a) oscillates with a period of 8.000 ms? b) At what angle should the knob be set so that the circuit What voltage battery do I need to charge up the capacitor to an initial charge of 87.60 µC? (assume the period is set to 8.000 ms) What is the maximum current in this LC Circuit? c) (assume the initial charge is 87.60 µC)
a) oscillates with a period of 8.000 ms? b) At what angle should the knob be set so that the circuit What voltage battery do I need to charge up the capacitor to an initial charge of 87.60 µC? (assume the period is set to 8.000 ms) What is the maximum current in this LC Circuit? c) (assume the initial charge is 87.60 µC)
Introductory Circuit Analysis (13th Edition)
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Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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Transcribed Image Text:2) An LC circuit is made by attaching a 0.2221 H inductor is connected across a
variable capacitor whose capacitance can be varied by turning a knob.
a) . At what angle should the knob be set so that the circuit
ocillates with a period of 8.000 ms?
SuF
OµF
- 10µF
b)
What voltage battery do I need to charge up the capacitor to an
initial charge of 87.60 µC? (assume the period is set to 8.000 ms)
Сарасiance
c)
What is the maximum current in this LC Circuit?
(all numbers 4 sig figs)
(assume the initial charge is 87.60 µC)
Expert Solution
Step 1
a)
The LC circuit’s frequency can be represented as,
Here, L and C represent the inductance and capacitance, respectively.
Also,
Here, T is the time period.
Thus,
Substitute the relevant values.
Now, the knob makes an angle of 900 for 5μf capacitance.
Thus, for 1μf, the knob’s angle can be found as,
Thus, for 7.299μf, the knob’s angle is,
Substitute the relevant values.
Step 2
b)
In part (a), the capacitor required to have a time period of 8 ms is found to be 7.299μf.
Now, the charge in a capacitor can be represented as,
Here, V represents the capacitor’s voltage.
Substitute the relevant values.
Thus, the capacitor’s voltage is about 12.00 V.
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