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

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$13.2$

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(B) Let $\varepsilon$ be the $emf$ and $r$ be the internal resistance of the battery.When current flows from negative to positive terminal (charging), the potential difference is given by $V = \varepsilon + Ir$. However, the problem states current flows from negative to positive terminal and gives $12\, V$. Let's analyze the terminal voltage equation: $V = \varepsilon - Ir$ (discharging) and $V = \varepsilon + Ir$ (charging).Case $1$: Current $I_1 = 2\, A$ flows from negative to positive terminal (charging), so $12 = \varepsilon + 2r$ ... $(i)$Case $2$: Current $I_2 = 3\, A$ flows in the opposite direction (discharging), so $15 = \varepsilon - 3r$ ... $(ii)$Wait, re-evaluating the problem statement: If $2\, A$ flows from negative to positive, it is charging: $V = \varepsilon + Ir$. If $3\, A$ flows in the opposite direction, it is discharging: $V = \varepsilon - Ir$.Solving the system:$12 = \varepsilon + 2r$ ... $(i)$$15 = \varepsilon - 3r$ ... $(ii)$Subtracting $(ii)$ from $(i)$:$(12 - 15) = (\varepsilon - \varepsilon) + (2r - (-3r))$$-3 = 5r \implies r = -0.6\, \Omega$. Since resistance cannot be negative, let's re-read: "When current flows from negative to positive terminal, potential difference is $12\, V$" implies discharging if the terminal is defined as the drop across the internal resistance. Let's assume the standard convention: $V = \varepsilon - Ir$ for discharging and $V = \varepsilon + Ir$ for charging.If $12 = \varepsilon - 2r$ and $15 = \varepsilon + 3r$:Subtracting $(i)$ from $(ii)$:$15 - 12 = (\varepsilon - \varepsilon) + (3r - (-2r))$$3 = 5r \implies r = 0.6\, \Omega$.Substituting $r = 0.6$ into $(i)$:$12 = \varepsilon - 2(0.6)$$12 = \varepsilon - 1.2$$\varepsilon = 13.2\, V$.

When a current of $2\, A$ flows in a battery from negative to positive terminal, the potential difference across it is $12\, V$. If a current of $3\, A$ flowing in the opposite direction produces a potential difference of $15\, V$, the $emf$ of the battery is .............. $V$.

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