Physics - Current Electricity Question with Solution | TestHub

The power of the battery, when charged, is given by $P_1=V_1{I}_1$ The electrical energy dissipated is given by $E_1=P_1{t}_1$
i. e., $E_1=P_1{I}_1{t}_1=15\times 10\times 8=1200\mathrm{Wh}$
Similarly, the electrical energy dissipated during the discharge of battery is given by$E_2=V_2{I}_2{t}_2=14\times 5\times 15=1050\mathrm{Wh}$

Hence, watt-hour efficiency of the battery is given by $\eta =\frac{E_2}{E_1}\times 100=0.875\times 100=87.5\%$