Calculate the emf and internal resistance of the equivalent cell to replace the below combination of cells

\frac{4}{7} V, \frac{1}{7} Ω

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4 V, 2 Ω

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\frac{4}{7} V, \frac{2}{7} Ω

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\frac{2}{7} V, \frac{1}{7} Ω

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Solution

The correct option is C $\frac{4}{7} V, \frac{2}{7} Ω$
Since $E_{e q} = \frac{\frac{E_{1}}{r_{1}} + \frac{E_{2}}{r_{2}} + \frac{E_{3}}{r_{3}} + \frac{E_{4}}{r_{4}}}{\frac{1}{r_{1}} + \frac{1}{r_{2}} + \frac{1}{r_{3}} + \frac{1}{r_{4}}}$

$\Rightarrow E_{e q} = \frac{\frac{1}{1} + \frac{2}{1} - \frac{3}{1} + \frac{4}{2}}{\frac{1}{1} + \frac{1}{1} + \frac{1}{1} + \frac{1}{2}} = \frac{2}{(\frac{7}{2})} = \frac{4}{7} V$
Also,
$\frac{1}{R_{e q}} = \frac{1}{r_{1}} + \frac{1}{r_{2}} + \frac{1}{r_{3}} + \frac{1}{r_{4}}$
$\Rightarrow R_{e q} = \frac{2}{7} Ω$
The equivalent cell arrangement between $A$ and $B$ is shown for reference.

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Calculate the emf and internal resistance of the equivalent cell to replace the below combination of cells

The correct option is C 47 V,27 Ω Since Eeq=E1r1+E2r2+E3r3+E4r41r1+1r2+1r3+1r4 ⇒Eeq=11+21−31+4211+11+11+12=2(72)=47 V Also, 1Req=1r1+1r2+1r3+1r4 ⇒Req=27 Ω The equivalent cell arrangement between A and B is shown for reference.