Electric Circuits

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Electric Circuits
Lectures 13, 14, 15
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George M. Fuller 427 SERF gfuller_at_ucsd.edu 822-1
214
Office Hours 200 PM - 330 PM Tuesday 329 SERF
PM
TH 700 PM - 800 PM WLH 2204
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problem assignment TBA
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E
Sources of Electromotive Force, or emf
Essentially these are devices that maintain a
constant potential difference (voltage) E across
their terminals.
Example a perfect battery
E
R1
R2
current I
Conservation of energy around the circuit E
- I R1 - I R2 0
Voltage
Location in circuit
battery increases potential to E
voltage drop I R1
voltage drop I R2
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Non-Ideal (Real) Batteries
These have an internal resistance. There is a
limit to the current they can supply. This limit
has its origin in the finite rates of the
chemical reactions in them which separate
charges.
We can model them as an ideal EMF plus an
internal resistance
-
RINT

E
RLOAD
E I RINT I RLOAD
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Resistors in series act like Voltage Dividers
Each has the same current through them.
Example a perfect battery and two resistors in
series, with an assumed steady state
current I as shown. If R1 5W represents a
load for which we desire a current
of 0.6 A through it, what should R2 be if E
12 V?
E
R1
R2
Conservation of energy around the circuit E
- I R1 - I R2 0
current I
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Adding resistances in parallel and in series
Parallel
Series
in both cases, potential drop from A to B is VAB
A
A
R1
R1
R2
R2
B
Parallel Case each resistor has the same
potential across it
B
Series Case each resistor has the same current
through it
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Adding capacitances in parallel and in series
in both cases, potential drop from A to B is VAB
Parallel
Series
A
A
C1
Note that this element is neutral
C1
C2
C2
B
B
Series Case each capacitor has the same charge
on it
Parallel Case each capacitor has the same
potential across it
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Kirchhoffs Laws for Analyzing (multi-loop)
Circuits

• The Loop Law (Energy Conservation)
• The sum of voltage changes across all
  circuit elements around
• any closed loop 0.

(2) The Node Law (Conservation of Electric
Charge no buildup of charge
in steady state current flow)
The sum of currents at any node (junction of
wires/elements) 0.
Procedure for any circuit
(a) Identify circuit loops and nodes.
(b) Assign variables and directions to each
current. (The directions you assign can be
arbitrary the correct signs will emerge
from the algebra.)
(c) Sum these assigned currents at each node and
set this sum to zero.
(d) Sum voltage gains and drops through each
element around each loop. The direction you
choose around each loop is arbitrary. It will all
work out as long as you follow these
conventions
(d.1.) For batteries traversed from the negative
to positive direction the voltage change is
E, while for battery traversal in the other
direction it is -E.
(d.2.) For resistors traversed in the direction
of your assigned current the voltage change is -I
R, while for resistors traversed in a
direction opposite to your assigned current, the
change is I R.
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Example In a circuit as shown three batteries
have EMFs E1 12 V, E2 3 V, and
E3 3 V, with the indicated directions,
and three resistors have
resistances R1 1 W, R2 5 W, and R3 5 W.
Find the current through
resistor R3.
Assign currents as shown.
R3
Identify loops
I1
R1
I3
R2
I2

