10 cells,each of emf E and internal resistanc | vedclass

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(D) For $n$ cells in series,the total $emf$ is $E_{eq} = nE$ and total internal resistance is $r_{eq} = nr$.The terminal potential difference $V$ is g

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$4.2$

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(D) For $n$ cells in series,the total $emf$ is $E_{eq} = nE$ and total internal resistance is $r_{eq} = nr$.The terminal potential difference $V$ is given by $V = E_{eq} - I r_{eq}$.Given $n = 10$,so $V = 10E - I(10r)$.From the graph,when $I = 12 \, A$,$V = 6 \, V$. Substituting these values:$6 = 10E - 12(10r) \implies 6 = 10E - 120r$ .......... $(1)$From the graph,when $V = 0 \, V$ (short circuit),$I = 14 \, A$. Substituting these values:$0 = 10E - 14(10r) \implies 10E = 140r \implies r = \frac{10E}{140} = \frac{E}{14}$ .......... $(2)$Substitute $10r = \frac{10E}{14}$ into equation $(1)$:$6 = 10E - 12 \left( \frac{10E}{14} \right)$$6 = 10E - \frac{120E}{14} = 10E - \frac{60E}{7}$$6 = \frac{70E - 60E}{7} = \frac{10E}{7}$$10E = 42 \implies E = 4.2 \, V$.

$10$ cells,each of $emf$ $E$ and internal resistance $r$,are connected in series to a variable external resistance. The figure shows the variation of the terminal potential difference of their combination with the current drawn from the combination. The $emf$ of each cell is ................ $V$.

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