Download PHY 184

PHY 184
Spring 2007
Lecture 19
Title: Kirchoff’s Rules for Circuits
2/12/07
184 Lecture 19
1
Results of Midterm 1 (Sec 2)
 Pretty high average; will even be higher after the
correction set!
Midterm Section 1- average 63%
(9 correct problems out of 10 gives 100%)
30
27
26
23
23
23
19
20
Result
19
30
27
15
25
9
10
6
5
5
2
0
0
1
2
3
4
5
6
7
8
9
10
Number of Students
24
23
24
23
22
20
18
15
9
10
6
Correct Answers
5
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00
%
10
0.
9%
88
.8
8%
77
.7
7%
66
.6
6%
55
.5
4%
44
.4
3%
33
.3
2%
22
.2
1%
0
11
.1
3/9 or 33.3% --> 53%
5/9 or 55.6% --> 69%
2
00
%
Correction set: Example
0.
Number of Students
25
Percent Correct
184 Lecture 19
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Announcements
 Corrections Set 1 will open soon
• Corrections Set 1 is just Midterm 1 with different
numbers
 You will get 30% credit for those problems you did
incorrectly on Midterm 1
• However, you must do all the problems in Corrections Set
1 to get credit
• After Corrections Set 1 is due, we will open Midterm 1 so
that you can see your particular questions and answers
 This week we will continue with circuits and start
magnetism.
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Resistivity of Wires in a Circuit
 The circuit on the right shows a resistor of
resistance R=6 connected to an ideal battery
providing 12V by means of two copper wires.
Each wire has the length 20 cm and radius 1
mm. We generally neglect their resistance, the
voltage drop across them and the energy
dissipated. Justify by calculating the voltage
across R accounting for the influence of the
wire!
 Idea: For each wire
 Total resistance of the circuit is R+2Rw
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184 Lecture 19
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Resistivity of Wires in a Circuit (2)
 The circuit on the right shows a resistor of
resistance R=6 connected to an ideal
battery providing 12V by means of two
copper wires. Each wire has the length
20.cm and radius 1 mm. We generally neglect
their resistance, the voltage drop across
them and the energy dissipated. Justify by
calculating the voltage across R accounting
for the influence of the wire!
 The voltage across R is V=Ri
to be compared to 12 V if we neglect the wires. Neglecting the
resistance of the wire introduces an error of only 0.03%
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184 Lecture 19
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Clicker Question
 The circuit on the right shows a resistor of
resistance R=6 connected to an ideal battery
providing 12V by means of two copper wires.
Each wire has the length 20.cm and radius 1
mm. We generally neglect their resistance, the
voltage drop across them and the energy
dissipated. What is the total voltage drop
across the two wires (i=1.9993 A, R=0.0011
each)?
A) 5.5 10-4 V
B) 0.01875 V
C) 4.4 mV
D) 2.2 mV
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One wire: Vw = iR = 2.2 mV
184 Lecture 19
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Circuits
 We have been working with simple circuits
containing either capacitors or resistors.
 Capacitors wired in parallel
n
Ceq   Ci
i 1
 Capacitors wired in series
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184 Lecture 19
n
1
1

Ceq i 1 Ci
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Circuits (2)
 Resistors wired in parallel
 Resistors wired in series
n
1
1

