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Homework Help: RC Circuit - Question

  1. Oct 26, 2009 #1
    SOLVED - RC Circuit - Question

    http://img291.imageshack.us/img291/5585/bulbd.th.jpg [Broken]

    This problem is giving me a headache. I feel certain that bulb B is the brightest when the capacitor has half its maximum charge. However, I am having a hard time coming up with equations for the circuit using Kirchhoff's voltage and current laws.

    I do know that if the capacitor had its maximum charge then there would be no current in the right hand circuit. But I am a little unsure what is going on under the declared condition. I suspect that there is no current in the closed circuit that includes bulb C. And the fact that the capacitor has half its maximum charge means that the potential difference is 55V across the capacitor.

    I would love a nudge in the right direction... I am stuck. Cheers!
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Oct 26, 2009 #2


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    Homework Helper

    Welcome to PF, Hawkbyte. You have an interesting question. Yes, ignore bulb C - it is shorted out. You'll have to figure out the charge on that capacitor to be sure of your answer. I see the capacitor taking a high current initially, gradually charging up and the current gradually diminishing to zero. At that point it has its max charge and it is easy to calculate because the capacitor acts like an infinite resistance. From the V, you can figure out the C. Then reverse that to get the V when the C is halved.
  4. Oct 26, 2009 #3
    Since the capacitor is at 50% of its maximum charge, then the potential difference across its plates is 55V. Does that mean that I would apply that 55V to the entire closed loop that contains bulb A and the capacitor? Would I treat bulb A as a resistor in series with C? Again, the geometry of this circuit is confusing me.

    I am not after exact values, I just want to see hard evidence of the fact that more current is moving through bulb B.
  5. Oct 26, 2009 #4


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    At FULL charge, zero current, the voltage across the capacitor is 55.
    That is Q = CV = 55C. Half charge is 27.5*C. So the voltage across the capacitor (and bulb A) is V = Q/C = 27.5*C/C = 27.5. That leaves a higher voltage across bulb B so your intuition was right on.
  6. Oct 26, 2009 #5
    Aye, aye... thanks very much for your assistance!
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