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Homework Help: Final voltage across the capacitor

  1. Jun 25, 2014 #1

    The final voltage across the capacitor is the same as the source voltage is 90 V since a capacitor behaves as an open circuit in presence of a DC voltage.

    But what about the resistor? He is placed in series with the capacitor and wouldn't it be:

    final voltage cap = 90 V - voltage across the resistor? There is a voltage drop over the resistor right?


    NEver mind. I was confusing the meaning of short circuit with open circuit and a capacitor with inductor. No current is flowing through the cap and so the voltage on its terminals like the source voltage.

    My excuses.
  2. jcsd
  3. Jun 25, 2014 #2
    I get -30 for the final voltage. Try treating the circuits as two different circuits. circuit b, and circuit a, and look at their steady state voltages. and remember, its not an open circuit in the presence of DC, its an open circuit after a long period of just DC. There is stored energy in the cap that is going to go places.
  4. Jun 25, 2014 #3


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    Ignore me. I missed a resistor.
  5. Jun 25, 2014 #4
    I was talking about when the switch stays long enough in position b, then the final voltage is 90 V (of the left side of the circuit) because it takes over the votlage of the source.

    The initial voltage of vc is indeed -30V when you turn from a to b and its the 'final voltage of the circuit if switch a' is locked for a long time (so before), so the cap is a open circuit, no current, the resistor of 60 is a voltage divider 60/(60+20) times negative 40V = -30 V

    Sorry, If i wasn't clear. Im using the Nilsson Electric Circuits

    My problem was, I messed up the meaning of open and short circuits(perhaps also inductor and cap) and I thought current was flowing through the cap when in position b in steady state. In that erroneous case You would get a voltage drop across your resistor of 4000kΩ and so the voltage across the cap couldn't be 90V. That was a silly mistake.
    Last edited: Jun 25, 2014
  6. Jun 25, 2014 #5
    ah, i misread the circuit. I had it backwards. yah its 90. initial value is -30.

    after a while, there will be no current through the resistor, and hence no voltage drop. so the voltage will be the same on both sides.
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