Charging a capacitor

  1. I am quite puzzled about charging a capacitor.

    Given a resistor, in series with a capacitor and a battery, when the switch is closed

    1. Why there is a current immediately

    The capacitor is separated by insulator, hence this is not a closed circuit. The appearance of current seems to be a bit confusing. I thought through this, and conclude that this current is caused by the electronics flow which is for sure due to potential difference. However, when I further think of potential difference, I cannot form a logic explanation to this. The two plates have no potential difference since at the very start it is not charged. How can the electrons flow from a plate to another? I know something has to do with the battery. Could you help me explain more about this?

    If there is a circuit with battery, switch, resistor, and the wire split at one place, will something happen like what happens in RC circuit

    Wire has its cross area also, and when they are separated apart, can I imagine it as a capacitor? If so, will it also be the case of charging a capacitor?

    Thank you for answering all my doubts!!
  2. jcsd
  3. Simon Bridge

    Simon Bridge 14,429
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    This is not correct - if you put a voltmeter across the capacitor, you will see a potential difference due to the battery.

    If you put an ammeter in the circuit you will be able to measure the current that flows.

    The electrons flow through the wires, through the battery. Think of a battery as a pump for electrons - it forces electrons to flow from a low potential to a high potential by using an internal supply of energy - usually chemical.

    If there is a circuit with battery, switch, resistor, and the wire split at one place, will something happen like what happens in RC circuit[/quote]Yes - the ends of the wire become charged. The maths is a bit more complicated than the plate-capacitor though.

    Yes. However the charging/discharging time is usually very short since both the effective capacitance and the resistance in the wire is very small.

    People who build compact electric circuits have to be careful to avoid accidentally adding too much unwanted capacitance when they put breaks in the wires or even when two wires run alongside each other.
  4. Drakkith

    Staff: Mentor

    1. The battery introduces a potential difference to the circuit when you close the switch. Current starts to flow through the circuit, but when it gets to the capacitor it encounters an open since it can't get through. However, a capacitor is designed to have a significant amount of capacitance, aka the ability to store charge. Current, in this case electrons, flows into one plate, making it more and more negatively charged, and out of the other plate, making that plate positively charged. This process continues until the charges on each plate build up enough to counteract the applied voltage. Since the insulator keeps current from flowing from one plate to the other, you now have two electrically charged plates and no more current flow.

    2. Yes, a cut wire is exactly like a very low capacitance capacitor. The principals are identical.
  5. Thank you for your answer. I just have one more doubt now:) according to C=Q/V, when the capacitor is initially not charged, is Q equal to zero here? If it is, potential difference will be zero too right?
  6. Nugatory

    Staff: Mentor

    As long as the battery is not hooked up, yes. But as soon as you hook up the battery the potential difference will become the battery voltage. Current will start flowing and Q will start to increase as the capacitor charges. Once Q becomes equal to VC the capacitor will be fully charged, Q will stop increasing, and the current will fall to zero.

  7. Thank you for your answer. now my doubt is this:
    If we apply V=Q/C here

    Initially Q=0

    Then it means V is also equal to zero?
  8. Integral

    Integral 7,341
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    No, why do you think that?
  9. Because when we want to know the time for charging, we actually apply V=Q/C for all infinitely small time period. So I think V=Q/C is applicable at all time
  10. I would think that is true.
    Th voltage across the terminals of the capacitor is dependant upon how much charge is held within the capacitor.
    If the capacitor is uncharged, then Q=0, and then V=0.

    By closing the circuit, charge will begin to flow into the capacitor. As more charge flows into the capacitor, the voltage across the capacitor terminals will increase to satisfy the equation V=Q/C.

    Since there is a resistor in series with the capacitor, at the moment of closing the circuit, the voltage across the resistor, Vr, equals the battery voltage E, and the voltage across the capacitor, Vc= 0. When the capacitor is fully charged, Vr = 0 and Vc = E.

    At all times E = Vr + Vc.
  11. Simon Bridge

    Simon Bridge 14,429
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    Only if there is no other voltage source besides the capacitor.

    It means that the voltage due to the charge on the capacitor is zero when there is no charge separation - which is what it would be if there was no battery. But there is a battery. Check it against real life - actually get a capacitor and a resistor and a battery and a voltmeter and see.

    Try to get values for R and C so that RC is a decent amount of time - like 5-10seconds.
    Hook the voltmeter up so the negative terminal is on the battery's negative terminal and the positive terminal is between the resistor and the capacitor. Go do it.
  12. + and - of capacitor

    Which is positive terminal and negative terminal when we make our own capacitor as shown in this video. Thank you for the help.
    Last edited by a moderator: Sep 25, 2014
  13. Simon Bridge

    Simon Bridge 14,429
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    Parallel-plate capacitors do not have positive and negative terminals.
    The positive terminal on a voltmeter is the one marked with a "+" sign.
    On a multimeter, it is usually color-coded red.
  14. Thank you for your reply. But I want to figure out that why when we start to charge a capacitor in a DC RC circuit, the current is equal to E/R. Does it imply that the potential difference on the two sides of capacitor is zero?
  15. sophiecentaur

    sophiecentaur 13,278
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    You are correct and there may have been some confusion earlier in the thread. Initially when there is no charge on the Capacitor, the PD across it will be 0V (Q=CV always). That means the whole of the supply volts E will be across the Resistor. As charge flows into the Capacitor, the PD across it will increase and the PD across the Resistor will decrease, exponentially.
    E = VR + VC (Kirchoff II)
    Because I = V/R, this means that the current will also decrease exponentially. The time constant (time to 1/e of any given value) will be RC.
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