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A battery is charged at a potential of \(15 \mathrm{~V}\) for 8 hours when the current flowing is \(10 \mathrm{~A}\). The battery on discharge supplies a current of \(5 \mathrm{~A}\) for 15 hour. The mean terminal voltage during discharge is 14V. The "Watt-hour" efficiency of the battery is
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\(87.5 \%\)
Efficiency is given by \(\eta=\frac{\text { output }}{\text { input }}\)
\(=\frac{5 \times 15 \times 14}{10 \times 8 \times 15}=0.875 \text { or } 87.5 \%\)
\(=\frac{5 \times 15 \times 14}{10 \times 8 \times 15}=0.875 \text { or } 87.5 \%\)
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