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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

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0.51 amp

5.1 amp

0.15 amp

1.5 amp

answer is D.

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Detailed Solution

$i = \frac{{{E_1} + {E_2} + {E_3} + ..... + {E_n}}}{{({r_1} + {r_2} + {r_3} + ....... + {r_n})}} = \frac{{1.5({r_1} + {r_2} + {r_3} + ...... + {r_n})}}{{({r_1} + {r_2} + {r_3} + ..... + {r_n})}} = 1.5A$

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