Download + R - Purdue Physics

EXAM2
 Wednesday, March 26, 8:00-10:00 pm
 No lecture on that day.
 room 112 for students in R21/22/23/24
 room 114 for students in R25/26/27
 Chapters 17, 18, 19 & 20
 Special needs:
 Students have been contacted for special arrangement. Please let me know if
you haven’t receive the e-mail.
 AOB
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–
•
•
•
•
multiple choice.
Prepare your own scratch paper, pens, pencils, erasers, etc.
Use only pencil for the answer sheet
Bring your own calculators
No cell phones, no text messaging which is considered cheating.
No crib sheet of any kind is allowed. Equation sheet will be provided.
1
Last time
•
•
•
•
•
Different configuration of circuit
Charge and discharge capacitors
Different way of connecting capacitors.
(Capacitance)
(More details on resistance)
2
Today
•
•
•
•
Capacitance
Different way of connecting resistors
Solving complicated circuit using Kirchhoff’s Rules
(Ammeters, Voltmeters and Ohmmeters)
3
Capacitance
-Q
+Q
Electric field in a capacitor: E  Q / A
0
 
V    E  dl
f
V  Es
i
V 
Q/ A
0
E
s
Q
0 A
s
V
In general: Q ~ V
s
Definition of capacitance:
Capacitance of a parallelQ  C V
Wiki: The SI unit of capacitance plate capacitor:
is the farad (symbol: F), named
0 A
C
Capacitance after the English physicist
s
Michael Faraday
4
Exercise
A capacitor is formed by two rectangular plates 50 cm by 30
cm, and the gap between the plates is 0.25 mm. What is its
capacitance?
C
C
0 A
s
0 A
s
9  10


12

C2 /N  m 2 0.15 m 2
2.5  10
4
m
  5.4  10
9
F  5.4 nF
5
A Capacitor With an Insulator Between the Plates
No insulator:
E
With insulator:
Q/A
Q/A
E
K 0
0
V  Es
V 
Q
V  Es
Q/ A
0
0 A
s
s
V
C
D
0 A
s
Q/ A
V 
s
K 0
Q
s
K 0 A
V
s
CK
0 A
s
6
Unit: J, A/m2; E, V/m; , (A/m2)/(V/m)
7
Conductivity with two kinds of charge carrier
  q1 n1u1  q2 n2u2
8
Resistance is in the unit of Ohm.
9
10
Kirchhoff’s Rules
Kirchhoff’s Rule 2: Loop Rule
 When any closed loop is traversed completely in a circuit,
the algebraic sum of the changes in potential is equal to zero.
 V
i
0
 Coulomb force is conservative
loop
Kirchhoff’s Rule 1: Junction Rule
 The sum of currents entering any junction in a circuit is equal to
the sum of currents leaving that junction.
I  I
i
in
out
j
 Conservation of charge
 In and Out branches
 Assign Ii to each branch
11
Series Resistance
Vbatt + V1 + V2 + V3 = 0
emf - R1I - R2I - R3I = 0
emf = R1I + R2I + R3I
emf = (R1 + R2 + R3) I
emf = Requivalent I , where Requivalent = R1 + R2 + R3
For resistors made of the same material and with the same A
it follows straight from the definition of resistance:
L
R
A
12
Parallel Resistance
I = I 1 + I2 + I3
emf emf emf
I


R1
R2
R3
 1
1
1
emf
I  
  emf 
Requivalent
 R1 R2 R3 
1
Requivalent
1
1
1
 

R1 R2 R3
For resistors made of the same material and with the same A
it follows straight from the definition resistance:
L
1 A
Aequivalent  A1  A2  A3
R

A
R
L
13
Real Batteries: Internal Resistance
Drift speed of ions
in chemical battery:
v ~ FNC  eEC 
In usual circuit elements:
J  E
rint - internal
resistance
In a battery:
I
J  
A
EC 
 force 
 FNC

