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# calculating voltage of parrallel cells watch

1. when i calculate the voltage of parallel cells, it seems that the total voltage, is the voltage of the higher cell always, does anyone have any physical(not mathematical) explanation for this?
2. The difference in cell voltage will produce large currents that flow into the cell of lower emf.

This produces a large current defined by I = Vhigh - Vlow / Rtot

where Rtot includes the internal cell resistance and interconnecting conductors.

The power produced by this current is dissipated in the internal resistance and it is this which causes the batteries to overheat, expand and even explode in extreme cases.
3. Conceptually, it's not so much that the total E.M.F (the 'voltage' of the cell) is equivalent to the higher emf battery. It's that ONLY the emf of the higher one is used.

Think of a circulatory river being similar to an electronic circuit. Now the battery's EMF is the amount of motive (kinetic) energy gained per unit charge. So think of a battery as a downhill section of a river (the steeper the downhill gradient, the more kinetic energy gained per unit volume of water).

In parallel, it is analogous to a river meeting two downhill junctions (like an upside down 'Y'). The steeper turn is the battery with the higher emf. Now, the flow of water (I.E current) will naturally flow down the quickest path and hence all the water falls down the steeper gradient! This implies the the higher EMF battery is the only one putting work into the system!

with respect, I don't think the above explanation answers your question with satisfaction.
4. (Original post by hecandothatfromran)
with respect, I don't think the above explanation answers your question with satisfaction.
When help is offered freely, it is highly patronising and not a little arrogant for a third party to pass comment and make a judgement on what the recipient will find satisfactory or not.

Kindly refrain from passing such comment in future - it's called social etiquette.
5. (Original post by uberteknik)
When help is offered freely, it is highly patronising and not a little arrogant for a third party to pass comment and make a judgement on what the recipient will find satisfactory or not.

Kindly refrain from passing such comment in future - it's called social etiquette.
Sorry dude, I really didn't mean any offence. I was just trying to help with the confusion. FTR, I thought you smashed that circuit question with a home run.
6. (Original post by hecandothatfromran)
Sorry dude, I really didn't mean any offence. I was just trying to help with the confusion. FTR, I thought you smashed that circuit question with a home run.
Hey bro, no worries. I need to go change my diapers! lol.

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