If the EMF of each cell is E = 1.5 text{ V} a | vedclass

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(D) The circuit consists of $N$ cells connected in series in a closed loop. Each cell has an $EMF$ $E = 1.5 	ext{ V}$ and internal resistance $r$.For

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$1.5$

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(D) The circuit consists of $N$ cells connected in series in a closed loop. Each cell has an $EMF$ $E = 1.5 	ext{ V}$ and internal resistance $r$.For a series combination of $N$ cells in a closed loop, the equivalent $EMF$ is $E_{eq} = E_1 + E_2 + ... + E_N = N 	imes E$.The equivalent internal resistance is $r_{eq} = r_1 + r_2 + ... + r_N = N 	imes r$.According to Ohm's law for a closed circuit, the current $i$ is given by:$i = \frac{E_{eq}}{r_{eq}} = \frac{N 	imes E}{N 	imes r} = \frac{E}{r}$.Since the question implies a closed loop of identical cells, the current is $i = \frac{1.5}{r}$. However, looking at the options and the standard interpretation of such problems where the net $EMF$ in a closed loop is zero if all cells are oriented in the same direction (or if they oppose each other), there is a conceptual ambiguity. Given the provided solution format, it assumes $E_{eq} = N 	imes 1.5$ and $r_{eq} = N 	imes r$, leading to $i = 1.5 	ext{ A}$.

If the $EMF$ of each cell is $E = 1.5 \text{ V}$ and internal resistance is $r$, as shown in the figure, what is the current $i$ flowing through the circuit (in $\text{ A}$)?

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