Transcript + + R

Chapter 32
Fundamentals of Circuits
L.A. Bumm
Schematic Symbols
ideal wire with
zero resistance
Basic Circuits and Kirchhoff’s Loop Law
Vloop   Vi  Vbat  VR  0
i
Vbat  E ; VR   IR
Vbat  VR  E  IR  0
E  IR
E
I  ; VR  E
R
Power in Electrical Circuits
Power supplied by batteries
Pbat  IE
U  qVbat ; Vbat  E
U  qE
dU dq
Pbat 
 E  IE
dt
dt
Power in Electrical Circuits
Power dissipated in resistors.
Electrical energy is converted in to heat.
W  F s  qEd
Emech  K  qEd
Ethermal  qEL  qVR
d
dq
Ethermal 
VR  I VR
dt
dt
PR  I VR
PR 
Because current and voltage difference for resistors are related by resistance (Ohm’s
Law), we can eliminate I or ΔV from the power equation. But this only works for
resistors!!.
 V 
PR  I VR  I  IR    R  VR ; VR  IR
 R 
VR2
PR  I VR  I R 
R
2
This is always true:
P  I V
Clicker Question
Which schematic is NOT equivalent to the cartoon picture?
A)
B)
C)
D)
Clicker Question
Which equation is not Ohm’s Law
V  I R
V
R
I
R
I
V
A)
B)
C)
V
I
R
D)
These are all
statements of
Ohm’s Law
E)
Resistors in Series and Parallel
•
Resistors in series
– The current is the same through each resistor.
– The voltage drops across each resistor sum to the
voltage drop across all the resistors.
– The equivalent resistance will be equal to or
greater than the largest resistance in the series
circuit.
•
Req  R1  R2  R3  
Resistors in Parallel
– The voltage drop is the same across each resistor.
– The current through each resistor sums to the
total current supplied to the parallel circuit.
– The equivalent resistance is less than or equal to
the smallest resistor in the parallel circuit.
1
1
1
1
 
 
Req R1 R2 R3
Combinations of series and parallel resistors.
Any network of resistors can be reduced to a single equivalent resistance.
When the network is not a simple series or parallel combination, it can
always be reduced to a single resistor by the sequential application of the
series and parallel resistor combination rules. Don’t be afraid to redraw
the circuit to make it look more familiar to you.
Reduce this circuit to its equivalent resistance.
1
1 
 1


  30 
 90  45  
10   30   40 
1
1 
 1


  15.4 
 40  25  
Example: A nichrome wire with diameter d and length L is connected to a
battery of EMF E by two lengths of copper wire with the same d and L as the
nichrome wire.
Draw a schematic of this circuit. Do not neglect the resistance of the copper
wire. What is the resistance of each segment of wire. What is the current in
the circuit. How much power is supplied by the battery. How much power is
dissipated in the nichrome wire. What percentage of the power is delivered
to the nichrome wire?
E = 1.5 V; d = 1.0 mm; L = 200 cm; ρnichrome = 1.5×10−6 Ωm; ρCu = 1.7×10−8 Ωm
We are ask to find RNC, RCu, I, Pbat, PNC, and PNC/Pbat.
R
RCu1
I
RNC 
L
A
 NC L
A
RNC
E
RCu 
RCu2
Cu L
A
I
V
R
1.5 10


6
P  I V  I 2 R

 m 200 102 m
  5 104 m 
1.7 10


8
2
RNC  3.8 

 m 200 102 m
  5 104 m 
  3.8197 
2
  0.0433 
RCu  4.3 102 
Example: A nichrome wire with diameter d and length L is connected to a
battery of EMF E by two lengths of copper wire with the same d and L as the
nichrome wire.
Draw a schematic of this circuit. Do not neglect the resistance of the copper
wire. What is the resistance of each segment of wire. What is the current in
the circuit. How much power is supplied by the battery. How much power is
dissipated in the nichrome wire. What percentage of the power is delivered
to the nichrome wire?
E = 1.5 V; d = 1.0 mm; L = 200 cm; ρnichrome = 1.5×10−6 Ωm; ρCu = 1.7×10−8 Ωm
We are ask to find RNC, RCu, I, Pbat, PNC, and PNC/Pbat.
I
RCu1
RNC
RCu2
P  I V  I 2 R
R  RCu  RNC  RCu  2 RCu  RNC
I
E
V
R


