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 terminal voltage during discharge is `14 V`. The 'Watt-hour" efficiency of the battery is.

Options

A. `80%`
B. `90%`
C. `87.5%`
D. `82.5%`

Explanation

To find the 'Watt-hour' efficiency of the battery, we need to calculate both the charging energy and the discharging energy, and then use these values to find the efficiency.

### Step-by-Step Solution:

1. **Calculate Charging Energy:**
- The formula for energy is given by:
\[ \text{Energy} = \text{Power} \times \text{Time} \]
- Power can be calculated using:
\[ \text{Power} = \text{Voltage} \times \text{Current} \]
- Given:
- Voltage during charging (V) = 15 V
- Current during charging (I) = 10 A
- Time (t) = 8 hours
- First, convert time from hours to seconds for consistency:
\[ t = 8 \text{ hours} \times 3600 \text{ seconds/hour} = 28800 \text{ seconds} \]
- Now calculate the power:
\[ \text{Power} = 15 \text{ V} \times 10 \text{ A} = 150 \text{ W} \]
- Now calculate the charging energy:
\[ \text{Charging Energy} = 150 \text{ W} \times 28800 \text{ s} = 4320000 \text{ J} \text{ or } 4320 \text{ kJ} \]

2. **Calculate Discharging Energy:**
- Given:
- Voltage during discharge (V) = 14 V
- Current during discharge (I) = 5 A
- Time (t) = 15 hours
- Convert time from hours to seconds:
\[ t = 15 \text{ hours} \times 3600 \text{ seconds/hour} = 54000 \text{ seconds} \]
- Calculate the power during discharge:
\[ \text{Power} = 14 \text{ V} \times 5 \text{ A} = 70 \text{ W} \]
- Now calculate the discharging energy:
\[ \text{Discharging Energy} = 70 \text{ W} \times 54000 \text{ s} = 3780000 \text{ J} \text{ or } 3780 \text{ kJ} \]

3. **Calculate Efficiency:**
- The efficiency (η) can be calculated using the formula:
\[ \eta = \frac{\text{Discharging Energy}}{\text{Charging Energy}} \times 100\% \]
- Substitute the values:
\[ \eta = \frac{3780000 \text{ J}}{4320000 \text{ J}} \times 100\% \]
- Calculate the efficiency:
\[ \eta = \frac{3780}{4320} \times 100\% \approx 87.5\% \]

### Final Answer:
The 'Watt-hour' efficiency of the battery is **87.5%**.