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Capacitor question

  1. Mar 22, 2008 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations
    V = IR
    CV = Q

    3. The attempt at a solution
    I found the equivalent resistance to be 68 Ohms, then I found the current, then I found that the current splits 4/5 of its total across the branch with the capacitor, but if I apply Kirchhoff's loop law, I just get that the capacitor has a potential difference of 0 V.
  2. jcsd
  3. Mar 23, 2008 #2
    "The switch has been closed for a long time."

    I don't know how this should be apparent an introductory level but...

    With the switch closed for a while, everything is static, the currents are done changing because the capacitor has as much charged as it's going to get. So...

    No current flows through the capacitor. It's, like, an open circuit.
  4. Mar 23, 2008 #3
    Oh...so then no current also flows through the 10 Ohm resistor?
  5. Mar 23, 2008 #4
    You've got it. well done
  6. Mar 23, 2008 #5
    Okay, so if the 2nd branch doesn't affect the circuit, the current is 100 / (60+40) = 1A, and so the voltage difference across the 40 Ohm resistor is equal to the voltage difference across the capacitor? :S
  7. Mar 23, 2008 #6
    That's correct.
  8. Mar 23, 2008 #7
    What's the point of that 10 Ohm resistor then?
  9. Mar 23, 2008 #8
    I dunno. It would effect the charge rate when the switch is closed. Ask your prof.
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