Class 10: Outline
Hour 1:
DC Circuits
Hour 2:
Kirchhoff’s Loop Rules
P10- 1
Last Time:
Capacitors & Dielectrics
P10- 2
Capacitors & Dielectrics
To calculate:
1) Put on arbitrary ±Q
2) Calculate E
3) Calculate ∆V
Capacitance
Q
C=
∆V
Energy
2
E
ε
Q2 1
1
2
U=
= Q ∆V = C ∆V = ∫∫∫ uE d 3 r = ∫∫∫ o d 3 r
2
2C 2
2
Dielectrics
free
G
G
q inside
⇒ CFilled with Dielectric = κ C0
d
κ
E
⋅
A
=
w
∫∫
S
ε0
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
G I
J ≡ Iˆ
A
Î points in direction of current
G
G G
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:
σ: conductivity
ρ: resistivity
P10-11
Microscopic Ohm’s Law
G
G
E = ρJ
ρ≡
or
G
G
J =σE
1
σ
ρ and σ depend only on the microscopic properties
of the material, not on its shape
P10-12
Demonstrations:
Temperature Effects on ρ
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?
G G
∆V = Vb − Va = − ∫ E ⋅ d s = EA
b
a
∆V / A «
J= =
ρ
ρ ⎪⎪
» ρA ¿
¬ ⇒ ∆V = I ¼
À ≡ IR
I
½ A Á
⎪
J=
⎪­
A
E
P10-16
Ohm’s Law
∆V = IR
R=
ρA
A
R has units of Ohms (Ω) = 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 = I1 R1 = 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.
G G
∆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 =
ε −Ir
(Even if you short the leads you don’t get infinite current)
P10-
Steps of Solving Circuit Problem
1. Straighten out circuit (make squares)
2. Simplify resistors in series/parallel
3. Assign current loops (arbitrary)
4. Write loop equations (1 per loop)
5. 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
−2ε − I1 R − ( I1 − I 2 ) R = 0
− ( I 2 − I1 ) R + ε = 0
−ε
Add these: −2ε − I1 R + ε = 0 → I1 =
R
We wanted I2:
( I 2 − I1 ) R = ε → I 2 =
I2 = 0
ε
R
+ I1
P10-33
Group Problem: Circuit
Find meters’ values. All resistors are R, batteries are ε
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 ε
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
ε
Q
− − IR = 0
C
Multiplying by I:
ε
2
»
¿
Q
dQ
d
1
Q
2
2
= I R+ ¼
I = I R+
À
C dt
dt ½ 2 C Á
(power delivered by battery) = (power dissipated through resistor)
+ (power absorbed by the capacitor)
P10-40
PRS Questions:
More Light Bulbs
P10-41
0
