Batteries for VNs: Why The Silence?

[ven]

Would the Kinoko IMR or the MKE be a good battery for the X10vn as well?

I am sure this Q was asked on another thread and vinh said it would improve or should improve output(not to them words but along those lines)

So yes

 

[Kevin1322]

I've been meaning to ask, and this seems the best place to do it. If a light is pulling, say 6 amps and is powered by 3 batteries, does 6 amps get pulled from each battery or 2 amps? I understand that when batteries are placed one behind the other it's called a series and the voltage goes up per battery, and that in parallel (side by side) the voltage stays the same, you are just increasing your power capacity; i.e. 3 by 2000mAh batteries is 6000mAh. This then implies that, in parallel, only 2 amps would be pulled from the three batteries when the LED is drawing 6 amps. Is this right?

 

[Mbehe87]

I think you are pretty close... For cells in series, the current is the same through each cell but the voltage adds. For cells in parallel, the current from each cell adds together but voltage is the same. It may be helpful to think of the power delivered by the cells (volts times amps). Using your examples:

For 3 cells in series providing 6 amps, (3.7 volts)x(3 cells)x(6 amps)=(11.1 volts)x(6 amps)=66 Watts

For 3 cells in parallel providing 2 amps each or 6 amps total, (3.7 volts)x(2 amps)x(3 cells)=(3.7 volts)x(6 amps)=22 Watts

So you can see in the second scenario, you are providing less power to the driver. To supply the same amount of power using parallel cells, you would need each cell to provide 6 amps each or 18 amps total. (3.7 volts)x(6 amps)x(3 cells)=(3.7 volts)x(18 amps)=66 Watts

 

[Kevin1322]

That's much easier to understand haha. Thanks!

I think you are pretty close... For cells in series, the current is the same through each cell but the voltage adds. For cells in parallel, the current from each cell adds together but voltage is the same. It may be helpful to think of the power delivered by the cells (volts times amps). Using your examples:

For 3 cells in series providing 6 amps, (3.7 volts)x(3 cells)x(6 amps)=(11.1 volts)x(6 amps)=66 Watts

For 3 cells in parallel providing 2 amps each or 6 amps total, (3.7 volts)x(2 amps)x(3 cells)=(3.7 volts)x(6 amps)=22 Watts

So you can see in the second scenario, you are providing less power to the driver. To supply the same amount of power using parallel cells, you would need each cell to provide 6 amps each or 18 amps total. (3.7 volts)x(6 amps)x(3 cells)=(3.7 volts)x(18 amps)=66 Watts