hw10.6 - fC C X C 2 1 1 fL I V C 2 max max fC I X I V C C 2...

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HW 10.6 Given: F i n d : A 26 max m I ? max L V H 50 m L ? max C V ? max V 59 max C V C U Hz f . 5500 ? max f ? C The peak current in an LC circuit is I max = 26 mA. The inductance is fixed at L = 50 mH. The capacitance is variable and has a rating (breakdown voltage) of V C max = 59 V. (a) If the circuit is oscillating at a frequency f = 5500 Hz, what is the peak voltage across L ? (b) For the same conditions, calculate the peak voltage across C . (c) Calculate the peak energy stored in the capacitor under the above conditions. (d) What is the highest frequency at which this circuit can operate safely for the given peak current? (e) Calculate the capacitance that produces the resonant frequency calculated in part (d). Solutions a.) To find max we can use the following formula: L V fL L X L 2 fL I X I V L L 2 max max max fL I V L 2 max max b.) To find we can use the following formula: max C V fL I V V L C 2 max max max since 0 loop V fL I V C 2 max max
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c.) To find max we can use the following formula: C U 2 max 2 1 CV U C we just need C
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Unformatted text preview: fC C X C 2 1 1 fL I V C 2 max max fC I X I V C C 2 1 max max max fL I fC I 2 2 1 max max L f fL fI I C 2 2 max max 4 1 2 2 2 2 4 4 2 4 1 2 1 2 1 2 max 2 2 2 2 2 2 max 2 max 2 2 2 max L I L f L f I fL I L f CV U C 2 max max 2 1 LI U C d.) To find we can use the following formula: max f L f I V V L C max max max max 2 max max max 2 LI V f C e.) To find C we can use the following formula L f C 2 2 4 1 from part c 2 max 2 max 2 max 2 2 max 2 2 2 max 2 2 2 max 2 2 max max 2 2 max 2 4 4 4 4 1 2 4 1 4 1 C C C C V LI L V I L I L L V L LI V L f C 2 max 2 max C V LI C...
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hw10.6 - fC C X C 2 1 1 fL I V C 2 max max fC I X I V C C 2...

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