Transcript Presentation_W04D2

Class 10: Outline
Hour 1:
DC Circuits
Hour 2:
Kirchhoff’s Loop Rules
P10- 1
Last Time:
Capacitors & Dielectrics
P10- 2
Capacitors & Dielectrics
Capacitance
To calculate:
1) Put on arbitrary ±Q
2) Calculate E
3) Calculate V
Q
C
V
Energy
2

E
Q2 1
1
2
U
 Q V  C V   uE d 3r   o d 3r
2
2C 2
2
Dielectrics
  E  dA 
S
free
qinside
0
 CFilled with Dielectric   C0
P10-
This Time:
DC Circuits
P10-
Examples of Circuits
P10- 5
Current: Flow Of Charge
Average current Iav: Charge Q
flowing across area A in time t
Q
I av 
t
Instantaneous current:
differential limit of Iav
dQ
I
dt
Units of Current: Coulombs/second = Ampere
P10- 6
Direction of The Current
Direction of current is direction of flow of pos. charge
or, opposite direction of flow of negative charge
P10- 7
Current Density J
J: current/unit area
 I
J  Iˆ
A
Î points in direction of current
I   J  nˆ dA   J  d A
S
S
P10- 8
Why Does Current Flow?
If an electric field is set up in a conductor, charge
will move (making a current in direction of E)
Note that when current is flowing, the conductor is
not an equipotential surface (and Einside ≠ 0)!
P10- 9
Microscopic Picture
Drift speed is velocity forced by applied electric field
in the presence of collisions.
It is typically 4x10-5 m/sec, or 0.04 mm/second!
To go one meter at this speed takes about 10 hours!
How Can This Be?
P10-10
Conductivity and Resistivity
Ability of current to
flow depends on
density of charges &
rate of scattering
Two quantities summarize this:
s: conductivity
r: resistivity
P10- 11
Microscopic Ohm’s Law
E  rJ
r
or
J sE
1
s
r and s depend only on the microscopic properties
of the material, not on its shape
P10-12
Demonstrations:
Temperature Effects on r
P10-13
PRS Questions:
Resistance?
P10-14
Why Does Current Flow?
Instead of thinking of Electric Field, think of potential
difference across the conductor
P10-15
Ohm’s Law
What is relationship between V and current?
b
V  Vb  Va   E  d s  E
a
E V /
J 
r
r
I
J
A


r 
  V  I 
  IR
 A 


P10-16
Ohm’s Law
V  IR
R
r
A
R has units of Ohms (W) = Volts/Amp
P10-17
Examples of Circuits
P10-18
Symbols for Circuit Elements
Battery
Resistor
Capacitor
Switch
P10-19
Sign Conventions - Battery
Moving from the negative to positive terminal of a
battery increases your potential
 V  Vb  Va
Think:
Ski Lift
P10-20
Sign Conventions - Resistor
Moving across a resistor in the direction of current
decreases your potential
 V  Vb  Va
Think:
Ski Slope
P10-21
Sign Conventions - Capacitor
Moving across a capacitor from the negatively to
positively charged plate increases your potential
 V  Vb  Va
Think:
Ski Lodge
P10-22
Series vs. Parallel
Series
Parallel
P10-23
Resistors In Series
The same current I must flow through both resistors
V  I R1  I R2  I ( R1  R2 )  I Req
Req  R1  R2
P10-24
Resistors In Parallel
Voltage drop across the resistors must be the same
V  V1  V2  I1R1  I 2 R2  IReq
V V V
I  I1  I 2 


R1
R2
Req
1
1
1
 
Req R1 R2
P10-25
PRS Questions:
Light Bulbs
P10-26
Kirchhoff’s Loop Rules
P10-27
Kirchhoff’s Rules
1. Sum of currents entering any junction in a circuit
must equal sum of currents leaving that junction.
I1  I 2  I 3
P10-28
Kirchhoff’s Rules
2. Sum of potential differences across all elements
around any closed circuit loop must be zero.
 
V    E  d s  0
Closed
Path
P10-29
Internal Resistance
Real batteries have an internal resistance, r, which is
small but non-zero
Terminal voltage:
V  Vb  Va    I r
(Even if you short the leads you don’t get infinite current)
P10-
Steps of Solving Circuit Problem
1.
2.
3.
4.
5.
Straighten out circuit (make squares)
Simplify resistors in series/parallel
Assign current loops (arbitrary)
Write loop equations (1 per loop)
Solve
P10-31
Example: Simple Circuit
You can simplify
resistors in series
(but don’t need to)
What is current
through the bottom
battery?
P10-32
Example: Simple Circuit
Start at a in both loops
Walk in direction of current
2  I1R   I1  I 2  R  0
  I 2  I1  R    0

Add these: 2  I1 R    0  I1 
R
We wanted I2:
 I 2  I1  R    I 2 
I2  0

R
 I1
P10-33
Group Problem: Circuit
Find meters’ values. All resistors are R, batteries are 
HARDER
EASIER
P10-34
Power
P10-35
Electrical Power
Power is change in energy per unit time
So power to move current through circuit elements:
d
d
dq
P  U   qV  
V
dt
dt
dt
P  I V
P10-36
Power - Battery
Moving from the negative to positive terminal of a
battery increases your potential. If current flows
in that direction the battery supplies power
I
Psupplied  I V  I 
P10-37
Power - Resistor
Moving across a resistor in the direction of current
decreases your potential. Resistors always
dissipate power
V
Pdissipated  I V  I R 
R
2
2
P10-38
Power - Capacitor
Moving across a capacitor from the positive to
negative plate decreases your potential. If current
flows in that direction the capacitor absorbs power
(stores charge)
2
dQ Q d Q
dU
Pabsorbed  I V 


dt C dt 2C dt
P10-39
Energy Balance

Q
  IR  0
C
Multiplying by I:

2


Q
dQ
d
1
Q
2
2
I  I R
 I R 

C dt
dt  2 C 
(power delivered by battery) = (power dissipated through resistor)
+ (power absorbed by the capacitor)
P10-40
PRS Questions:
More Light Bulbs
P10-41