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So I know when current is divided among parallel wires, you can combine them to one current at the ends. My question is can I combine a circuit with 3 sources in parallel to get an increase current?

I have googled, and I am hearing conflicting answers saying "yes if you add sources" but also "no, because charge does not accumulate." Can anybody give any insight.

Example of circuit I am asking about:enter image description here

Does my logic follow?

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2 Answers 2

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You are finding conflicting answers on Google because both "yes" and "no" are correct depending on the context.

Why no is correct: If you have a fixed resistor that represents your load, the current through the resistor is governed by Ohm's Law: \$I=\frac{V}{R}\$. Current through the resistor is the same no matter how many power supplies you have in parallel. This is, of course, assuming the power supplies are all set to the same voltage. The only way to increase current through the load is to decrease the load's resistance and/or increase the voltage if the power supplies.

Why yes is correct: If you have a circuit that is trying to draw more current than the power supply is capable of supplying, then adding more power supplies will increase the current. Think of attaching a small battery to a DC motor. Maybe just one battery isn't enough to fully power the motor. But add a few more batteries in parallel and you can get the motor to rotate at full power. However, after a point, adding more batteries won't make the motor turn faster for the reason stayed in the paragraph above.

The circuit you drew is correct if the resistor is 2.67\$\Omega\$. Each voltage supply will contribute 1/3rd of the .75A. If you added a 4th supply, the current through the resistor wouldn't change (V and R haven't changed), but each supply would now contribute 1/4th of the .75A.

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  • \$\begingroup\$ The second case is what I am trying to achieve. I am trying to power a low wattage light bulb using a combination of sources that are about 5Vs each with .28A of current provided from them. How do I know once I have reached a point of no more increasing current possible? \$\endgroup\$
    – user2967087
    Mar 10, 2015 at 7:58
  • \$\begingroup\$ Ultimately it depends on the resistance of the lightbulb's filament when it's hot. What's the significance of .28A? \$\endgroup\$
    – Dan Laks
    Mar 10, 2015 at 8:05
  • \$\begingroup\$ I am using solar panel with rated .28A as max current out. Test with one panel, and it is accurate. Now I am just trying to combine more to get the desired wattage. \$\endgroup\$
    – user2967087
    Mar 10, 2015 at 8:10
  • \$\begingroup\$ Ah, solving this in terms if wattage makes this much easier. If each panel is 5V and can deliver .28A, then they are 5*.28=1.4Watt panels. Do you know the wattage of the bulb? \$\endgroup\$
    – Dan Laks
    Mar 10, 2015 at 8:13
  • \$\begingroup\$ I was thinking to try and power at least a 20 watt. My initial plan was to use an LM380 op amp, but after realizing the effort to make another larger solar panel to power it I am thinking to use a bunch of smaller ones I have. My logic is I should not need so many if I can increase amps enough to be above 1. \$\endgroup\$
    – user2967087
    Mar 10, 2015 at 8:20
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From your circuit, by adding sources in parallel, you didn't increase the current going through the load but you DID decrease the current being pulled out of the batteries(compared to having a single battery). This is because the three sources, assuming they all have equal voltages when in parallel, does not increase the output voltage and with your load which we consider to be constant in resistance thus, it will not increase the current going through the load.

Basically, if you look at the math it looks like the current from the sources add up but actually you didn't add them up, you split them up into three.

If you want to increase current, either you increase voltage or lessen the resistance of the load.

Maybe this will help to understand adding sources in parallel and what happens to the current.

Single source: enter image description here

Multiple sources in parallel: enter image description here

Multiple sources in series: enter image description here

Hope this helps! Goodluck with your project :D

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  • \$\begingroup\$ This also supports Dan Laks answer regarding the motor. The motor can't run because it needs higher current than what the battery can supply. By adding more batteries they are.. let's say "helping each other" by splitting the current to each battery so that they can supply the needed current of the motor. :) \$\endgroup\$
    – Alexander Sabiona
    Mar 10, 2015 at 9:19
  • \$\begingroup\$ I thought having two power sources in series does not add current? It is why in my solar panels, the solar cells in series add voltages but not current? \$\endgroup\$
    – user2967087
    Mar 10, 2015 at 9:22
  • \$\begingroup\$ not current, but it adds voltage :) \$\endgroup\$
    – Alexander Sabiona
    Mar 10, 2015 at 9:24
  • \$\begingroup\$ but it does increase current as a result when we consider that your resistance is constant. As already mentioned Voltage = current x resistance or V=IR \$\endgroup\$
    – Alexander Sabiona
    Mar 10, 2015 at 9:26

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