0

when cells of same emf are joined in parallel then the resultant emf is taken to be the emf of the cells but what if the emfs are different??

Kaustav K.Patro
  • 1
  • Two cells of different EMFs in parallel = a fire hazard... (They source/sink as much current as they can trying to bring themselves to the same potential; the potential between the terminals in the period before they catch fire depends on their internal resistance.) – HTNW Apr 16 '22 at 05:06
  • See Millman's theorem. – Puk Apr 16 '22 at 05:15

1 Answers1

2

It depends on the internal resistances of the cells. If we include the internal resistances the circuit looks like this:

Circuit

I've assumed $V_1 \gt V_2$ so there is a current $I$ flowing clockwise. This current is given by:

$$ I = \frac{V_1 - V_2}{r_1 + r_2} $$

The voltage of the cells in parallel is then the potential difference that I've labelled $V$ in the diagram. We can calculate this by considering the voltages dropped across the internal resistances, and these are just $\Delta V = Ir$ as given by Ohm's law. So the voltage at V is $V_1 - I r_1$, or we can also write is as $V_2 + I r_2$, which will give the same result. If we do this calculation we get:

$$ V = V_1 - \frac{r_1(V_1 - V_2)}{r_1 + r_2} = \frac{r_2 V_1 + r_1 V_2}{r_1 + r_2} $$

As a sanity check let's see what happens if the cells are identical i.e. if $V_1 = V_2 = V$, then the equation gives us:

$$ \frac{r_2 V + r_1 V}{r_1 + r_2} = \frac{V(r_2 + r_1)}{r_1 + r_2} = V $$

So as expected when the cells are identical the parallel voltage is just the cell voltage.

The obvious next question is what happens if the internal resistances are zero, which is obviously not physically possible as all cells have some resistance, but is nevertheless an interesting question. And the answer is that if all resistances are zero the current never settles to a steady value so the EMF cannot be defined.

John Rennie
  • 355,118