Class 10: Outline
Hour 1:
DC Circuits
Hour 2:
Kirchhoff’s Loop Rules
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Capacitors & Dielectrics
Capacitance
C
=
Q
∆
V
To calculate:
1) Put on arbitrary ±Q
2) Calculate E
3) Calculate ∆ V
U
Energy
=
Q
2
=
2 C
1
2
Q
∆
V
=
1
2
C
∆
V
2 =
3 u d r
E
=
ε o
E
2
3 d r
2
Dielectrics w κ
G G
⋅ d
=
S
E A free q
ε inside
0
⇒
C
Filled with Dielectric
= κ
C
0
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This Time:
DC Circuits
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Examples of Circuits
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Current: Flow Of Charge
Average current I av
: Charge ∆ Q flowing across area A in time ∆ t
I av
=
∆
Q
∆ t
Instantaneous current: differential limit of
I av
I
dQ dt
Units of Current: Coulombs/second = Ampere
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Direction of The Current
Direction of current is direction of flow of pos. charge or, opposite direction of flow of negative charge
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Current Density J
J: current/unit area
G
J
I
A
I
ˆ
I points in direction of current
I
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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 E inside
≠ 0)!
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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?
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Conductivity and Resistivity
Ability of current to flow depends on density of charges & rate of scattering
Two quantities summarize this:
σ : conductivity
ρ : resistivity
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Microscopic Ohm’s Law
G
E
=
G
J or
G
J
=
G
E
ρ and σ depend only on the microscopic properties of the material, not on its shape
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Demonstrations:
Temperature Effects on
ρ
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PRS Questions:
Resistance?
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Why Does Current Flow?
Instead of thinking of Electric Field, think of potential difference across the conductor
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Ohm’s Law
What is relationship between ∆
V and current?
V V b
−
V a
= −
a b
G
E
⋅ d
G s
=
E
A
=
E
ρ
=
∆
V
ρ
/ /
I I
⎬
⎪
⎫
⎪⎪
⎪⎭
V I
⎝
⎛ ρ A
⎜
A
⎟
⎠
⎞
≡
IR
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Ohm’s Law
∆
V
=
IR
R
=
ρ A
A
R has units of Ohms ( Ω ) = Volts/Amp
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Examples of Circuits
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Symbols for Circuit Elements
Battery
Resistor
Capacitor
Switch
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Sign Conventions - Battery
Moving from the negative to positive terminal of a battery increases your potential
∆ =
V b
−
V a
Think:
Ski Lift
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Sign Conventions - Resistor
Moving across a resistor in the direction of current decreases your potential
∆ =
V b
−
V a
Think:
Ski Slope
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Sign Conventions - Capacitor
Moving across a capacitor from the negatively to positively charged plate increases your potential
∆ =
V b
−
V a
Think:
Ski Lodge
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Series vs. Parallel
Series Parallel
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Resistors In Series
The same current I must flow through both resistors
∆ =
I R
1
+
I R
2
=
(
1
+
R
2
)
=
I R eq
R eq
=
R
1
+
R
2
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Resistors In Parallel
Voltage drop across the resistors must be the same
∆ = ∆ = ∆ =
1 2
I R
1 1
=
I R
2 2
=
IR eq
I
= +
1
I
2
=
∆
V
+
∆
V
=
∆
V
R R R
1 2 eq
1
=
1
+
1
R R R eq 1 2
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PRS Questions:
Light Bulbs
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Kirchhoff’s Loop Rules
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Kirchhoff’s Rules
1. Sum of currents entering any junction in a circuit must equal sum of currents leaving that junction.
I
1
= +
2
I
3
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Kirchhoff’s Rules
2. Sum of potential differences across all elements
∆ around any closed circuit loop must be zero.
V
= −
E
Closed
G
⋅ d
G s
=
0
Path
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Internal Resistance
Real batteries have an internal resistance, r
, which is small but non-zero
Terminal voltage: ∆ =
V b
−
V a
I r
(Even if you short the leads you don’t get infinite current)
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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
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Example: Simple Circuit
You can simplify resistors in series
(but don’t need to)
What is current through the bottom battery?
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Example: Simple Circuit
Start at a in both loops
Walk in direction of current
2
ε
I R
1
− (
I
1
−
I
2
)
R
=
0
− (
I
2
−
1
)
0
Add these:
We wanted
I
2
:
2
(
ε
I
2
I R
−
1
1
)
ε
I
=
2
0
= ε
0
→ =
1
− ε
ε
R
→
I
2
= +
R
I
1
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Group Problem: Circuit
Find meters’ values. All resistors are R , batteries are
ε
HARDER EASIER
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Electrical Power
Power is change in energy per unit time
So power to move current through circuit elements:
P
= d dt
U
= d dt
( q V
) = dq dt
∆
V
P
=
I
∆
V
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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
P supplied
=
I
∆
V
=
I
ε
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Power - Resistor
Moving across a resistor in the direction of current decreases your potential. Resistors always dissipate power
P dissipated
=
I
∆
V
=
I
2
R
=
∆
V
2
R
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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)
P absorbed
=
I
∆
V
= dQ Q dt C
= d dt
Q
2
2 C
= dU dt
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Energy Balance
−
Q
−
IR
=
C
0
Multiplying by I:
I
= 2
I R
+
Q dQ
C dt
= 2
I R d
⎛
1 Q
2
⎞
⎟ dt ⎝ 2 C ⎠
(power delivered by battery) = (power dissipated through resistor)
+ (power absorbed by the capacitor)
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PRS Questions:
More Light Bulbs
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