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RC Circuit under charging

  1. Oct 6, 2009 #1
    1. The problem statement, all variables and given/known data

    In an RC circuit under charging, does the current actually flow across the plates of a parallel plate capacitor?

    Does charge emanate from both terminals of the battery and move to the plates of a capacitor or does charge only emanate from positive terminal and get stored in one of the plates and manage to polarize the other? How is it that both plates have equal magnitude of charge?

    Under steady state, does no current pass through the wires connected to the capacitor even if a resistor is connected in series with it? In such case how do we find the current through the circuit,do we take into account the resistor connected in series with capacitor or just the other ones?

    Just wanted to clear all errors in perception. Your help is appreciated.
  2. jcsd
  3. Oct 6, 2009 #2


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    No, but if you hook up an ammeter to the circuit, it will detect a decaying current as charges accumulate on the plates.

    The entire circuit is neutrally charged and remains so at all times. As soon as you connect the capacitor, the positive terminal of the battery is at higher potential than the plate to which it is connected. The free electrons in the conductors will move from the positive plate, through the battery and onto the negative plate. When the positive plate plus connecting wire is at the same potential as the positive terminal, electrons stop flowing. The electrons on the negative plate leave an absence of electrons on the positive plate because the whole thing is neutral. That is why the charges o the plate are equal and opposite.

    Remember these simple rules:

    1. When an uncharged capacitor is suddenly connected to a battery, it initially (that means an infinitesimal amount of time after the connection) behaves as if it were a short (straight wire).
    2. When a fully charged capacitor is suddenly connected to a resistor, it initially behaves as a battery of voltage Q/C.
    3. A long time after a capacitor is connected to a circuit (steady state), it behaves as if it were an open switch.

    By "behaves" I mean that, if one of the conditions as described above is met, you can replace the capacitor in the given circuit with a short or battery or open switch and analyze the new circuit instead as far as currents in resistors, etc. are concerned.
  4. Oct 6, 2009 #3
    I have one more question.

    If a resistor is connected parallel to a straight wire joining the two terminals of a battery, does any current pass through the resistor. Is there any current in the circuit?
  5. Oct 6, 2009 #4


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    A straight wire in a circuit diagram is, by convention, an equipotential line and has zero resistance. If such a wire (a short) is across a resistor, the two ends of the resistor will be at the same potential, therefore the potential difference across the resistor will be zero. According to Ohm's Law (V = IR), there will be no current through the resistor because V=0. All the current will flow through the short.

    If there is no other resistor in the circuit, mathematically the current in the short will be infinite (I = emf/R) since the resistance of the short is zero. Physically there will be a very large current in the short because the short has a small but non-zero resistance. In that case either the short will vaporize or the battery will explode. That's why people often use fuses to protect circuits and current-limiting resistors to protect batteries.
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