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Fuse Wire

A group of N ...

Question

A group of N cells whose emf varies directly with the internal resistance as per the equation $E_{n}$ = 1.5 $r_{N}$ are connected as shown in the figure below. The current I in the circuit is

A

0.51 amp

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B

5.1 amp

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C

0.15 amp

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D

1.5 amp

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Solution

The correct option is D 1.5 amp
$i = \frac{E_{1} + E_{2} + E_{3} + . . . . . . . . + E_{n}}{(r_{1} + r_{2} + r_{3} + . . . . . . . . . + r_{n})}$
$= \frac{1.5 (r_{1} + r_{2} + r_{3} + . . . . . . . . + r_{n})}{(r_{1} + r_{2} + r_{3} + . . . . + r_{n})}$ = 1.5A.

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