When a current of 2, A flows in a battery fro | vedclass

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

Vedclass · Accepted Answer

$13.2$

Vedclass · Answer

(B) Let $E$ be the $emf$ and $r$ be the internal resistance of the battery.When current flows from negative to positive terminal (charging),the terminal potential difference is given by $V = E + Ir$.Given $I_1 = 2\, A$ and $V_1 = 12\, V$,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 $I_2 = 3\, A$ and $V_2 = 15\, V$,we have: $15 = E - 3r$ ... $(2)$Wait,the problem states current flows from negative to positive terminal when $V=12\, V$. This is charging,so $V = E + Ir$. If current flows in the opposite direction (positive to negative),it is discharging,so $V = E - Ir$. Let's re-evaluate: Case $1$: $12 = E + 2r$ (Charging)Case $2$: $15 = E - 3r$ (Discharging)This leads to $E - 2r = 12$ and $E + 3r = 15$ is not possible as $E$ would be negative. Correct interpretation: When current flows from negative to positive,it is charging $(V = E + Ir)$. When current flows from positive to negative,it is discharging $(V = E - Ir)$.Actually,the question implies: $12 = E - 2r$ (Discharging at $2\, A$)$15 = E + 3r$ (Charging at $3\, A$)Adding $3 	imes (1)$ and $2 	imes (2)$: $3(12) + 2(15) = 3(E - 2r) + 2(E + 3r) \Rightarrow 36 + 30 = 3E - 6r + 2E + 6r \Rightarrow 66 = 5E \Rightarrow E = 13.2\, V$.

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