A battery is charged at potential of 15 V for 8 h by means of a current of 10A. While discharging it supplies a current 5A for 15h at a potential difference of 14 V. Calculate the watt-hour efficiency of the battery.

Options

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

Answer

To calculate the watt-hour efficiency of the battery, we need to find the energy consumed during charging and the energy supplied during discharging. Then, we can use these values to find the efficiency.

### Step-by-Step Solution:

1. **Calculate Energy Consumed During Charging:**
- The formula for energy consumed is given by:
\[ \text{Energy}_{\text{charging}} = V \times I \times T \]
- Here, the potential difference \( V_1 = 15 \, \text{V} \), current \( I_1 = 10 \, \text{A} \), and time \( T_1 = 8 \, \text{h} \).
- Convert time from hours to seconds (1 hour = 3600 seconds):
\[ T_1 = 8 \times 3600 = 28800 \, \text{s} \]
- Now, calculate the energy:
\[ \text{Energy}_{\text{charging}} = 15 \, \text{V} \times 10 \, \text{A} \times 28800 \, \text{s} = 4320000 \, \text{J} \]

2. **Calculate Energy Supplied During Discharging:**
- The formula for energy supplied is:
\[ \text{Energy}_{\text{discharging}} = V \times I \times T \]
- Here, the potential difference \( V_2 = 14 \, \text{V} \), current \( I_2 = 5 \, \text{A} \), and time \( T_2 = 15 \, \text{h} \).
- Convert time from hours to seconds:
\[ T_2 = 15 \times 3600 = 54000 \, \text{s} \]
- Now, calculate the energy:
\[ \text{Energy}_{\text{discharging}} = 14 \, \text{V} \times 5 \, \text{A} \times 54000 \, \text{s} = 3780000 \, \text{J} \]

3. **Calculate Watt-Hour Efficiency:**
- The efficiency can be calculated using the formula:
\[ \text{Efficiency} = \left( \frac{\text{Energy}_{\text{discharging}}}{\text{Energy}_{\text{charging}}} \right) \times 100 \]
- Substitute the values:
\[ \text{Efficiency} = \left( \frac{3780000 \, \text{J}}{4320000 \, \text{J}} \right) \times 100 \]
- Simplifying this gives:
\[ \text{Efficiency} = \left( \frac{378}{432} \right) \times 100 \approx 87.5\% \]

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