Obtain the expression for the effective emf and the effective internal resistance of two cells connected in parallel such that the currents are flowing in the same direction.

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Solution

Let $ε_{1}$ and $ε_{2}$ be the emfs of the two cells connected in parallel.
Let $r_{1}$ and $r_{2}$ be their intermal resistance respectively
$I_{1}$ and $I_{2}$ are the currents leaving the positive electrodes of the cells.
At $B_{1}$ , currents $I_{1}$ and $I_{2}$ flow in whereas the current $I$ flows out. Therefore $I = I_{1} + I_{2}$
Let $V_{1}$ and $V_{2}$ be the potentials at $B_{1}$ and $B_{2}$ respectively.
P.d across the first cell is $V_{1} - V_{2} = V = ε_{1} - I_{1} r_{1}$
P.d across the second cell is also $V_{1} - V_{2} = V = ε_{1} - I_{1} r_{1}$
Hence $I_{1} = \frac{ε_{1} - V}{r_{1}}$ and $I_{2} = \frac{ε_{2} - V}{r_{2}}$
$I = I_{1} + I_{2} = \frac{ε_{1} - V}{r_{1}} + \frac{ε_{2} - V}{r_{2}}$
$I = \frac{ε_{1}}{r_{1}} - \frac{V}{r_{1}} + \frac{ε_{2}}{r_{2}} - \frac{V}{r_{2}}$
$I = (\frac{ε_{1}}{r_{1}} + \frac{ε_{2}}{r_{1}}) - V (\frac{V}{r_{1}} + \frac{V}{r_{2}})$
$I = \frac{ε_{1} r_{2} + ε_{2} r_{1}}{r_{1} r_{2}} - V (\frac{r_{1} + r_{2}}{r_{1} r_{2}})$
$V (\frac{r_{1} + r_{2}}{r_{1} r_{2}}) = \frac{ε_{1} r_{2} + ε_{2} r_{1}}{r_{1} r_{2}} - I$
$V (r_{1} + r_{2}) = ε_{1} r_{2} + ε_{2} r_{1} - I r_{1} r_{2}$
$V = \frac{ε_{1} r_{2} + ε_{2} r_{1}}{r_{1} + r_{2}} - I (\frac{r_{1} r_{2}}{r_{1} + r_{2}})$
If the parallel combination of cell is replaced by a single cell between $B_{1}$ and $B_{2}$ of emf $ε_{P}$ and intermal resistance $r_{p}$ then, $V = ε_{P} - I r_{p}$
Comparing the above equations
$ε_{P} = \frac{ε_{1} r_{2} + ε_{2} r_{1}}{r_{1} + r_{2}}$ and $r_{P} = \frac{r_{1} r_{2}}{r_{1} + r_{2}} O R \frac{1}{r_{P}} = \frac{1}{r_{1}} + \frac{1}{r_{2}}$

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