 2 4.3 102    3.8    3.9063 
1.5 V 

E
I 
 0.3840 A
R  3.9063  m 
R  3.9 
I  0.38 A
Example: A nichrome wire with diameter d and length L is connected to a
battery of EMF E by two lengths of copper wire with the same d and L as the
nichrome wire.
Draw a schematic of this circuit. Do not neglect the resistance of the copper
wire. What is the resistance of each segment of wire. What is the current in
the circuit. How much power is supplied by the battery. How much power is
dissipated in the nichrome wire. What percentage of the power is delivered
to the nichrome wire?
E = 1.5 V; d = 1.0 mm; L = 200 cm; ρnichrome = 1.5×10−6 Ωm; ρCu = 1.7×10−8 Ωm
P  I V  I 2 R
We are ask to find RNC, RCu, I, Pbat, PNC, and PNC/Pbat.
RCu1
Pbat  IE   0.3840 A 1.5 V   0.5760 W
I
PNC  I 2 RNC   0.3840 A   3.8197    0.5630 W
PNC  0.56 W
PNC
PNC  0.5630 W 
 98 %

 0.977836
P
Pbat  0.5760 W 
bat
2
RNC
E
Pbat  0.58 W
RCu2
We could have ignored the resistance of the copper wire, depending on our application.
Kirchhoff’s Laws
Kirchhoff’s Junction Law (Conservation of charge)
At any junction point, the sum of all the currents
entering the junction must equal the sum of all
currents leaving the junction.
Kirchhoff’s Loop Law (Conservation of energy)
The sum of the changes in potential around any
closed path of a circuit must be zero.
How to Use the Kirchhoff’s Laws
•
Draw the circuit and draw currents with an arrow in every separate branch of the
circuit. (A branch is a section where the current does not change.)
•
Apply the junction law to enough junctions so that every current is used at least
once.
•
Apply the loop law to enough closed loops so that each current appears at least
once. Remember the sign convention for the potential changes:
Sign Conventions
Batteries
+ if moving around the loop you pass from the − terminal to the + terminal.
− if moving around the loop you pass from the + terminal to the − terminal.
Resistors
− if moving around the loop you are moving in the direction of the defined
current (your arrows).
+ if you are moving against the defines current.
Caution: If your circuit has multiple branches your loop path may go with the
defined current in one branch and against it in another.
Example: Analyzing circuits with more that one loop
330 Ω
19 V
150 Ω
+
12 V
270 Ω
+
Find the current and its direction through
the 150 Ω resistor and the potential
difference across it.
Example: Analyzing circuits with more that one loop
• Draw the circuit and draw currents with an arrow in every separate branch of
the circuit. (A branch is a section where the current does not change.)
• Apply the junction law to enough junctions so that every current is used at
least once.
junction eq.
I
in
  I out
I1  I 2  I 3
I1
330 Ω
150 Ω
270 Ω
I3
19 V
+
12 V
+
I2
We could also have used this junction.
We do not need to use both, because
the information is redundant.
I
in
  I out
I 2  I 3  I1
Example: Analyzing circuits with more that one loop
• Apply the loop law to enough closed loops so that each current appears at least
once. Remember the sign convention for the potential changes:
Sign Conventions
Batteries
+ if moving around the loop you pass from the − terminal to the + terminal.
− if moving around the loop you pass from the + terminal to the − terminal.
Resistors
− if moving around the loop you are moving in the direction of the defined
current (your arrows).
+ if you are moving against the defines current.
Caution: If your circuit has multiple branches your loop path may go with the
defined current in one branch and against it in another.
loop A: 0    V i
I1
i
330 Ω
19 V
150 Ω
+
A
12 V
270 Ω
I3
+
I2
  19 V    330   I1  150   I 3  12 V 
units check:
 V      A      A    V 
V  IR  V  A 
A Ω is the same as a V so we can add them.
Example: Analyzing circuits with more that one loop
loop equations
I1
330 Ω
150 Ω
A
+
19 V
12 V
270 Ω
I3
loop A: 0    V i
i
+
I2
  19 V    330   I1  150   I 3  12 V 
I1
330 Ω
19 V
150 Ω
+
12 V
I1
330 Ω
19 V
+
I2
loop B: 0    V i
i
  12 V   150   I 3   270   I 2
The third loop is redundant and provides no new information.
There are n−1 unique loops. Here, we can chose any two loop.
270 Ω
+
12 V
I3 B
C
150 Ω
+
270 Ω
I3
I2
loop C: 0    V i
i
  19 V    330   I1   270   I 2
Example: Analyzing circuits with more that one loop
I1
330 Ω
We have e equations and 3 unknowns.
150 Ω
270 Ω
I3
19 V
+
12 V
+
I2
I1  I 2  I 3
0   19 V    330   I1  150   I 3  12 V 
0   12 V   150   I 3   270   I 2
Eliminate I1 first, let’s keep I3 because that is what we are asked to find.
I1  I 2  I 3
0   19 V    330   I 2  I 3   150   I 3  12 V 
  19 V  12 V    330   I 2   330   150   I 3
   7 V    330   I 2   480   I 3
  7 V    330   I 2   480   I 3
0   12 V   150   I 3   270   I 2
 12 V     270   I 2  150   I 3
Example: Analyzing circuits with more that one loop
I1
330 Ω
I1  I 2  I 3
150 Ω
0   19 V    330   I1  150   I 3  12 V 
270 Ω
I3
19 V
+
12 V
0   12 V   150   I 3   270   I 2
+
I2
  7 V     330   I 2   480   I 3
330
 12 V     270   I 2  150   I 3 

