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woleraymond

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?

  • one year ago
  • one year ago

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  1. radar
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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)

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

    • one year ago
  3. Algebraic!
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    which part?

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

    • one year ago
  5. RaphaelFilgueiras
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    @woleraymond are you there?

    • one year ago
  6. woleraymond
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    yep im here

    • one year ago
  7. RaphaelFilgueiras
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    what you didnt understood?

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

    • one year ago
  9. RaphaelFilgueiras
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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

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

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

    • one year ago
  12. RaphaelFilgueiras
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    14.4 microjoules

    • one year ago
  13. woleraymond
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    interesting....

    • one year ago
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