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(B) Let $E$ be the $emf$ of the battery and $r$ be its internal resistance.When current flows from negative to positive terminal (charging), the termi

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$13.2$

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(B) Let $E$ be the $emf$ of the battery and $r$ be its internal resistance.When current flows from negative to positive terminal (charging), the terminal potential difference is given by $V = E + Ir$.Given $I = 2\, A$ and $V = 12\, V$, we have:$12 = E + 2r$ ..........$(1)$When current flows in the opposite direction (discharging), the terminal potential difference is given by $V = E - Ir$.Given $I = 3\, A$ and $V = 15\, V$, we have:$15 = E - 3r$ ..........$(2)$Wait, re-evaluating the problem statement: If current flows from negative to positive, it is charging $(V = E + Ir)$. If current flows from positive to negative, it is discharging $(V = E - Ir)$.Let's re-read: "When a current of $2\, A$ flows in a battery from negative to positive terminal, the potential difference across it is $12\, V$." This implies $12 = E + 2r$."If a current of $3\, A$ flows in the opposite direction, the potential difference across the terminals of the battery is $15\, V$." This implies $15 = E - 3r$.Adding $(1)$ and $(2)$ does not yield a simple solution. Let's re-examine the physics: If $V = 12$ when $I=2$ (charging) and $V=15$ when $I=3$ (discharging), this is physically impossible as $V$ should be less than $E$ during discharge. Correct interpretation: The first case is discharging $(V = E - Ir)$ and the second is charging $(V = E + Ir)$.$E - 2r = 12$ ..........$(1)$$E + 3r = 15$ ..........$(2)$Subtracting $(1)$ from $(2)$:$(E + 3r) - (E - 2r) = 15 - 12$$5r = 3 \Rightarrow r = 0.6\, \Omega$Substitute $r$ in $(1)$:$E - 2(0.6) = 12$$E - 1.2 = 12$$E = 13.2\, V$.

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