270
 7.667 V  
0
  663.3   I 3
  7 V     330   I 2   480   I 3

7.667 V 

I3 
 11.56 mA
663.3



480
 12 V     270   I 2  150   I 3 

150
 45.40 V    1194   I 2 
45.40 V 

I2 
 38.02 mA
1194  
0
I1  I 2  I 3   38.02 mA    11.56 mA   26.46 mA
I1  26 mA
I 2  38 mA
I 3  12 mA
Example: Analyzing circuits with more that one loop
I1
330 Ω
150 Ω
270 Ω
I3
19 V
+
12 V
I1  26 mA
I 2  38 mA
I 3  12 mA
+
Find the current and its direction through
the 150 Ω resistor and the potential
difference across it.
I2
V150   IR   11.56 mA 150    1.734 V
V150   1.7 V
Clicker Question
Which is not a loop equation for the following circuit?
R1
E1
+
IB
R3
+
IA
R2
IC
E2
A) E1  I A R1  I A R2  I B R3
B) E1  E2  I A R1  I A R2
C) E1  I A R1  I A R2  I B R3  0
D) I B R3  E2
E) E2  I A R1  I A R2  E1
Ideal and Real Batteries
The ideal battery is a voltage source.
• It maintains a constant voltage across its terminals.
• It can supply any amount of current required to maintain its terminal voltage.
• The current the battery supplies is determined by the rest of the circuit.
A real batteries have internal resistance.
• Real batteries are simply an ideal voltage source (battery) with a series
resistor representing its internal resistance.
• The internal resistance limits the maximum current the battery can supply.
E
E
+
+
ideal battery
r
real battery
Ammeters
Ammeter.
• An ammeter measures current.
• It must be placed in series with the circuit in which the current is to be measured.
• The ideal ammeter has zero internal resistance.
• The ideal ammeter has zero voltage drop (the same as zero series resistance).
• A real ammeter has internal resistance and is modeled as an ideal ammeter in series
with a resistor, hence does contribute a voltage drop in the circuit it is measuring.
comments
• The ammeter symbol is ALWAYS an ideal ammeter.
• In problems always assume an ideal ammeter unless you are explicitly told otherwise.
A
ideal ammeter
A
r
real ammeter
Voltmeters
Voltmeter.
• A voltmeter measures voltage (potential difference).
• It must be placed in parallel with the circuit in which the voltage is to be measured.
• The ideal voltmeter has infinite internal resistance.
• The ideal voltmeter has draws no current (the same as infinite resistance).
• A real voltmeter has internal resistance and is modeled as an ideal voltmeter in
parallel with a resistor, hence does draw current from the circuit it is measuring.
comments
• The voltmeter symbol is ALWAYS an ideal voltmeter.
• In problems always assume an ideal voltmeter unless you are explicitly told
otherwise.
V
V
r
ideal voltmeter
real voltmeter
Example: A real battery with EMF E is connected to an external resistor R.
A voltmeter measures the potential difference across the resistor ΔVR.
What is the internal resistance of the battery?
What would the short circuit current of this battery be?
E = 9.0 V; R = 17 Ω; ΔVR = 8.5 V
VR
I
0    V   E  Ir  IR  E  Ir  VR
R
i
i
E  VR  Ir
  9.0 V  
 E

