Find the emf (V) and internal resistance (r) of a single battery which is equivalent to a parallel combination of two batteries of emfs $V_{1}$ and $V_{2}$ and internal resistances $r_{1}$ and $r_{2}$ respectively, with polarities as shown in figure

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Solution

$V = \frac{\frac{V_{1}}{r_{1}} - \frac{V_{2}}{r_{2}}}{\frac{1}{r_{1}} - \frac{1}{r_{2}}} = \frac{V_{1} r_{2} - V_{2} r_{1 ‘}}{r_{1} r_{2} \frac{(r_{1} - r_{1})}{r_{1} r_{2}}}$
we get $,$
$V = \frac{V_{1} r_{2} - V_{2} r_{1 ‘}}{r_{2} - r_{1}}$
$r = \frac{r_{1} r_{2}}{r_{1} + r_{2}}$

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Q. The emf (V) and internal resistance (

r

) of a single battery which is equivalent to a parallel combination of two batteries of emf's

V_{1}

and

V_{2}

and internal resistance

r_{1}

and

r_{2}

, respectively, with polarities as shown in the figure are given by

\frac{V_{1} r_{2} - V_{2} r_{1}}{r_{1} + r_{2}}, \frac{r_{1} r_{2}}{r_{1} + r_{2}}

A single battery fo emf E and internal resistance r is equivalent to a parallel combination of two batteries of emfs $E_{1} = 2 V$ and $E_{2} = 1.5 V$ and internal resistances $r_{1} = 1 Ω$ and $r_{2} = 2 Ω$ respectively, with polarities as shown in the figure.

(IIT-JEE 1997)