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RC Circuit help

  1. Aug 5, 2008 #1
    1. The problem statement, all variables and given/known data
    (a) What is the potential of point a with respect to point b in Fig. 26.61 (V = 16.0 V, R1 = 7.00 , R2 = 4.00 ) when switch S is open?

    (b) Which point, a or b, is at the higher potential?

    (c) What is the final potential of point b with respect to ground when switch S is closed?

    (d) How much does the charge on each capacitor change when S is closed?



    2. Relevant equations

    C=q/V, V=IR, Req=R1+R2

    3. The attempt at a solution

    I'm having trouble getting started on this one. I understand capacitors in a circuit, and I understand resistors in a circuit. I also understand RC circuits for the most part. Where I seem to get lost is when there are multiple capacitors with resistors.

    As for an attempt at the answer...
    a/b) I know that when the capacitors are fully charged there will be no current flowing. Is this the case at points a and b? If there is no current then V=IR at point b will be 0? Same at point a? If so, how can one be at a higher potential than the other? If not, then I'm not sure how to start part a.
    c) When the switch is closed, then the V source will have a direct route to ground through R1 and R2. Potential at point b is 16V?
    d) The capacitors would then discharge through the ground. So I assume the change of charge would be equal to the total charge of each capacitor. Q=CV the total charge would be Q1=C1(16V) and Q2=C2(16V).

    Any help would be much appreciated!
     

    Attached Files:

  2. jcsd
  3. Aug 6, 2008 #2

    LowlyPion

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    The V = IR describes the voltage drop across a resistor. So yes the first part of your answer is correct with respect to "b", because there is no voltage difference from the ground state at b.

    That said what voltage drop would you expect at point "a" if no current is flowing? How different from the voltage "V" would that be?
     
  4. Aug 6, 2008 #3
    So the voltage drop between 'a' and 'b' would be 16V with b at the higher potential.

    Am I on the right track for part c and d of the question?
     
  5. Aug 6, 2008 #4

    LowlyPion

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    Why would "b" be at the higher potential if it had a direct connection through the resistor R2 to ground?

    No. Part c is asking what the voltage potential is at "b". But the current is now flowing through "b". The point "b" must necessarily be at a voltage between V and ground - a voltage you can calculate with Ohm's law V = IR, if I am not mistaken.

    No. Part d is asking what the change was. What was the initial charge on each capacitor? What is the final charge on each? They are asking for that difference.

    From Q = CV and the given capacitor values and the known voltage potentials you provided in parts a and b, you know how to calculate that now.
     
  6. Aug 6, 2008 #5
    After the switch is closed, I can figure out how much current is flowing with I=V/Req. To find the potential at b, I thought it would just be I*R1, but that isn't right. The 6uF capacitor is discharging through 'b' now too, right? Wouldn't that effect my potential at 'b'?

    I know that the initial charge on each C is Qmax=CV. So Q1max = C1*V and Q2max = C2*V. This wouldn't be Ceq, right? Capacitors discharge as a function of time, but there is no amount of time given in the problem. After the switch is closed, the current will only flow through R1 and R2 to ground (path of least resistance), right? The capacitors will eventually discharge through ground completely. Which made me think that the change in Qs would be equal to the initial Qs. Either this is wrong, or my initial Qs are wrong.
     
  7. Aug 6, 2008 #6

    LowlyPion

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    First of all when the switch closes the charge on the capacitors will dissipate to the new steady state charge in a very brief time. You should ignore the transient changes for this problem, because they are asking for "final" charges on the capacitors.

    So not to get side tracked here - what is the current that will be flowing through "b" - steady state - after the charges on the capacitors have settled to their final value?

    By "Req" I am presuming that you mean "R1 + R2" and if so then yes your current is V/Req. However, the voltage across R2 looks to me to be the Voltage at "b". That is the step up from ground. The voltage across R1 is the additional voltage step to get you to the 16v.

    Now that you know what the voltage at "a" and "b" is you can calculate the FINAL charge on C1 and C2. You know what the INITIAL charges are because there was 16V across both C1 and C2 before the switch was closed. Simply subtract Final from Initial to get your changes.

    Edit: To avoid confusion in calculating the FINAL charge on the 6 uf capacitor that would be the voltage across R1. The voltage across the 3 uf capacitor would be the voltage across R2
     
    Last edited: Aug 6, 2008
  8. Aug 6, 2008 #7

    LowlyPion

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    To straighten this out the answer is no.

    Q = V * C

    If there is a voltage - there is a charge. Until you disconnect the voltage and the voltage across C is zero the charge will remain constant.
     
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