω
250 Ω
12 V
a
b
250 Ω
400Ω
d
Physics 1E03 Lecture 16
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Summary
• an emf converts electrical potential energy from another
source, like chemical potential energy in a battery
• the currents and potential drops across resistors in a
circuit can be found using the loop equations:
1) sum of currents at a junction is zero
2) sum of potential drops about a circuit is zero
Practice problems: Chapter 28, #4, 6, 15, 19
Next lecture: read sections 28.4
Physics 1E03 Lecture 16
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