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woleraymond

  • 2 years ago

The capacitance of a variable radio capacitor can be changed from 50PF to 95PF by turning the dial from 0 to 180 degrees. With the dial set at 180 degrees , the capacitor is connected to a 400V battery. After charging the capacitor is disconnected from the battery and the dial is turned to 0 degrees. a - What is the charge on the capacitor b- What is the p.d across the capacitor when the dial reads 0 degrees. c - WHat is the energy of the capacitor in this position?

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  1. radar
    • 2 years ago
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    a. Q=CV=(95 10^-12)(400) = 38 10^-9 coulomb b.760 volts c. Use approriate equation (I don't have it memorized)

  2. woleraymond
    • 2 years ago
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    @radar how did you arrive at your answer pls explain @experimentX @RaphaelFilgueiras @Algebraic! any info

  3. Algebraic!
    • 2 years ago
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    which part?

  4. RaphaelFilgueiras
    • 2 years ago
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    c) E=1/2 C V²

  5. RaphaelFilgueiras
    • 2 years ago
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    @woleraymond are you there?

  6. woleraymond
    • 2 years ago
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    yep im here

  7. RaphaelFilgueiras
    • 2 years ago
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    what you didnt understood?

  8. woleraymond
    • 2 years ago
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    how did @radar arrive at 760 volts for question b..also what impact does the turning have on the capacitor

  9. RaphaelFilgueiras
    • 2 years ago
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    the charge will be the same,when you turn the dial to zero degrees ,you will change the capacitance for 50 PF

  10. RaphaelFilgueiras
    • 2 years ago
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    then q=c.v-> 38E-9=50E-12.v->v=760

  11. RaphaelFilgueiras
    • 2 years ago
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    now energy will be 1/2.c.v²=1/2. 50E-12.(760²)

  12. RaphaelFilgueiras
    • 2 years ago
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    14.4 microjoules

  13. woleraymond
    • 2 years ago
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    interesting....

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