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Let's say you have a basic circuit with a battery power source that has 12 volts. You also have a load of some sort with 3 ohms of resistance. Therefore you'll have 4 amps of current throughout the circuit.

Is it possible to add some item within or along the circuit (and before the load) the could increase the amps on the output side of that item (whatever it is)?

In my question, the current coming from the power source is 4 amps, then after it passes through that item (with some unknown ohms of resistance), it increases to 5 amps (the input amps stays at 4 amps), and then stays 5 amps through the rest of the circuit.

Is this possible?

If it is possible, what would happen to voltage and what would the ohms likely be?

If not can you briefly explain why?

Basic circuit example

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  • \$\begingroup\$ Briefly: it's impossible cannot change the amps without creating charge from nothing (creating electrons, if metal wires.) It's like wanting to put one gallon into a pipe and get two gallons out of the other end. Don't forget that Amperes aren't quite like abstract properties, instead they are a flow of a "substance-like" entity: the coulombs. On the other hand, if using two wires, then you can speed up the current (more amps) while decreasing the voltage. Just chop the 12V into AC, and use a step-down transformer. \$\endgroup\$ – wbeaty Oct 18 '15 at 1:00
  • \$\begingroup\$ Have you been reading too many perpetual motion or free energy web sites? \$\endgroup\$ – Michael Karas Oct 18 '15 at 3:37
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Is this possible?

Yes and no. The way you drew it, it would violate Kirchhoff's current law. However, there is a similar way, it just needs one more connection: A Buck converter or much simpler, a Transformer in an AC circuit.

If it is possible, what would happen to voltage and what would the ohms likely be?

As mentioned, the circuit above does not exactly match your drawing. If you were viewing the resistance from the power supply side, there are interesting things happening. In the case of the (ideal) transformer, the resistance would be greater than the original resistance. For the buck converter it is even more complicated, as these often operate at a fixed output voltage, which leads to a current decrease as the voltage increases.

If not can you briefly explain why?

The "intuitive" way to explain why this is only possible with an additonal connection and not the way you drew it, is that current always flows in circuits. Along such a circuit the current can split into several paths or merge from several paths into one, but you can not lose or gain any current along the way. An easy view of this is that the charge carriers (i.e. electrons) responsible for the current must be available for that to happen. If you had a point in your circuit where the current suddenly increased, you would suck all the electrons out of it. ;)

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  • \$\begingroup\$ Thanks for the response! If I've understood you, would those options draw 5 amps from the battery source? I'm looking to see if possible that 4 amps can be drawn from the battery but yet somehow deliver 5 amps to the load. \$\endgroup\$ – ptownbro Oct 18 '15 at 0:35
  • \$\begingroup\$ Note, this solution isn't 'creating' extra amperage for the load. In the example, a "Buck Converter" can be used to increase the voltage across the load, which will increase the current. However, this also increases the current draw from the battery. Power out WILL ALWAYS be <= Power in, with emphasis on the < part. (The = part is theoretically possible, absolutely impractical). \$\endgroup\$ – Kurt E. Clothier Oct 18 '15 at 4:15
  • \$\begingroup\$ A buck converter won't do it since the load impedance is fixed. But a boost converter could, if it and the 12V battery are up to the job. NB if it's supplying 5A into 3 ohms (75W) it'll draw over 6A, probably 7A if it's reasonably efficient, from the battery. \$\endgroup\$ – Brian Drummond Oct 18 '15 at 10:45
  • \$\begingroup\$ Just realized I got more responses on this. Thanks guys. \$\endgroup\$ – ptownbro Oct 21 '15 at 5:47
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No, because the battery would be violating KCL. If the device has 4A on one side and 5A on the other, then it would either need a second pair of connections where it takes 5A on one side and puts out 4A on the other, or another single connection where it would absorb 1A.

Incidentally, since this would require an increase in voltage, the device would be a boost regulator.

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  • \$\begingroup\$ Thanks for the response. If I've understood you, I think you've completely changed my example and made it backwards. The input amps (from battery to unknown item) would be 5A, then the output amps would split to 4A in one wire and 1A in a second. I''m looking to see if it's possible to have it be 4A on the input end (from battery to uknown item) and get a 5A output. In that strict example, I think you're saying there's nothing that can do that? \$\endgroup\$ – ptownbro Oct 18 '15 at 0:24
  • \$\begingroup\$ Both scenarios (both mine and the one you've just described) are the same, even if the words have been rearranged. Neither KCL nor Ohm's law can be violated. \$\endgroup\$ – Ignacio Vazquez-Abrams Oct 18 '15 at 0:33
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In my question, the current coming from the power source is 4 amps, then after it passes through that item (with some unknown ohms of resistance), it increases to 5 amps (the input amps stays at 4 amps), and then stays 5 amps through the rest of the circuit.

So, let's examine this. The power source is 12 volts and you want to take 4 A to the thing that resides before the load. That thing can put out 5A into a 3 ohm load, yes?

Input power = 12 volts x 4 amps = 48 watts

Output power is 5 amps into 3 ohms = 75 watts (power = \$I^2R\$)

No, forget it because it breaks the laws of physics.

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A boost converter could do this, if it and the 12V battery are up to the job.

NB if it's supplying 5A into 3 ohms (75W) it'll draw over 6A, probably about 7A (84W) if it's reasonably efficient, from the battery.

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