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Let us say we have a simple purely resistive DC circuit. When we flip the switch a current starts flowing which after some time becomes constant. Now, I know that the electric field across difference components (wire and resistance in this case) is different and results in a net uniform current across the circuit. My questions concern the situation before everything is nice and uniform (before steady state).I have seen some answers that say that say the charges far from the terminals are acted upon by an electromagnetic field when we complete the circuit (switch).

How can these charges go through resistor and drop in potential when they haven't gained any energy in the first place (going through cell)? If charges could lose potential without going through cell wouldn't the circuit achieve steady state as soon as we flip thr switch?

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With the switch open, you have wires at two different potentials. The voltage source is holding one wire at a high potential and one wire low. The voltage source can't push any more charge onto the high wire.

When the circuit is closed, you have charges on one wire with a high potential adjacent to charges on a wire with a low potential. There must therefore be an electric field between them. This field will accelerate charges in the vicinity of the switch at first, but the pulse of changes moves down both wires at very high speed.

The voltage source was already pushing charges to different potentials before the switch was closed. It is this difference in potential that provides the energy to rearrange other charges immediately.

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  • $\begingroup$ What about the energy consideration. The charges close to the resistor (let's say a bulb) do light it up even before steady state right? How do these charges have energy when they haven't even gone through the voltage source yet? $\endgroup$ Commented Aug 19, 2020 at 8:02
  • $\begingroup$ Charges don't have to "go through" the voltage source to have potential energy. All the charges on positive wire have higher potential energy than the charges on the negative wire when the circuit is open. $\endgroup$
    – BowlOfRed
    Commented Aug 19, 2020 at 15:42
  • $\begingroup$ My professor told me that the potential at a point has no meaning and that the potential drop will still occur regardless. He told me that the circuits in DC state acquire steady state almost instantaneously and that the loop rule guarantees a potential drop (equal to the voltage of the cell) across the circuit. But this confused me since before steady state the Electric Field is not uniform right? So is KVL even applicable? $\endgroup$ Commented Aug 20, 2020 at 12:43

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