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Why is the current constant in a series circuit? I am confused as I thought that the charges closer to the negative terminal move with a greater velocity than the charges closer to the positive terminal at the moment the voltage is applied?

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The motion of electrons in wires is not that different from the motion of water in a circuit of pipes. In fact this analogy has a formal title - the hydraulic analogy.

Anyhow, if you have some circuit of pipes then there will be places where the water pressure gradient is high and other places where it is low. However the water can't move at different speeds because any particular bit of water has other bits of water behind it and other bits of water in front of it and it effectively stuck in a queue. Since water is effectively incompressible any particular bit of water is constrained to move at the same speed as all the other bits of water.

The electrons in a metal are actually more like a gas than a liquid, but it is a very incompressible gas so for our purposes it moves like a liquid does. The electrons in one part of the circuit can't move faster than the rest because they would pull away from the electrons behind them and bump into the electrons in front of them.

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I am confused as I would have thought that the charges closer to the negative terminal move with a greater velocity than the charges closer to the positive terminal at the moment the voltage is applied?

If what you claimed was true, it could cause a charge build up near the battery terminals. This cannot happen.

The amount of charge that leaves the negative terminal is equal to the amount of charge that reaches the positive terminal.

There can never be any charge build up anywhere in the circuit (capacitors and similar components are exceptions, but they store an equal amount of charge on both sides; hence ensures that equal current flows on both sides).

If charge kept building up at a particular location, after a long time, there will be so much charge that it would stop the current. In reality, we know that this is not true.

Pick an arbitrary point in the circuit. If there must be no charge buildup at that point, then the charge entering the point must be equal to the charge leaving the point. Therefore, the current on both sides must be the same.

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  • $\begingroup$ this answer varifies that current is constant but does not tell us why it is constant and answer is given in Griffiths. $\endgroup$ – Paul Jan 31 '17 at 17:18
  • $\begingroup$ @Theasgardian Added some new information. To be honest, the answer is not correct because some charge does build up but is negligible. I intend not to complicate the answer since the OP seems to be new to electrical circuits. $\endgroup$ – Yashas Jan 31 '17 at 17:24

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