JEE Physics 2025 Question with Solution

Given: Voltage of the battery is $V = 4.2 \, \text{V}$ and capacity is $5800 \, \text{mAh}$.

Find: The energy stored in the battery when fully charged.

Use the relation between electrical energy, voltage, and charge:

$$E = Vq$$

Convert the battery capacity into charge in coulombs:

$$q = 5800 \times 3600 \times 10^{-3} \, \text{C}$$ $$q = 5800 \times 3.6 \, \text{C} = 20880 \, \text{C}$$

Now calculate the energy stored:

$$E = 4.2 \times 20880$$ $$E = 87696 \, \text{J}$$

Converting into kilojoules:

$$E = 87.696 \, \text{kJ}$$

Therefore, the energy stored in the battery when fully charged is approximately $87.7 \, \text{kJ}$. The correct option is C.

Given: Voltage is $4.2 \, \text{V}$ and battery capacity is $5800 \, \text{mAh}$.

Find: Total energy stored in the battery.

The electrical energy stored in a battery is the product of voltage and total charge:

$$E = V \times Q$$

First convert milliampere-hour into ampere-hour:

$$5800 \, \text{mAh} = 5.8 \, \text{Ah}$$

Now convert ampere-hour into coulomb using $1 \, \text{Ah} = 3600 \, \text{C}$:

$$Q = 5.8 \, \text{Ah} \times 3600 \, \frac{\text{C}}{\text{Ah}}$$ $$Q = 20880 \, \text{C}$$

Substitute in the energy formula:

$$E = 4.2 \, \text{V} \times 20880 \, \text{C}$$ $$E = 87696 \, \text{J}$$

Expressing in kilojoules:

$$E = \frac{87696}{1000} \, \text{kJ} = 87.696 \, \text{kJ}$$

Hence, the total energy stored is $87696 \, \text{J}$ or approximately $87.7 \, \text{kJ}$.