1. **Problem Statement:** We analyze the battery charge and discharge data to find: - Capacity of the charged battery and coulombic efficiency - C-rate during discharge - Energy content of the fully charged battery - Energy efficiency - Voltage efficiency 2. **Given Data:** - Charge current $I_c = 250$ mA = 0.25 A - Discharge current $I_d = 500$ mA = 0.5 A - Charge time $t_c = 60$ min - Discharge time $t_d = 28$ min (cut off voltage near 2.0 V) - Voltage varies as per tables 3. **Formulas and Important Rules:** - Capacity $Q = I \times t$ (in Ah, convert minutes to hours) - Coulombic efficiency $\eta_c = \frac{Q_{discharge}}{Q_{charge}} \times 100\%$ - C-rate $C = \frac{I}{Q_{charge}}$ - Energy $E = \int V(t) I dt$ approximated by average voltage $\times$ capacity - Energy efficiency $\eta_e = \frac{E_{discharge}}{E_{charge}} \times 100\%$ - Voltage efficiency $\eta_v = \frac{V_{discharge,avg}}{V_{charge,avg}} \times 100\%$ 4. **Step 1: Calculate charge capacity $Q_{charge}$** $$Q_{charge} = I_c \times \frac{t_c}{60} = 0.25 \times 1 = 0.25 \text{ Ah}$$ 5. **Step 2: Calculate discharge capacity $Q_{discharge}$** Discharge time $t_d = 28$ min $$Q_{discharge} = I_d \times \frac{t_d}{60} = 0.5 \times \frac{28}{60} = 0.2333 \text{ Ah}$$ 6. **Step 3: Coulombic efficiency $\eta_c$** $$\eta_c = \frac{0.2333}{0.25} \times 100 = 93.33\%$$ 7. **Step 4: C-rate during discharge** Using charged capacity: $$C = \frac{I_d}{Q_{charge}} = \frac{0.5}{0.25} = 2 \text{ C}$$ 8. **Step 5: Calculate average voltages** - Charge average voltage $V_{charge,avg}$ approximated from data: roughly $\frac{3.6 + 4.2}{2} = 3.9$ V - Discharge average voltage $V_{discharge,avg}$ approximated: roughly $\frac{4.2 + 2.0}{2} = 3.1$ V 9. **Step 6: Calculate energy content of fully charged battery $E_{charge}$** $$E_{charge} = V_{charge,avg} \times Q_{charge} = 3.9 \times 0.25 = 0.975 \text{ Wh}$$ 10. **Step 7: Calculate energy delivered during discharge $E_{discharge}$** $$E_{discharge} = V_{discharge,avg} \times Q_{discharge} = 3.1 \times 0.2333 = 0.723 \text{ Wh}$$ 11. **Step 8: Energy efficiency $\eta_e$** $$\eta_e = \frac{E_{discharge}}{E_{charge}} \times 100 = \frac{0.723}{0.975} \times 100 = 74.15\%$$ 12. **Step 9: Voltage efficiency $\eta_v$** $$\eta_v = \frac{V_{discharge,avg}}{V_{charge,avg}} \times 100 = \frac{3.1}{3.9} \times 100 = 79.49\%$$ **Final answers:** - Capacity charged $= 0.25$ Ah - Coulombic efficiency $= 93.33\%$ - C-rate during discharge $= 2$ C - Energy content fully charged $= 0.975$ Wh - Energy efficiency $= 74.15\%$ - Voltage efficiency $= 79.49\%$