A battery is charged at a potential of 15 V for 8 hours when the current flowing is 10 A. The battery on discharge supplies a current of 5 A for 15 hours. The mean terminal voltage during discharge is 14 V. The "Watt-hour" efficiency of the battery is
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The power of the battery, when charged, is given by $\mathrm{P}_1 = \mathrm{V}_1 \mathrm{I}_1$ The electrical energy dissipated is given by $E_{1} = P_{1} t_{1}$
i. e., $E_1 = P_1 I_1 t_1 = 15 \times 10 \times 8 = 1200 \text{ Wh}$
Similarly, the electrical energy dissipated during the discharge of battery is given by $E_2 = V_2 I_2 t_2 = 14 \times 5 \times 15 = 1050 \text{ Wh}$
Hence, watt-hour efficiency of the battery is given by $\eta = \frac{E_2}{E_1} \times 100 = 0.875 \times 100 = 87.5\%$
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