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Circuit question

  1. Jul 27, 2011 #1
    Please see the attached word document for the circuit and further explanation. any help would be appreciated.

    Attached Files:

  2. jcsd
  3. Jul 27, 2011 #2


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    Staff: Mentor

    The voltage at A before the switch is [STRIKE]closed[/STRIKE] opened is not 5V. The two resistors form a voltage divider, so V(A) is less than 5V. Have you learned about voltage dividers yet?

    If not, just use V=IR as you mention in the document, and figure out what the current is through the two resistors. Then use that current to figure out the voltage drops across the individual resistors.

    Makes sense?

    EDIT -- I see that I misread the question -- it asks for the steady state voltage when the switch has been closed for a long time. The rest of my comments still hold though.
    Last edited: Jul 27, 2011
  4. Jul 27, 2011 #3


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    Staff: Mentor

    The 800 and 3200 Ohm resistors form a voltage divider.

    At steady state, when the capacitor has charged to whatever its final value will be (presuming some relatively long time has elapsed during which the switch at A was closed), the capacitor will effectively "look like" an open circuit -- no current will be flowing into or out of it -- the circuit will present a voltage at node A as though the capacitor was not present. What might that voltage be?

    You might start by considering what current will flow through the two series-connected resistors.
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