Transcript PHY 184 - MSU Department of Physics and Astronomy

PHY 184
Spring 2007
Lecture 20
Title:
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184 Lecture 20
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Announcements
 We hope to open the correction set tonight
You will have a week to complete the problems
• You can re-do all the problems from the exam
• You will receive 30% credit for the problems you missed
• To get credit, you must do all the problems in
Corrections Set 1, not just the ones you missed
 Homework Set 5 is due on Tuesday, February 20 at 8 am
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Kirchhoff’s Law, Multi-loop Circuits
 One can create multi-loop circuits that cannot be
resolved into simple circuits containing parallel or
series resistors.
 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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Circuit Analysis Conventions
Element
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Analysis Direct ion
Current Direct ion
Volt age Drop


iR


iR


iR


iR

Vemf

Vemf

Vemf

Vemf
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i is the
magnitude of
the assumed
current
4
Multi-Loop Circuits
 To analyze multi-loop circuits, we must apply both
the Loop Rule and the Junction Rule.
 To analyze a multi-loop circuit, identify complete
loops and junction points in the circuit and apply
Kirchhoff’s Rules to these parts of the circuit
separately.
 At each junction in a multi-loop circuit, the current
flowing into the junction must equal the current
flowing out of the circuit.
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Multi-Loop Circuits (2)
 Assume we have a junction point a
 We define a current i1 entering junction a and two
currents i2 and i3 leaving junction a
 Kirchhoff’s Junction Rule tells us that
i1  i2  i3
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Multi-Loop Circuits (3)
 By analyzing the single loops in a multi-loop circuit
with Kirchhoff’s Loop Rule and the junctions with
Kirchhoff’s Junction Rule, we can obtain a system
of coupled equations in several unknown variables.
 These coupled equations can be solved in several
ways
• Solution with matrices and determinants
• Direct substitution
 Next: Example of a multi-loop circuit solved with
Kirchhoff’s Rules
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Example - Kirchhoff’s Rules
 The circuit here has three resistors, R1, R2, and R3 and two
sources of emf, Vemf,1 and Vemf,2
 This circuit cannot be resolved into simple series or parallel
structures
 To analyze this circuit, we need to assign currents flowing
through the resistors.
 We can choose the directions of these currents arbitrarily.
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Example - Kirchhoff’s Laws (2)
 At junction b the incoming current must equal the outgoing
current
i2  i1  i3
 At junction a we again equate the incoming current and the
outgoing current
i1  i3  i2
 But this equation gives us the
same information as the
previous equation!
 We need more information
to determine the three currents – 2 more independent
equations
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Example - Kirchhoff’s Laws (3)
 To get the other equations we must apply Kirchhoff’s Loop
Rule.
 This circuit has three loops.
• Left
• R1, R2, Vemf,1
• Right
• R2, R3, Vemf,2
• Outer
• R1, R3, Vemf,1, Vemf,2
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Example - Kirchhoff’s Laws (4)
 Going around the left loop counterclockwise starting at
point b we get
i1R1  Vemf ,1  i2 R2  0  i1R1  Vemf ,1  i2 R2  0
 Going around the right loop clockwise starting at point b we
get
i3 R3  Vemf ,2  i2 R2  0  i3 R3  Vemf ,2  i2 R2  0
 Going around the outer loop
clockwise starting
at point b we get
i3 R3  Vemf ,2  Vemf ,1  i1R1  0
 But this equation gives us no new
information!
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Example - Kirchhoff’s Laws (5)
 We now have three equations
i1  i3  i2
i1R1  Vemf ,1  i2 R2  0
i3 R3  Vemf ,2  i2 R2  0
 And we have three unknowns i1, i2, and i3
 We can solve these three equations in a variety of ways
i1  
i2  
i3  
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(R2  R3 )Vemf ,1  R2Vemf ,2
R1 R2  R1 R3  R2 R3
R3Vemf ,1  R1Vemf ,2
R1 R2  R1 R3  R2 R3
R2Vemf ,1  (R1  R2 )Vemf ,2
R1 R2  R1 R3  R2 R3
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Clicker Question
 Given is the multi-loop circuit on
the right. Which of the
following statements cannot be
true:
 A)
 B)
 C)
 D)
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Clicker Question
 Given is the multi-loop circuit on
the right. Which of the
following statements cannot be
true:
 A)
Junction rule
 B)
 C)
Not a loop!
Upper right loop
 D)
Left loop
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Ammeter and Voltmeters
 A device used to measure current is called an ammeter
 A device used to measure voltage is called a voltmeter
 To measure the current, the ammeter must be placed in the
circuit in series
 To measure the voltage, the voltmeter must be wired in
parallel with the component across which the voltage is to
be measured
Voltmeter in parallel
High resistance
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Ammeter in series
Low resistance
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RC Circuits
 So far we have dealt with circuits containing
sources of emf and resistors.
 The currents in these circuits did not vary in time.
 Now we will study circuits that contain capacitors
as well as sources of emf and resistors.
 These circuits have currents that vary with time.
 Consider a circuit with
• a source of emf, Vemf,
• a resistor R,
• a capacitor C
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RC Circuits (2)
 We then close the switch, and current begins to
flow in the circuit, charging the capacitor.
 The current is provided by the
source of emf, which maintains
a constant voltage.
 When the capacitor is fully charged,
no more current flows in the circuit.
 When the capacitor is fully charged, the voltage
across the plates will be equal to the voltage
provided by the source of emf and the total charge
qtot on the capacitor will be qtot = CVemf.
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Capacitor Charging
 Going around the circuit in a counterclockwise direction we can
write
Vemf  VR  VC  Vemf
q
 iR   0
C
 We can rewrite this equation
remembering that i = dq/dt
Vemf
dq q
dq
q
R   Vemf 


