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Finding charge of a capacitor in RC circuit

  1. Mar 27, 2013 #1
    Thanks, solved!
     
    Last edited: Mar 28, 2013
  2. jcsd
  3. Mar 28, 2013 #2

    ehild

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    Yes, after a short time that the circuit was connected the capacitor is charged and does not take current any more.

    The capacitor is connected parallel to the 4 ohm resistor. What do you know about the voltage across parallel connected circuit elements?

    ehild
     
  4. Mar 28, 2013 #3
    Okay so I can calculate current I=V/R = 19/16 = 1.1875 A

    V total = (V of Capacitor) + (V of 11Ω Resistor)

    V of 11Ω Resistor = IR = (1.1875)(11) = 13.0625

    19 = (V of Capacitor) + 13.0625
    V of Capacitor = 5.9375

    Then plugging into the capacitance formula:
    Q = CV = (4µ)(5.9375) = 23.75µ

    Does this work look good?
     
  5. Mar 28, 2013 #4

    ehild

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    The current is correct.

    No, what do you mean of Vtotal? And you do not need it.



    You know the voltage across the 11 Ω resistor. Look at the figure. Where are the plates of the capacitor connected to?

    ehild
     
  6. Mar 28, 2013 #5
    Is the voltage I want then just the voltage across the 4Ω resistor?
     
  7. Mar 28, 2013 #6

    ehild

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    Yes. And what current flows though it?


    ehild
     
  8. Mar 28, 2013 #7
    Ah is it the same current as throughout, since the capacitor is quickly charged?

    (I = 1.1875 A so V = (1.1875)(4) and then this voltage is the same that would go through the capacitor?)
     
  9. Mar 28, 2013 #8

    ehild

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    Yes, they are connected to the same points, so the voltage must be the same across both. (The voltage does not "go through", it is across two points)

    ehild
     
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