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Homework Help: Simple Capacitance Question.

  1. Feb 9, 2012 #1
    1. The problem statement, all variables and given/known data

    An electronic engineer is designing a circuit in which a capacitor of capacitance of 4700 microFarads is to connected across a potential difference of 9.0V. He has four 4700 microFarads, 6V capacitors available. Draw a diagram to show how the four capacitors could be used for this purpose.

    2. The attempt at a solution

    I've got the answer (see the attachment), and I get it that the capacitance equals 4700 this way, but how does the voltage equal 9V like this? I don't get it. Can someone help? :(

    Attached Files:

  2. jcsd
  3. Feb 9, 2012 #2


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    Gold Member

    hmmm, by 6V capacitors I assume you mean voltage rating, the max voltage across the capacitor without it going pop.

    Let's say my assumption is right, how much voltage are there across each individual capacitor in your setup? Will the capacitor blow up?
  4. Feb 9, 2012 #3
    Your assumption IS right. Each capacitor has a voltage rating of 6 volts. And I'm not supposed to think about capacitors blowing up so I'm positive that no the capacitors are not going to blow up.
  5. Feb 9, 2012 #4


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    actually, the purpose of voltage ratings is to prevent the capacitor from breaking down.

    What the question is asking is: if that dude just stick a single capacitor in there, he will get the 4.7uF capacitance he wanted, but he will be exceeding the cap's voltage rating.
    With 4 of the same capacitors, how can he use them to obtain the same capacitance without exceeding the voltage rating?

    in other words, you are not supposed to make a system with overall voltage rating of 9V, otherwise a slight fluctuation in your voltage source will let you see fireworks.

    (Basically you have worked out the right answer but I'm just telling you why this is right. :biggrin: )
  6. Feb 9, 2012 #5
    Yes but see, this is 12th grade physics I'm doing, and I don't need to consider all that.

    What I HAVE to do is: Use those four capacitors to get that specific p.d. and capacitance.

    I get the capacitance.. but I don't understand how the voltage is supposed to be 9V. I mean, how do I calculate 9V or how do I read 9V from that circuit I attached to the question?
  7. Feb 9, 2012 #6


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    Gold Member

    The voltage is 9v because you are supplying 9V to the circuit.

    The important thing is your capacitors arranged this way can take over 9V without dying, because if you measure the voltage across each individual capacitor, they will have less voltage than their rating (<6V).

    If you want to know the voltage across each individual capacitor, treat them like resistors
  8. Feb 9, 2012 #7


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    Homework Helper

    What have you learnt about the relation among voltage, charge and capacitance? About series and parallel connected capacitors?

  9. Feb 10, 2012 #8


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    Staff: Mentor

    Maybe you do.
    No you don't. You are misunderstanding the requirements.
    It isn't. You just have to make sure no individual capacitor will be subjected to more than 6v in your arrangement. Under 6v -- good; over 6v -- bad. :wink:
  10. Feb 10, 2012 #9
    Well wukunlin says I've done the question right, and I think I got it! :D

    So thanks a lot for the help!
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