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Homework Help: RC circuit problem

  1. Nov 16, 2012 #1
    I have a series rc circuit.

    80 v battery, capacitor1, resistor1: 10M ohm, capacitor 2, then back to the negative end of the battery.

    Capacitor 1 and 2 are equal, after a certain time, the voltage on each capacitor is 40.0v. What is the capacitance of each capacitor?

  2. jcsd
  3. Nov 16, 2012 #2
    You should make an attempt to solve the problem first before posting here
  4. Nov 16, 2012 #3
    My whole physics class could not colve this, we used the V = Vmax(1-e^(-t/RC)) equation, and Q= CV, but the Q could not be found to find C. Is there a trick to this?
  5. Nov 16, 2012 #4
    Probably you made a mistake with the time constant. Not sure but it could be.
  6. Nov 16, 2012 #5
    We used t = 5RC for time, that's the only thing we had.
  7. Nov 16, 2012 #6
    I asked about the time constant. It is 5C here.

    Lets forget about the resistor for a moment and consider only the capacitors and battery. What can be the maximum potential difference across both the capacitors?
  8. Nov 16, 2012 #7
    Would that be 2q/C ? From Ctot = C/2 and q = CV
  9. Nov 16, 2012 #8
    It is but what's the value of it? :smile:

    You have a 80 V battery.
  10. Nov 16, 2012 #9
    I don't know how to find q in that. Is it 40v?
  11. Nov 16, 2012 #10


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    Staff: Mentor

    Is this the full statement of the problem? Is there any information missing (such as the "certain time" having a specified value)?
  12. Nov 16, 2012 #11
    That's everything, our teacher said that we should use t = 5RC when the problems say 'after sufficient time' to charge the resistors fully
  13. Nov 16, 2012 #12


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    Staff: Mentor

    Then there is insufficient information to calculate specific capacitor values. Any two equal valued capacitors in the circuit will eventually end with a 40V potential across each of them. The only difference will be the time constant RC that applies, that is, what the value of t = 5RC will be when the circuit has essentially reached steady state.
  14. Nov 16, 2012 #13


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    Gold Member

    I think you need more help ...

    The equation t=5RC is an approximation. It means that after t (=5RC) the voltage will be 99% of the way towards 80V. In reality it never quite gets to 80V.


    So if you measure or are given t you can solve t=5RC to give C.

    Then you need to remember that the calculated value for C is is the total capacitance. eg for two capacitors in series. The question asks you for the individual values so you have to work those out.

    That should be sufficiently accurate for this problem. However for the future...

    This method is pretty inaccurate because the voltage curve is very flat as it approaches the supply voltage 80V. Any error in measuring the voltage changes t a lot and hence C. If doing this for real you would do better to measure the time to charge to say 8V or 16V (but still using the 80V supply) and then use the more general equation for the voltage on a charging capacitor...

    Vc = VCC (1-e-t/RC)

    There is another potential problem with the method as well. Many large capacitors have a high leakage current. You may find that this leakage current through the 10MOhm resistor produces a noticable voltage drop...so it never gets to even 99% of 80V.
  15. Nov 16, 2012 #14
    Thanks everyone! I'll let the class know.
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