Find the emf and internal resistance of the equivalent cell between A and B. put $ε_{1}$ =3V, $r_{1} = 2 Ω$ $ε_{2} = 2 V, r_{2} = 1 Ω$ . $ε_{3} = 6 V$ , $r_{3} = 1 Ω$

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Solution

$ε_{1}$ and $ε_{2}$ aare in parallel and they are supporting.In this type of cases:
$E_{e q} = \frac{E_{1} r_{2} + E_{2} r_{1}}{r_{1} + r_{2}}$
$λ r_{e q} = \frac{r_{1} r_{2}}{r_{1} + r_{2}}$
Since $r_{1} = 2 Ω, E_{1} = 3 V$
$r_{1} = 1 Ω, E_{1} = 2 V$
$_{e q} = \frac{3 \times 1 + 2 \times 2}{1 + 2}$
$= \frac{3 + 4}{3} = \frac{7}{3} V$
Now $E_{3}$ and $E_{e q}$ are in series combination and in opposite.
So equivalent $E^{'}$ and $R^{'}$ will be
$E^{'} = E_{3} - E_{e q}$
$E^{'} = 6 - \frac{7}{3}$

$= \frac{11}{3}$
and net equivalent $r^{'} = r_{3} - r_{e q}$
$= 1 - \frac{2}{3}$
$= \frac{1}{3} Ω$
hence answers are , $E m f = \frac{11}{3} V a n d$ Resistance= $\frac{1}{3} Ω$

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(IIT-JEE 1997)