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Hint: Resistances connected in parallel have net resistance as the sum of inverses of each resistance.
Complete answer:
Consider \[{R_1},{R_2},{R_3},...\] be the internal resistances of each cell. Final resistance, ${R_f}$ formed by connecting cells in series and parallel have different relationships with the individual resistances. When the cells are connected in parallel, then ${R_f}$is calculated as follows:
$\dfrac{1}{{{R_f}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}} + ...$
This way ${R_f}$ estimated is lesser than when connected in series, where the final resistance is simply numerical addition of all the resistances. So the purpose of connecting Primary cells in parallel is to decrease the final resistance.
So, Option C is the correct answer.
Note: Generally, when such connections are done, cells of different ampere ratings are not connected or else there will be exchange of charge between the cells and may cause complications for instance heating of the cells.
Complete answer:
Consider \[{R_1},{R_2},{R_3},...\] be the internal resistances of each cell. Final resistance, ${R_f}$ formed by connecting cells in series and parallel have different relationships with the individual resistances. When the cells are connected in parallel, then ${R_f}$is calculated as follows:
$\dfrac{1}{{{R_f}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}} + ...$
This way ${R_f}$ estimated is lesser than when connected in series, where the final resistance is simply numerical addition of all the resistances. So the purpose of connecting Primary cells in parallel is to decrease the final resistance.
So, Option C is the correct answer.
Note: Generally, when such connections are done, cells of different ampere ratings are not connected or else there will be exchange of charge between the cells and may cause complications for instance heating of the cells.
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