 
 EC 
 ~  
 e

 unit charge 
emf
FNC
I
FNC s s


V

E
s


I
, assuming uniform field:
C
e
A
e
A
Vbattery  emf  rint I
14
Circuit Analysis Tips
• Simplify using equivalent resistors
• Label currents with arbitary directions
•If the calculated current is negative, the real direction is opposite to the one
defined by you.
• Apply Junction Rule to all the labeled currents.
•Useful when having multiple loops in a circuit.
• Choose independent loops and define loop direction
•Imagine your following the loop and it’s direction to walk around the circuit.
• Use Loop Rule for each single loop
•If current I direction across a resistor R is the same as the loop direction,
potential drop across R is ∆V = −I×R, otherwise, ∆V = I×R
•For a device, e.g. battery or capacitor, rely on the direction of the electric
field in the device and the loop direction to determine the Potential drop
across the device
• Solve simultaneous linear equations
15
Loop Example with Two EMF Devices
 V
i
0
loop
 IR1  IR2   2  Ir2  IR3  1  Ir1  0
 I
1   2
R1  R2  R3  r1  r2
If 1 <2, we have I<0 !?
This just means the actual current flows reverse to the assumed
direction. No problem!
16
Finding Potential and Power in a Circuit
Va  0  12  I  1 V 
But what is I? Must
solve for I first!
I
12  4
 0.5 ( A)  0
1 5  5 1 4
Va  12  0.5  1  11.5(V )
Vb  Va  I  5  9(V )
P12V
The rest?
Just means
0 V here
supplied by
 12  0.5  6(W ) 12V battery
PR  0.52  16  4 (W ) dissipated by
into 4V battery
(charging)
resistors
P4V  4  0.5  2(W )
17
Charging a Battery
• Positive terminal to positive terminal
• Charging EMF > EMF of charged device
good
battery
(12V)
Say, R+r1+r2=0.05 (R is for jumper cables).
Then,
12  11(V )
I
 20( A)
0.05 ()
battery being
charged (11V)
P 2  11  20  220 (W )
power into battery 2
• If connected backward,
12  11
I
 460 ( A)
0.05
 Large amount of gas produced
 Huge power dissipation in wires
18
Using Kirchhoff’s Laws in Multiple
Loop Circuits
• Identify nodes and use Junction Rule:
i3  i1  i2
• Identify independent loops and use Loop Rule:
1  i1R1  i2 R2   2  i1R1  0
 2  i1  i2  R1  i2 R2   2  i1  i2  R1  0
Only two are
independent.
 2  i1  i2  R1  i1R1  1  i1R1  i1  i2  R1  0
19
Example
I1+I2
I2
• What’s the current I1 ?
I1
(a). 2.0A
(b). 1.0A
(c). -2.0A
(d). -1.0A
(e). Need more information to
calculate the value.
20
I1+I2
I2
• Sketch the diagram
• Simplify using equivalent resistors
I1
• Label currents with directions
• Use Junction Rule in labeling
• Choose independent loops
• Use Loop Rule
Replace by equivalent
R=2 first.
• Solve simultaneous linear equations
18  12( I1  I 2 )  6 I1  0
 3I 1  2 I 2  3
3I 2  21  2 I 2  6 I1  0
6 I1  5I 2  21
 I 2  3( A), I1  1( A)
21
iClicker Question
What is the current through R1 ?
30
30
a. 0.575A
b. 0.5A
c. 0.75A
45V
R1
R3
R2
45V
30
d. 0.33A
e. 1.5A
22
Ammeters, Voltmeters and Ohmmeters
Ammeter: measures current I
Voltmeter: measures voltage difference V
Ohmmeter: measures resistance R
23
Ammeter Design: rint
Ammeter is inserted in series into a circuit – measured current
flows through it.
A
rint
R
Process of measuring requires emf
charges to do some work:
Internal resistance
No ammeter: emf  RI  0
With ammeter: emf  rint I  RI  0
emf
I
R
I
emf
R  rint
Internal resistance of an ammeter must be very small
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Voltmeter
Voltmeters measure potential difference
VAB – add a series resistor to ammeter
V
I
R
Measure I and convert to VAB=IR
Connecting Voltmeter:
Higher potential must be connected to
the ‘+’ socket and lower one to the ‘-’
socket to result in positive reading.
25
Voltmeter: Internal Resistance
VAB in absence of a voltmeter
R1
emf
VAB 
B
rint
R2
A
A
R2
emf
R1  R2
VAB in presence of a voltmeter
VAB 
R2||int 
R2||int
R1  R2||int
emf
R2 rint
R2  rint
Internal resistance of a voltmeter must be very large
26
Ohmmeter
How would you measure R?
A
emf
I
R
R
emf
I
Ohmmeter
Ammeter with a small voltage source
27
Quantitative Analysis of an RC Circuit
Vround _ trip  emf  RI  VC  0
Q
0
C
dQ emf  Q / C
I

dt
R
emf
I

Initial situation: Q=0
0
R
emf  RI 
Q and I are changing in time
dI d  emf  d  Q 
 
 

dt dt  R  dt  RC 
Q
VC 
C
d
dt
dI
1 dQ

dt
RC dt
dI
1

I
dt
RC
28
RC Circuit: Current
dI
1

I
dt
RC
1
1
dI  
dt
I
RC
I
t
1
1
dI


dt
I I

RC 0
0
t
ln I  ln I 0  
RC
I
t
ln  
I0
RC
I
e
I0
t

RC
Current in an RC circuit
I  I 0e  t / RC
What is I0 ?
Current in an RC circuit
I
emf t / RC
e
R
29
RC Circuit: Charge and Voltage
What about charge Q?
I
dQ
dt
dQ  Idt
Current in an RC circuit
I  I 0e
 t / RC
Current in an RC circuit
emf t / RC
I
e
R
t
emf
Q   Idt 
R
0
t
t / RC
e
dt

0
Q  C emf 1  et / RC 
Q
V 
C
Check: t=0, Q=0, t--> inf, Q=C*emf
30
RC Circuit: Summary
Current in an RC circuit
I
emf t / RC
e
R
Charge in an RC circuit
Q  C emf 1  et / RC 
Voltage in an RC circuit
V  emf 1  et / RC 
31
The RC Time Constant
Current in an RC circuit
emf t / RC
I
e
R
When time t = RC, the current I drops by a factor of e.
RC is the ‘time constant’ of an RC circuit.
e t / RC  e 1 
1
 0.37
2.718
A rough measurement of how long it takes to reach final equilibrium
32