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A = Amperes as symbol.

Take a simple DC current that has a battery, a light bulb, and to copper conductors. If the current coming out of the negative terminal of the battery.

let's say 1 amp, but the light only uses/needs is 5 A, does the current coming out from the light back to the positive terminal of the battery also has 5 A? This is what I would assume.

However, I just saw a video relative to typical homes AC circuits that state if you have a 20 A circuit with a light on it then 20 A would be returning on the neutral wire from the light. Why is this? My understanding is that the resistance of a household load, like a light, will resist the current flow based on what the load needs.

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  • $\begingroup$ What do you mean by "uses/need 0.5 amps"? $\endgroup$
    – Dan
    Dec 18, 2021 at 4:50
  • $\begingroup$ Imagine electrons are little worker people. Each one of them leaves the battery with fresh high energy. Then it does some work in the lamp, and comes back tired (low energy). The number of workers that go to the lamp and come out is the same (therefore current is the same), but their energy is lost in the lamp (so electrons' voltage will be lower when they leave the lamp). $\endgroup$ Dec 18, 2021 at 8:47

2 Answers 2

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If the current coming out of the negative terminal of the battery is let's say 1 amp, but the light only uses/needs .5 amps, does the current coming out from the light back to the positive terminal of the battery also .5 amps?

These sentences are in conflict. In this simple circuit there is nowhere for the current to go but between the battery and the bulb. If $I$ current is leaving the battery, then $I$ current is entering the bulb.

We model the battery as a voltage source. That means the voltage difference across the battery terminals is nearly constant (for small loads) and the current depends on the load. The resistance in the light bulb will limit the current that flows.

If the bulb is designed to work with that battery and to use 0.5 A, then that's how much current will flow from the battery when attached. The situation might be different as you add more elements to the circuit (like parallel loads).

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. . . . the light only uses/needs .5 amps, does the current coming out from the light back to the positive terminal of the battery also .5 amps?

is better written as,

. . . . the light has .5 amps passing through it, does the current coming out from the light back to the positive terminal of the battery also .5 amps?

The answers is yes as there are no "sources" or "sinks" of electrical charge and this is really an application of Kirchhoff's current law.

Think of a tap.
Open it a certain amount and water flows out of the tap.
Without changing the settings on the tap attach a hose to the tap.
The rate of flow of water through the tap (and the hose) will decrease but the rate of flow of water at both ends of the hose will be the same.

What a resistor does is affect the rate of flow of charge in a circuit.

The $0.5\,\rm A$ is the rated current for the bulb to work as per specification ie give out a certain amount of light without burning out.
If the current through the bulb is less than its rated value then the bulb will be less bright and if the current through the bulb is more than its rated value then the bulb will be brighter and it will be more likely to blow.
In all cases although the rate of flow of charge (current) through the bulb is altered the current passing into the bulb is the same as the current flowing out of the bulb.

So perhaps, The resistance in the light bulb will limit the current that flows, should read, The resistance in the light bulb affect the current that flows,?

. . . . if you have a 20 amp circuit with a light on it then 20 amps would be returning on the neutral wire from the light.
The value of $20\,\rm A$ is the rated value of the circuit.
That means that for correct usage the current in that part of the home circuit should not exceed $20\,\rm A$ otherwise the circuit might overheat and cause a fire.
If a $110\,\rm V$ circuit rated at $20\,\rm A$ has only one $110\,\rm V,\,0.5\,\rm A$ bulb connected in it then a current of $0.5\,\rm A$ will flow throughout the whole of the circuit.
Connecting ten $0.5\,\rm A$ bulbs in parallel in that circuit will result in a total current of $10\,\rm A$ flowing in the circuit.
Connecting two bulbs with a $0.5\,\rm A$ rating, in series will result in the current in the circuit being reduced but being the same throughout the whole circuit.

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