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Standard XII

Physics

Mixed Combination of Cells

Two cells of ...

Question

Two cells of emfs $E_{1} and E_{2}$ and internal resistances $r_{1} and r_{2}$ when connected in series and across an external resistance R give a current of 2A. When the polarity of one of the cells is reversed, the current through R is 1A. The ratio $E_{1} / E_{2}$ =

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Solution

The correct option is B
3

We have,

$2 = \frac{E_{1} + E_{2}}{R + r_{1} + r_{2}} And 1 = \frac{E_{1} + E_{2}}{R + r_{1} + r_{2}} \Rightarrow 2 (E_{1} - E_{2}) = E_{1} + E_{2} E_{1} = 3 E_{2} \frac{E_{1}}{E_{2}} = 3$

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Two cells of emfs $E_{1} and E_{2}$ and internal resistances $r_{1} and r_{2}$ when connected in series and across an external resistance R give a current of 2A. When the polarity of one of the cells is reversed, the current through R is 1A. The ratio $E_{1} / E_{2}$ =

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$2 = \frac{E_{1} + E_{2}}{R + r_{1} + r_{2}} And 1 = \frac{E_{1} + E_{2}}{R + r_{1} + r_{2}} \Rightarrow 2 (E_{1} - E_{2}) = E_{1} + E_{2} E_{1} = 3 E_{2} \frac{E_{1}}{E_{2}} = 3$