Req i 1 Ri
n
Req   Ri
i 1
 We dealt with circuits containing either capacitors or
resistors that could be resolved in systems of parallel
or series components.
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184 Lecture 19
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Complex Circuits
 Circuits can be constructed that cannot be
resolved into series or parallel systems of
capacitors or resistors
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184 Lecture 19
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Kirchhoff’s Rules for Multi-loop Circuits
 To handle these types of circuits, we must
apply Kirchhoff’s Rules.
 Kirchhoff’s Rules can be stated as
• Kirchhoff’s Junction Rule
• The sum of the currents entering a junction must
equal the sum of the currents leaving a junction.
• Kirchhoff’s Loop Rule
• The sum of voltage drops around a complete circuit
loop must sum to zero.
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Kirchhoff’s Junction Rule
 Kirchhoff’s Junction Rule is a direct
consequence of the conservation of charge.
 In a conductor, charge cannot be created or
destroyed.
 At a junction: all charges streaming into the
junction must also leave the junction
i1
i2
i3
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184 Lecture 19
i1=i2+i3
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Kirchhoff’s Loop Rule
 Kirchhoff’s Loop Rule is a direct consequence of
the conservation of electric potential energy.
 Suppose that this rule was not valid
• we could construct a way around a loop in such a way that
each turn would increase the potential of a charge
traveling around the loop
• we would always increase the energy of this charge, in
obvious contradiction to energy conservation
 Kirchhoff’s Loop Rule is equivalent to the law of
energy conservation.
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184 Lecture 19
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EMF Devices - Directions
 An emf device (e.g., a battery) keeps the
positive terminal (labeled +) at a higher
electrical potential than the negative
terminal (labeled -).
 When a battery is connected in a circuit, its
internal chemistry causes a net current
inside the battery : positive charge
carriers move from the negative to the
positive terminal, in direction of the emf
arrow.
 Or: Positive charge carriers move from a
region of low electric potential (negative
terminal) to a region of high electric
potential (positive terminal)
 This flow is part of the current that is set
up around the circuit in that same direction.
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184 Lecture 19
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The half-reactions are:
At the cathode…
2 MnO2 + H2O + 2 e- —>Mn2O3 + 2 OHAt the anode…
Zn + 2 OH- —> ZnO + H2O + 2 e The overall reaction is:
Zn + 2MnO2 —> ZnO + Mn2O3 + [E=1.5 V]
Anode (negative terminal): Zinc powder
Cathode (positive terminal): Manganese dioxide (MnO2) powder
Electrolyte: Potassium hydroxide (KOH)
The flow of electrons is always from anode—to--cathode outside of the cell (i.e., in the
circuit) and from cathode—to--anode inside the cell. Inside a chemical cell, ions are
carrying the electrons from cathode—to--anode inside the cell.
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Alkaline battery
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Al Kaline batter
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Single Loop Circuits
 We begin our study of more complicated circuits by
analyzing circuits with several sources of emf and resistors
connected in series in a single loop.
 We will apply Kirchhoff’s Rules to these circuits.
 To apply these rules: establish conventions for determining
the voltage drop across each element of the circuit
depending on the assumed direction of current and the
direction of the analysis of the circuit.
 Because we do not know the direction of the current in the
circuit before we start, we must choose an arbitrary
direction for the current.
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Single Loop Circuits (2)
 We can determine if our assumption for the
direction of the current is correct after the
analysis is complete.
 If our assumed current is negative, then the
current is flowing in the direction opposite to
the direction we chose.
 We can also choose the direction in which we
analyze the circuit arbitrarily.
 Any direction we choose will give us the same
information.
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Single Loop Circuits (3)
 If we move around the circuit in the same direction as the
current, the voltage drop across a resistor will be negative.
 If we move around the circuit in the direction opposite to
the current, the voltage drop across the resistors will be
positive.
 If we move around the circuit and encounter a source of
emf pointing in the same direction, we assume that this
source of emf contributes a positive voltage.
 If we encounter a source of emf pointing in the opposite
direction, we consider that component to contribute a
negative voltage.
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184 Lecture 19
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Circuit Analysis Conventions
Element
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Analysis Direction
Current Direction
Voltage Drop


iR


iR


iR


iR

Vemf

Vemf

Vemf

Vemf
184 Lecture 19
i is the
magnitude of the
assumed current
20
Single Loop Circuits
 We have studied circuits with various networks of
resistors but only one source of emf.
 Circuits can contain multiple sources of emf as well as
multiple resistors.
 We begin our study of more complicated circuits by
analyzing a circuit with two sources of emf (Vemf,1 and
Vemf,2 ) and two resistors ( R1 and R2 ) connected in series
in a single loop
 We will assume that the two sources of emf have
opposite polarity.
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Single Loop Circuits (2)
 We assume that the current is flowing around the circuit in a
clockwise direction.
 Starting at point a with V=0, we analyze around the circuit in a
clockwise direction.
 Because the components of the circuit are in series, all components
have the same current, i
i
i
i
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Single Loop Circuits (3)
 Start at point a.
 The first circuit component is a source of emf Vemf,1, which
produces a positive voltage gain of +Vemf,1
 Next we find resistor R1, which produces a voltage drop V1 given
by -iR1
 Continuing around the circuit we find resistor R2, which
produces a voltage drop V2 given by -iR2
 Next we meet a second source of emf, Vemf,2
 This source of emf is wired into the circuit
with a polarity opposite that of Vemf,1
 We treat this component as a voltage drop
of -Vemf,2 rather than a voltage gain
 We now have completed the circuit
and we are back at point a.
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Single Loop Circuits (4)
 Kirchoff’s loop rule for the
voltage drops states
Vemf ,1  V1  V2  Vemf ,2  Vemf ,1  iR1  iR2  Vemf ,2  0
 Generalization: the voltage
drops across components in a
single loop circuit must sum to
zero.
 This statement must be
qualified with conventions for
assigning the sign of the voltage
drops around the circuit.
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Single Loop Circuits (5)
 Now let’s analyze the same circuit in the counterclockwise direction starting at point a
i
 The first circuit element is Vemf,2 which is a positive
i
voltage gain
 The next element is R2
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Single Loop Circuits (6)
i
 Because we have assumed that the
current is in the clockwise direction
and we are analyzing the loop in the
counter-clockwise direction, this
i
voltage drop is +iR2
 Proceeding to the next element in the loop, R1, we use a
similar argument to designate the voltage drop as +iR1
 The final element in the circuit is Vemf,1, which is aligned in a
direction opposite to our analysis direction, so the voltage
drop across this element is -Vemf,1
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Single Loop Circuits (7)
 Kirchhoff’s Loop Rule then gives us
Vemf ,2  iR2  iR1  Vemf ,1  0
 Comparing this result with the result we obtained
by analyzing the circuit in the clockwise direction
Vemf ,1  iR1  iR2  Vemf ,2  0
 … we see that they are equivalent.
 The direction that we choose to analyze the circuit
does not matter.
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Clicker Question
 The figure shows the
current i in a single-loop
circuit with a battery B
and a resistance R. How
should the emf arrow be
drawn?
A) pointing to the right
B) pointing to the left
C) doesn’t matter
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Clicker Question
 The figure shows the
current i in a single-loop
circuit with a battery B
and a resistance R. How
should the emf arrow be
drawn?
A) pointing to the right
in direction of the
current
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