Why do the amps in a circuit decrease as you add more wire?

They don't. the voltage decreases & the amps will increase. The longer the wire the more resistance the less voltage. In order for you power, KW or VA to remain the same, if your voltage drops your amps go up, if your voltage goes up your amps will drop. (I=P/E),(P=ExI)(E=P/I)

I=amps

E=volts

P=power or watts or better yet VA

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The answer above presumes that KW or VA remains the same in a circuit regardless of what you do to it. They do not. I do not intend to be argumentative but here is my answer.

Adding resistance IN SERIES with a circuit decreases the amps that would otherwise flow because it has to overcome more resistance. Since conductors have resistance and are by their very nature IN SERIES with a load, adding length to conductors can, does, and will decrease the amps across the entire circuit.

If you aren't into reading technical descriptions, stop reading now. The simple answer is in my second paragraph above, and is all you probably needed or wanted to know.

What confuses people is how the circuit behaves when you add another load IN PARALLEL, which is how most circuits you and I use are wired. Adding a load IN PARALLEL will increase the amps of the overall circuit because it decreases the net resistance of the circuit. The best comparison that will help you understand this is a water line. If a water line has two faucets with only one faucet open, you have a certain amount of pressure inside to water pipe. If you open both faucets you now have more water flowing and a net decrease in pressure within the pipe. Comparing a circuit to a water pipe in this manner is not far fetched. It is exactly how and why you get more amps when you add lights to a circuit. They are connected IN PARALLEL to each other.

Getting back to our conductors, however, you can see that the first leg of conductors from the breaker to the first load is carrying the amps of the whole circuit. Each leg from one load to another carries decreasing amps because it is carrying amps to fewer loads. The amount of power lost in these conductors is equal to I2R. Plugging in this value to the other formulas allows you to calculate how much voltage is "dropped" across any leg of the circuit, knowing the voltage applied and getting the resistance of the conductors from the code book.

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As always, if you are in doubt about what to do, the best advice anyone should give you is to call a licensed electrician to advise what work is needed.

Before you do any work yourself,

on electrical circuits, equipment or appliances,

always use a test meter to ensure the circuit is, in fact, de-energized.

IF YOU ARE NOT ALREADY SURE YOU CAN DO THIS JOB

SAFELY AND COMPETENTLY

REFER THIS WORK TO QUALIFIED PROFESSIONALS.