OSCILLATIONS IN AN LC CIRCUIT ε
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1.
Switch
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In the circuit shown in the figure, the battery emf is 55 V, the resistance R is 200 , and the capacitance C is 0.500 µF. The switch is closed for a very long time. a.
At this point in time what is the current i trough the inductor L? Solution: L
R
0.275 b.
Now the switch is opened and after a short time the potential difference across the capacitor reaches a maximum value of 150 V. What is the value of the inductance L? Hint: use energy conservation – where is the energy stored at the instant when the switch is opened? Where is the energy stored when the potential difference across the capacitor reaches its maximum value? Solution: At the time the switch is opened the energy is stored in the inductor (there is no voltage drop across the capacitor, so no energy is stored in the capacitor at this point in time). So we have 0
. When the potential difference across the capacitor reaches its ,
maximum there is no current flowing through the inductor, thus the total energy of the system is now stored in the capacitor for which we had found: . As there is no dissipation in ,
which one can solve for L: the circuit (no current flows through the resistor R) we have: 0.5 · 10
c.
0.149 .
What is the frequency f of the oscillations in this LC circuit? Solution: Here one can use that: 2.
2
√
or √
583
A variable capacitor Cvar with a range from 10 pF to 365 pF is used with a coil to form a variable frequency LC circuit to tune the input of a radio. a. What is the ratio of maximum frequency to minimum frequency that can be obtained with this variable capacitor? Solution: We can use √
to calculate the frequency of the circuit. From this one can see that the minimum frequency is obtained for the maximum capacitance and vice versa (independent of the value of L): and from which we get for the ratio: C
6 b.
If one would like to achieve a tunable frequency range from 0.54 MHz to 1.60 MHz the range calculated in a turns out to be too large. Suppose you want to modify the circuit by adding a capacitor CP in parallel to the variable capacitor to achieve this tuning range. What capacitor do you have to choose? Solution: and thus we For capacitors in parallel the capacitance adds, so the new capacitance is: have for the frequency ratio: .
which we can solve for CP: .
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35.6
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c.
1 Finally what is the value of the inductor one has to choose to obtain a tunable frequency range from 0.54 MHz to 1.60 MHz using the capacitor combination calculated in b? Solution: For the maximum frequency we have: 2
3.
which one can solve for L: or . ·
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0.217
. A series RLC circuit has an inductance L=20mH, a capacitance C=2μF and a resistance R=2Ω. At time t=0 oscillations start with a charge Q0 on the capacitor. At what time will the amplitude of the charge oscillation drop to 10% of the initial value? Solution: The amplitude of the charge oscillations is given by: , we are looking for the time for which:
0.1
thus: eliminating Q0 and taking the natural logarithm on both sides we have: ln 0.1 or 0.1
ln 0.1
0.046