E  VR E  VR
R
r


 1  17   
 1  1.0 
E  VR   R 



V
I
VR
R
 VR 
  8.5 V  
R
r  1.0 
I short 
E 9.0 V

 9.0 A
r 1.0 
I short  9.0 A
Clicker Question
What four identical resistor values R will have an equivalent
resistance of 100 Ω when connected as shown?
A)
25 Ω
R
R
Req = 100 Ω
B)
50 Ω
C)
100 Ω
D)
200 Ω
E)
none of the above
R
R
Light Bulbs and Brightness
Light Bulbs.
• Electrical devices that emit light.
• Incandescent lamps are resistors, that get hot enough to emit black body radiation.
• The brightness of a lamp is proportional to the power dissipated in the lamp.
comments
• The relative brightness of a light bulb in a circuit may be determined by finding the
power dissipated in each bulb.
• Well will ignore the changes in resistance with temperature of incandescent lamps. The
filament in an incandescent lamp is typically made from tungsten because of its very
high melting point. The resistivity of metals increases with increasing temperature.
Because the filament operates at thousands of degrees above room temperature, the
change in resistance from on to off can be very significant.
Light bulb
Light Bulb Brightness
The following circuit has 3 identical bulbs,
a battery, and a switch.
Assumptions:
The voltage across the battery will be constant.
The resistances R of all three bulbs are the same.
Switch open:
When the switch is open, obviously bulb C is off.
We can simplify the circuit.
What can we say about the brightness of bulbs A and B?
Bulbs A & B are in series, thus the same current flows
through each. Because the two bulbs also have the
same resistance, the power (and therefore their
brightness) will be the same.
What is the actual power in each bulb?
2
I
E
E
E
E 

; P  I 2R  
R


R  R 2R
4R
 2R 
2
I
PA  PB 
2
1E
4 R
Light Bulb Brightness
Switch closed:
We can simplify the circuit.
What can we say about the brightness of
bulbs A, B, and C?
• Bulbs B & C are in parallel, thus the same voltage
across each. Because the two bulbs also have the
same resistance, the power (and therefore their
brightness) will be the same.
• Bulb A is in series with the B & C. All the current
flowing in the circuit passes through A. From
Kirchhoff’s junction law we know that IA = IB + IC.
Because the bulbs all have the same resistance, the
current through A is twice that though B and C. Thus
the power (and therefore their brightness) will be
four times that of bulbs B and C.
IA
IB
IC
Light Bulb Brightness
Switch closed:
We can simplify the circuit.
What can we say about the brightness of
bulbs A, B, and C?
What is the actual power in each bulb?
IA
Equivalent resistance of the circuit:
1
 1
1 
3
1
Req  RA  

  R 2 R  2 R
 RB RC 
Power in bulb A.
IB
2
 E 
E
E
E2
2
4
IA 
 3 ; PA  I R   3  R  9
 R
Req 2 R
R
2 
4E2
PA 
9 R
Power in bulbs B and C.
I B  IC  I A 
1
2


1 E

2
 3 R 
2 
2
E
E2
E 
2
; PB  PC  I R  
 R
3R
9R
 3R 
1E2
PB  PC 
9 R
IC
Light Bulb Brightness
The following circuit has 3 identical bulbs,
a battery, and a switch.
What can we say about the change in brightness of bulbs A and B
when the switch is closed?
Qualitatively we can say:
• The brightness of A increases because the current in the circuit increases as C
is brought in parallel with B.
• The brightness of B decreases because more voltage is dropped across A as
the current in the circuit increases.
Quantitatively, we simply ratio the powers
PAclosed
PAopen
4E2 4
16
 9 R2  9 
1 9
1E
4
4 R
PAclosed 16

open
PA
9
PBclosed
PBopen
1E2 1
4
 9 R2  9 
1 9
1E
4
4 R
PBclosed 4

open
PB
9
Clicker Question
In the circuit to the right. The switch open.
Compare the brightness of bulbs A and B.
A)
A>B
B)
A=B
C)
A<B
D)
There is no way to tell.
Clicker Question
In the circuit to the right. The switch open.
Compare the brightness of bulbs B and C.
A)
B>C
B)
B=C
C)
B<C
D)
There is no way to tell.
Clicker Question
In the circuit to the right. When the switch is closed,
the brightness of bulb B ______ .
A)
decreases
B)
increases
C)
does not change
D)
There is no way to tell.
Clicker Question
In the circuit to the right. The switch is now closed.
Compare the brightness of bulbs B and C.
A)
B>C
B)
B=C
C)
B<C
D)
There is no way to tell.
Grounding
Grounding provides a convenient reference point for potential differences.
All potentials we measure are potential differences. However when we say a
point in a circuit has a certain potential, we mean it has that potential with
respect to a common reference point. Typically a ground connection defines
that point.
V  IR  4 V
V  IR  6 V
Example: Adding a ground wire does not change the function of the circuit. It
only changes the potentials with respect to ground. This is true when there is
only ONE connection to ground so that no current can flow between the
circuit and ground.
In this example, only the point of grounding is changed. Note that the
currents and the potential differences with in the circuit are unchanged.
RC Circuits and Transients
So far we have considered capacitors that were charged and discharged.
Now we will described how circuits with capacitors evolve from the
charged state to the discharged state and vice versa.
The capacitor initially charged to Q0 = VC.
While the switch is open no current flows in the circuit.
The switch is closed at t = 0
current begins to flow and the capacitor begins to discharge.
I0 = V/R
We can use Kirchhoff’s Loop Law
I 
before switch is closed
I=0
ΔVC = Q0/C
dQ
Q
Q  dQ 
; 0  VC  VR   IR    
R
dt
C
C  dt 
dQ
1 t
Q
t