dt C
dt RC
R
 The solution of this differential
equation is
t
 

q(t)  q0  1  e  


 … where q0 = CVemf and  = RC
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The term Vc is negative since
the top plate of the capacitor
is connected to the positive higher potential - terminal of
the battery. Thus analyzing
counter-clockwise leads to a
drop in voltage across the
capacitor!
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Capacitor Charging (2)
 We can get the current flowing in the circuit by
differentiating the charge with respect to time



q(t)  q0  1  e 



t
 t 
V
dq  emf    RC 
i

e

dt  R 
 The charge and current as a function of time are shown
Math Reminder:
here ( = RC)
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Capacitor Discharging
 Now let’s take a resistor R and a fully charged capacitor C
with charge q0 and connect them together by moving the
switch from position 1 to position 2
 In this case current will flow in the circuit until the
capacitor is completely discharged.
 While the capacitor is discharging we can apply the Loop
Rule around the circuit and obtain
q
dq q
iR  VC  iR   0  R   0
C
dt C
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Capacitor Discharging (2)
 The solution of this differential equation for the charge is
 t 
 

RC 
q  q0 e
 Differentiating charge we get the current
 t 
 

RC 
dq  q0 
i

e

dt  RC 
 The equations describing the time dependence of the
charging and discharging of capacitors all involve the
exponential factor e-t/RC
 The product of the resistance times the capacitance is
defined as the time constant  of a RC circuit.
 We can characterize an RC circuit by specifying the time
constant of the circuit.
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Example: Time to Charge a Capacitor
 Consider a circuit consisting of a 12.0 V battery, a
50.0  resistor, and a 100.0 F capacitor wired in
series.
 The capacitor is initially uncharged.
 Question:
• How long will it take to charge the capacitor in this
circuit to 90% of its maximum charge?
 Answer:
• The charge on the capacitor as a function of time is
t



q t   q0  1  e RC 


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Example: Time to Charge a Capacitor (2)
t



RC
q t   q0  1  e 


 We need to know the time corresponding to
q t  / q0  0.90
 We can rearrange the equation for the charge on the
capacitor as a function of time to get
0.10  e
t

RC
Math Reminder: ln(ex)=x
t  RCln(0.10)  11.5 ms
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Example: More RC Circuits
A 15.0 k resistor and a capacitor are connected in
series and a 12V battery is suddenly applied. The
potential difference across the capacitor rises to
5V in 1.3 s. What is the time constant of the
circuit?
Answer: 2.41 s
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Clicker Question
A 15.0 k resistor and a capacitor are connected in
series and a 12V battery is suddenly applied. The
potential difference across the capacitor rises to
5V in 1.3 s. What is the capacitance C of the
capacitor?
A) 161 pF
B) 6.5 pF
C) 0
D) 49 pF
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