Videos
SP212
Ch. 27 – Circuits
Add PHet Demonstrations for Circuits (at end of class)
Do Capacitor Circuit Demonstration.
Find Circuit from Wolfram
Maj Jeremy Best USMC
Physics Department, U.S. Naval Academy
February 23, 2016
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SP212
February 23, 2016
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Find the Physics
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SP212
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Circuits
In order to make charges move through a conductor, we
need to maintain a potential difference across it.
We do this with an emf device , which is said to provide
an emf (E) , like a “charge pump”.
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SP212
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SP212
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emf
emf is defined as the work per unit charge that the
device does in moving charge from its negative to
positive terminals.
E=
dW
Joules
=
dQ
Coulomb
Thus, the SI unit for emf is the volt.
Some other Emf Devices?? Generators, Solar Panels,
thermopiles, nuclear batteries
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SP212
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Battery Resistance
Real emf devices also have an internal resistance which
dissipates some energy in the form of useless heat
E = iR
Therefore:
i=
E
R
Think of how a battery heats up. The chemical reactions
are not instantaneous. What happens when you plug in
your phone? Or Calculator...
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SP212
February 23, 2016
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SP212
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Resistors in Circuits
Problem: Equivalent Resistance
Resistors follow the same two algorithms as capacitors,
but differently ( Opposite ):
Resistors in series add “simple”:
n
X
Req =
Ri
R6
R2
R5
i=1
Resistors in parallel add “weird”
n
X
1
1
=
Req
Ri
R1
R3
R4
i=1
Think of water where velocity = i, volume = V, and
pumps are like batteries. Draw.
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SP212
February 23, 2016
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Maj Jeremy Best USMC (Physics Department, U.S. Naval Academy)
SP212
Solution:
1
Imagine a battery connected between the open ends.
2
Find equivalent resistance of parallel Resistors.
1
1
1
+
=
R1 R2
R12eq
1
1
1
1
+
+
=
R4 R5 R6
R456eq
3
Find equivalent resistance of series.
R12eq + R − 3 + R456eq = Req
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SP212
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Kirchoff’s Laws
Determining the current through a circuit and/or the
voltage drop across individual components is the main
topic of this chapter. We will do this using Kirchoff’s
Laws
Voltage Law (KVL): The algebraic sum of the
changes in potential around any complete traverse
of a loop in a circuit must equal 0.
Current Law (KCL): The sum of all currents
entering a node must equal the sum of all currents
leaving the node.
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SP212
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Circuit Analysis
X
Vall = 0
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SP212
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SP212
Circuit Analysis
Example
To apply these, do the following:
Assume a direction for current through each branch
in the circuit
Label the polarity of every circuit element, positive
on the “upstream” side, except for batteries, which
are labeled based on their inherent polarity
Traverse the loop (in any direction), writing down
the algebraic sum of the potential drops/emfs you
encounter, using the second polarity symbol. Set
this equal to 0.
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SP212
February 23, 2016
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R1
+
Assume current goes
clockwise, and walk
around the loop:
i
v −
+
v
E
R2
−
− v
i
i
+
R3
Maj Jeremy Best USMC (Physics Department, U.S. Naval Academy)
SP212
E
E
E
E
E
E
E
− iR1
− iR1 − iR2
− iR1 − iR2 − iR3 = 0
= iR1 + iR2 + iR3
= i(R1 + R2 + R3 )
= iReq
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Problem: Double Loop
The components in the following circuit have the values
E1 = 3.0 V, E2 = 6.0 V , E3 = 12.0 V, R1 = 38.0Ω, and R2 = 22.0Ω.
Find the magnitude and direction of the current.
R1
R1
R2
Solution:
On Board
E3
E1
E2
R1
R1
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SP212
February 23, 2016
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Maj Jeremy Best USMC (Physics Department, U.S. Naval Academy)
SP212
RC Circuits
RC Circuits
This is a first order differential equation to which we
know solutions:
Apply Kirchoff’s
Voltage Law
R
q(t) = C E(1 − e −t/RC )
dq
E
i(t) =
=
e −t/RC
dt
R
q
VC (t) = = E(1 − e −t/RC )
C
q
=0
C
dq
i=
dt
dq
q
R
+ =E
dt
C
E − iR −
E
C
These apply for a *Charging* Capacitor!!!
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SP212
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Problem: Charging Capacitor
The switch in the figure is closed at t=0, to begin
charging an initially uncharged capacitor of capacitance
C=17.7 µF through a resistor of resistance R=20.0 Ω.
At what time (in ms) is the potential across the
capacitor equal to that across the resistor?
R
E
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C
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Solution:
If we want the Voltage (potential) to be equal, we have
to look at 3 equations and combine them to make it
happen:
Lets try Ohms Law.
V = iR
RC Circuits - Discharging
To discharge the capacitor, we flip the switch and take
the battery out of the circuit. Now the charge steadily
decreases and the current flows the opposite direction,
while getting smaller in magnitude:
VC (t) = E(1 − e −t/RC )
E
i(t) =
e −t/RC
R
−t/RC
E(1 − e
) = Ee −t/RC
q(t) = q0 (e −t/RC )
q dq
0
i(t) =
e −t/RC
=−
dt
RC
q
VC (t) = = V0 e −t/RC
C
e −t/RC = (1 − e −t/RC )
t = RC ln 2
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SP212
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Problem: Finding τ
What multiple of the time constant τ gives the time
taken by an initialy uncharged capacitor in an RC circuit
to be charged to 85% of its final charge?
The Time Constant
The time constant is an often used shorthand notation:
Solution:
τ = RC
This represents a “characteristic time” for the circuit
(many other systems as well). It is the time it takes for
the system to change by a factor of e.
q(t) = C E(1 − e −t/τ )
where qeq = C E and τ = RC
.85C E = C E(1 − e −t/RC )
e −t/RC = 1 − .85
t
= − ln .15 = 1.90
τ
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SP212
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