dt

ln


Q0 Q RC 0
Q0
RC
Q
R
C
after switch is closed
 t 
 Q  Q0 exp  

 RC 
I
C
 t
ΔVR = 0
R
  RC ; Q  Q0 exp   
 
τ is the time constant, the rate the
capacitor charges and discharges
units:
  RC  s =  F
ΔVC = Q/C
ΔVR = IR
RC Circuits and Transients
The time constant τ is determined by R and C only. It is the rate the circuit evolves
from the initial state (charged) to the final state (discharged).
 t
  RC ; Q  Q0 exp   
 
The exponential decay of the charge decreases by
a factor e−1 over each time interval τ.
The current also decays exponentially with time.
I 

dQ
d 
 t
   Q0 exp  
dt
dt 
 
VC C
 t
exp  
RC
 
 t
I  I 0 exp   
 
  Q0
 t

exp

 
 
 
 VC
 t

exp


R

 

 t

I
exp
 0
 

 
RC Circuits and Transients
The charging process also follows exponential behavior with time constant τ.
The capacitor initially discharged with Q0 = 0.
While the switch is open no current flows in the circuit.
The switch is closed at t = 0.
I0 = V/R
The capacitor charges to Qmax following a
complimentary exponential.
 t

 t 
  RC ; Q  Qmax  Qmax exp     Qmax 1  exp    
 
  

The exponential decay of the charge difference
(Qmax − Q) decreases by a factor e−1 for each time
interval τ.

 t 
  RC ; Q  Qmax 1  exp    
  

Example: The switch has been in position a for a long time. It is changed to
position b at t = 0. What is the time constant of the circuit? What is the charge
on the capacitor and the current through the circuit after time t = 4.7 μs? How
long does it take for the capacitor to loose half its charge?
E = 9.0 V; R = 17 Ω; C = 1.5 μF
  RC  17   1.5 10 F   2.55 10 s
6
5
E
C
R
  26 μs
 t
Q  Q0 exp  
 

 t 


V
C
exp
bat




 RC 


4.7 106 s
6
  1.1228 10 5 C
  9.0 V  1.5 10 F exp  
 17   1.5 106 F 







Q  11 μC



6


4.7

10
s
9.0
V



V
t
t




bat
  0.4403 A
I  I 0 exp    
exp  
exp  

6
 17   1.5 10 F 
R
 
 RC  17  




I  0.44 A
Example: The switch has been in position a for a long time. It is changed to
position b at t = 0. What is the time constant of the circuit? What is the charge
on the capacitor and the current through the circuit after time t = 4.7 μs? How
long does it take for the capacitor to loose half its charge?
E = 9.0 V; R = 17 Ω; C = 1.5 μF
 t
Q  Q0 exp   
 
1 Q
 t 

 exp  

2 Q0
 RC 
t
1
ln    
  ln  2 
2
RC
 
E


C
t  RC ln  2   17   1.5 106 F ln  2   1.7675 10 5 s
t  18 μs
R
Clicker Question
Which of the following are not units of RC (resistance × capacitance)?
A)
ΩF
B)
C A 1
C)
kg m s 1 N 1
D)
s
E)
C V 1
Stop here
Ammeters and Voltmeters
Ammeter.
• An ammeter measures current.
• It must be placed in series with the circuit in which the current is to be measured.
• The ideal ammeter has zero internal resistance.
Voltmeter.
• A voltmeter measures voltage (potential difference).
• It must be placed in parallel with the circuit in which the voltage is to be measured.
• The ideal voltmeter has infinite internal resistance.
A
V
ideal ammeter
ideal voltmeter