E1
-
-
Loop 1
Loop 2

E2
E3
-

Use the Node Law relation in the Loop 1 equation
to eliminate I2, then do same in the Loop 2
equation.
Eliminate I1 between these equations to find I3.
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Example In a circuit as shown three batteries
have EMFs E1 12 V, E2 3 V, and
E3 3 V, with the indicated directions,
and three resistors have
resistances R1 1 W, R2 5 W, and R3 5 W.
Find the current through
resistor R3. Continued . . .
Notice that our result is negative. What does
that mean?
It means that the current flows in the opposite
direction from the direction that we had chosen
for I3 on the previous slide!
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Example The Wheatstone Bridge Problem 31. What
is the equivalent resistance
measured
between points A and B in the diagram?
Note the symmetry in the problem
A
R
If we assign currents I1 and I2 as shown then
I1
I I1 I2
2R
I2
E
R
C
D
I1 - I2
R
2R
B
Loop ACDA
Loop ACBA
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Voltmeters
A voltmeter measures the potential difference
across its terminals.
For example, if we want the potential difference
across resistor R2 in the circuit shown, we would
attach the terminals on either side of
the resistor, and in parallel with it.
R1
E
R2
V
Clearly, to get an accurate reading of the
potential difference across this circuit
element we need the voltmeter to have an internal
resistance RINT gtgt R2.
The total resistance through this circuit element
is now RTOT
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Ammeters
An ammeter measures the current flowing through
itself.
For example, if we want the current flowing
through resistors R1 and R2 in the circuit shown,
we would attach the ammeter in series as shown.
R1
E
R2
A
Clearly, to get an accurate reading of the
current through this part of the circuit we need
the ammeter to have an internal resistance RINT
ltlt R1 R2.
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Example Problem 30. What is the current
measured through the ammeter in the diagram
below if R
2.0 W and E 6.0 V?
If we assign currents as shown, then by the node
law and symmetry
A
R
I1
I I1 I2
R
I2
E
I3
A
C
I4
D
I5
2R
R
B
Consider the node law at point C
So now we need to know I and for that we need
the resistance
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Problem 30 solution continued . . .
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Circuits with Capacitors Case 1 the charging
RC circuit
Imagine that you have an open circuit as shown
and that at time t0 you close the circuit. What
happens?
R
Current I begins to flow through the resistor
and the capacitor begins accumulating charge.
Loop law gives
E
C Q/V
Note that the capacitor is charging so
Our loop law equation can then be written as
Now multiply both sides of this equation by
Note that we Can re-write the left hand side
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Circuits with Capacitors Case 1 the charging
RC circuit, Continued . . .
Now integrate both sides of this equation from
t0 to t
First the left hand side then the right hand side
Equating these again and rearranging terms
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Circuits with Capacitors Case 1 the charging
RC circuit, Continued . . .
For which the limiting cases are
Now calculate the current as a function of time
Note the limiting cases for the current
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Dimensions of RC
The product RC is called the time constant for
the circuit. Clearly, the bigger its value the
longer it will take for the capacitor to charge
and the longer it will take for the current to
die away.
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Circuits with Capacitors Case 1 the charging
RC circuit, Continued . . .
How does the potential difference V(t) Q/C
across the the capacitor plates change with time?
Note the limiting cases here the potential
across the capacitor is initially zero, and only
when t gtgt RC does it reach the batterys potential
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Circuits with Capacitors Case 1 the charging
RC circuit, Continued . . .
Our result for the current as a function of time
was
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Circuits with Capacitors Case 2 discharging
the capacitor in an RC circuit
Imagine that you have an open circuit as shown
with a charged capacitor and that at time t0 you
close the circuit. What happens?
Current I begins to flow through the resistor
and the capacitor begins to lose charge.
R
loop law gives

C Q/V
-
Note capacitor is dis-charging so
Our loop law equation can then be written as
Re-write as
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Short and long time behavior of an RC-circuit
R1
Close the switch
R2
C
R1
At very short times the capacitor is not charged
and it therefore has zero potential across it so
it behaves like a short circuit
R2
R1
At very long times the capacitor is charged and
it therefore has a potential across it which has
magnitude V IR2 (E R2)/(R1R2). In this case,
the capacitor acts like an open circuit, as no
current passes through it.
R2
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Example Problem 75. Use the loop and node laws
for the circuit below to find
its time constant.
R1
If we assign currents as shown, then by the node
law gives
IC
I1
I2
E
C
R2
Choose the obvious two loops to find
Use these and the node relation to
eliminate first I1 and then I2 in the loop
equations to get
This is equation is of the same form as our
original capacitor charging equation
from which we found time constant