# Two cell emf and internal resistance $\varepsilon _{1}, r_{1}$ and $\varepsilon _{2}, r_{2}$ are connected in parallel. Derive the expressions for the emf and internal resistance of a cell which can replace this combination.

Emf are $\varepsilon _{1}and \varepsilon _{2}$

Internal resistance are $r _{1}andr _{2}$

Here, $I=I _{1}+I_{2}$            ........(1)

Let V is the potential differnace between B1 and B2

Then $V = \varepsilon _{1}-I_{1}r_{1}$

Therefore , $I_{1}= \frac{\varepsilon _{1}-V}{r_{1}}$                ......(2)

Similarly. $I_{2}= \frac{\varepsilon _{2}-V}{r_{2}}$                   .......(3)

Putting equation (2) and (3) in (1)

$I=\frac{\varepsilon _{1}-V}{r_{1}}+ \frac{\varepsilon _{2}-V}{r_{2}}$

$I=\left ( \frac{\varepsilon _{1}}{r_{1}}+\frac{\varepsilon _{2} }{r_{2}} \right )-V\left ( \frac{1}{r_{1}}+\frac{1}{r_{2}} \right )$

or $V = \left ( \frac{\varepsilon _{1}r_{2}+\varepsilon _{2}r_{1}}{r_{1}+r_{2}} \right )- I\left ( \frac{r_{1}r_{2}}{r_{1}+r_{2}} \right )$

If we replace this with a single cell it can be written as

$V = \varepsilon _{equivalent}-Ir_{eqivalent}$

Then,

$\varepsilon _{eqivalent} = \left ( \frac{\varepsilon _{1}r_{2}+\varepsilon _{2}r_{1}}{r_{1}+r_{2}} \right )$

$r_{equivalent} =\left ( \frac{r_{1}r_{2}}{r_{1}+r_{2}} \right )$

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