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(C) In a series circuit,the total $emf$ $(E_{eq})$ is the sum of the individual $emfs$,and the total internal resistance $(R_{eq})$ is the sum of the

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$1.5$

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(C) In a series circuit,the total $emf$ $(E_{eq})$ is the sum of the individual $emfs$,and the total internal resistance $(R_{eq})$ is the sum of the individual internal resistances.Given $E_n = 1.5 r_n$ for each cell $n = 1, 2, ..., N$.The total $emf$ is $E_{eq} = \sum_{n=1}^{N} E_n = \sum_{n=1}^{N} 1.5 r_n = 1.5 \sum_{n=1}^{N} r_n$.The total internal resistance is $R_{eq} = \sum_{n=1}^{N} r_n$.According to Ohm's law,the current $I$ in the circuit is given by $I = \frac{E_{eq}}{R_{eq}}$.Substituting the values,$I = \frac{1.5 \sum_{n=1}^{N} r_n}{\sum_{n=1}^{N} r_n} = 1.5 \ A$.

As shown in the figure,a group of $N$ cells is connected in series,where the $emf$ of each cell $E_n$ varies with its internal resistance $r_n$ according to the relation $E_n = 1.5 r_n$. The current $I$ in the circuit is ................ $A$.

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