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Homework Help: Capacitor in Circuit

  1. Aug 20, 2005 #1
    Is it true that when a capacitor is connected in series with a globe, and is charged with a battery that while the charge on the capacitor is being increased, the globe will glow brightly. But when the capcitor reaches its maximum charge, the globe will get dimmer, because of the loss of current through the capacitor?
  2. jcsd
  3. Aug 20, 2005 #2
    It starts off full brightness and starts dimming more and more as the charge is stored on the capacitor.
  4. Aug 20, 2005 #3
    But it starts dimming because of the loss in current, right?
  5. Aug 20, 2005 #4
    Right, the current drops as the charge is stored on the capacitor.
  6. Aug 20, 2005 #5
    thankyou whozum
  7. Aug 21, 2005 #6
    By the way, with a capacitor that is dishcharging in a similar circuit, what formula can you use to find the current through the globe after a certain time, and the charge that is on the capacitor after a certain time?
  8. Aug 26, 2005 #7


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    Homework Helper

    Basically you have a series RC circuit (where R is the globe's resistance) using the discharge equation V = Vo e^-t/τ (where τ = RC)
    and substituting q/C for V and qo/C for Vo where qo is the initial charge on the capacitor you get an equation of the same form q = qo e^-t/τ. Therefore the charge also decreases exponentially. By differentiating the previous equation w.r.t time (i = dq/dt) you get i = -qo/τ e^-t/τ same with the current! (- implies current decreases with time)
  9. Aug 26, 2005 #8
    Capacitors store charges.When connected in a circuit , it takes care of its own job irrespective of others do or not. So when you connect it with a battery , current starts flowing in the circuit , through the capacitor and the globe(which is a type of resistor in true sense) , so with time capacitor keeps on charging and the current keeps falling with time.As the current falls , the brightness in the globe keeps on lowering down , eventually leading to the dusk of the globe.

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