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(B) Let $E$ be the emf and $r$ be the internal resistance of the battery.When current $I = 2\,A$ flows from negative to positive terminal (charging),t

Vedclass · Accepted Answer

$13.2$

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(B) Let $E$ be the emf and $r$ be the internal resistance of the battery.When current $I = 2\,A$ flows from negative to positive terminal (charging),the terminal potential difference is given by $V = E + Ir$.$12 = E + 2r$  ..........$(1)$When current $I = 3\,A$ flows in the opposite direction (discharging),the terminal potential difference is given by $V = E - Ir$.$15 = E - 3r$  ..........$(2)$Wait,let's re-evaluate the conditions: If current flows from negative to positive,it is charging: $V = E + Ir = 12$.If current flows from positive to negative,it is discharging: $V = E - Ir = 15$.Subtracting $(2)$ from $(1)$: $(E + 2r) - (E - 3r) = 12 - 15 \Rightarrow 5r = -3$. This implies a sign convention issue in the problem statement. Let's assume standard convention: Case $1$: $V = E - Ir = 12$ for $I = 2\,A \Rightarrow E - 2r = 12$.Case $2$: $V = E + Ir = 15$ for $I = 3\,A \Rightarrow E + 3r = 15$.Adding the two equations: $(E - 2r) + (E + 3r) = 12 + 15 \Rightarrow 2E + r = 27$.Multiplying $(1)$ by $3$ and $(2)$ by $2$: $3E - 6r = 36$ and $2E + 6r = 30$.Adding these: $5E = 66 \Rightarrow E = 13.2\,V$.

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 $3\,A$ flows in the opposite direction,the potential difference across the terminals of the battery is $15\,V$. The emf of the battery is ............... $V$.

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