MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
8.02
Fall 2009
Problem Set 3
Solutions
Problem 1:
LC
Circuit
(a) Initially, the capacitor in a series
LC
circuit is charged.
A switch is closed, allowing
the capacitor to discharge, and after time
T
the energy stored in the capacitor is one
quarter its initial value.
Determine
L
if
C
and
T
are known.
If the energy is at ¼ of it’s initial value then the voltage has fallen in half.
That is,
() ( )
2
0
0
2
19
cos
23
3
V
T
VT V
T
T
L
TC
LC
ππ
ωω
ω
π
==
⇒
=
⇒
=
=
⇒
=
(b) An
LC
circuit consists of a 90.0mH inductor and a
1.000 F
μ
capacitor.
If the
maximum instantaneous current is 0.100 A, what is the greatest potential difference
across the capacitor?
By conservation of energy,
22
11
max
max
LI
CV
=
, so
()
max
max
90 mH
0.1 A
30 V
1.0
μ
F
L
VI
C
=
Problem 2: Filters
One of the very useful things that you can do now that you understand AC circuits and
impedance is design filters.
A filter is a circuit that takes an input and returns an output
which, for a certain range of frequencies, is attenuated.
There are four basic filter types:
low pass
(in which low frequencies are let through unattenuated, but high frequencies are
blocked),
high pass
(the opposite),
band pass
(essentially a combination of a low and
high pass filter, letting through signals in a range, or band, of frequencies, and blocking
those outside), and
stop band
(the opposite).
There are also
notch
or
trap
filters which
are similar to stop band filters but try to notch out at a single frequency (such as 60 Hz,
which has an annoying habit of creeping into every signal and is frequently eliminated
using a notch filter).
When trying to analyze filters there are several simple steps to take.
First of all, in
thinking about what a filter will do, first look at the low and high frequency limits.
Recall that at low frequencies inductors look like wires and capacitors like large resistors
(or open circuits) while at high frequencies the opposite is true.
Problem Set 3 Solutions
p. 1 of 8
Fall 2009
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Fall 2009
Problem 2: Filters
continued
…
(a) This is a
high pass
filter.
At low frequencies the capacitor
will fill up and look like an open circuit, while at high
frequencies it will look like a wire.
The output voltage is given
by:
()
0,Out
0
0,in
0,in
22
2
0,Out
0
0,in
0
1
1
11
1
Transfer Func.
where
is the 3dB point
1
C
C
R
VI
R
V
V
RX
XR
V
VR
C
RC
ω
ωω
==
=
+
+
=
=
++
To find the slope we go to very low frequencies where our function simplifies:
2
0
0
1
Transfer Func.=
1
≈
+
, so clearly the slope is 20 log 2 = 6 dB per octave.
Here is a plot of the transfer function:
0.01
0.1
1
10
100
0.01
0.1
1
40 dB
20 dB
0 dB
slope = 6 dB/octave
= 20 dB/decade
3dB freq =
ω
0
Transfer Func.
Reduced Frequency:
ω
/
ω
0
(b) This is a
notch
filter.
At high frequencies the inductor
impedance will dominate the circuit and the output will essentially
see all of the voltage from that.
At low frequencies the capacitor
will dominate and again the output will get everything.
On
resonance the combined impedance of L & C in series will be zero
so the output voltage will fall to zero.
Let’s calculate:
( )
0,Out
0
0,in
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 Fall '07
 Hudson
 Frequency, LC circuit, high frequencies, low frequencies

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