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Will connecting 2 lipo batteries of 3.7 V of 1500 mAh in parallel for a power source to a drone fry it? I understand the voltage will continue to be 3.7 V because it's in parallel. The current or Amps is expected to double. Is this doubling just in capacity, and will the current or Amps pulled just be as needed by the motors and controller - same as before? Or will the drone fry?

Let's say the weight of the battery here is not an issue at all.

R

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TLDR; The drone will work the same

Given that weight is not an issue here (which might be the biggest concern in the real world), the only things we need to take into account are voltage and current.

As you can find in this question and this one, it is okay to use a battery that can supply more current than needed by your circuit. Due to Ohm's Law, $$I = \frac VR$$ the current $(I)$ that flows through a circuit is equal to voltage $(V)$ divided by resistance $(R)$.

If your drone remains the same, it will have the same resistance. And if your battery of increased size is still 3.7V, Ohm's law states that the current going through will remain the same.

Because the current is the same, your drone will function just like it did before, and will not fry.

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  • $\begingroup$ Thanks for such a nice answer. Had totally forgotten the equation. Now it makes so much sense $\endgroup$
    – Raster R
    Oct 15 at 5:10
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    $\begingroup$ Just to round out the answer. You will get more flight time due to extra power capacity, but not double due to the extra weight. $\endgroup$
    – Kevin
    Oct 15 at 13:02
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Putting LiPo batteries in parallel is fine so long as they all begin at a similar charge voltage. Read more about it in my answer here: How can I safely parallel charge my LiPo batteries?

When you put two equal batteries in parallel, their capacity doubles, their combined internal equivalent series resistance (ESR) gets cut in half, and their max current capability doubles, NOT their max current. In other words, since the ESR is cut in half, the ability to deliver current doubles, but that doesn't mean the current will double. The load will draw wha the load draws based on V = IR --> I = V/R.

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