Question

The battery of a mobile phone is rated as 4.2 V, 5800 mAh. How much energy is stored in it when fully charged?

• A. 43.8 kJ
• B. 48.7 kJ
• C. 87.7 kJ
• D. 24.4 kJ

Hint:

To calculate energy stored in a battery, use the formula \( E = Vq \), where \( V \) is the voltage and \( q \) is the charge in coulombs.

Answer 1

The Correct Option is C

Given the voltage \( V = 4.2 \) volts and the battery capacity \( 5800 \) mAh, we can calculate the energy stored in the battery using the formula: \[ \text{Energy supplied by battery} = Vq \] where \( q \) is the charge in coulombs. Converting \( 5800 \) mAh to coulombs: \[ q = 5800 \times 3600 \times 10^{-3} \, \text{C} = 5800 \times 3.6 \, \text{C} = 20880 \, \text{C} \] Thus, the energy supplied by the battery is: \[ \text{Energy} = 4.2 \times 5800 \times 3600 \times 10^{-3} = 87.696 \, \text{kJ} \] Therefore, the energy stored in the battery when fully charged is approximately \( 87.7 \, \text{kJ} \).

Answer 2

This problem asks for the total energy stored in a fully charged mobile phone battery, given its voltage and capacity rating.

Concept Used:

The electrical energy (\(E\)) stored in a battery is the product of its voltage (\(V\)) and the total charge (\(Q\)) it can deliver.

\[ E = V \times Q \]

The capacity of the battery is given in milliampere-hours (mAh), which is a unit of electric charge. To calculate the energy in the SI unit, Joules (J), we must first convert the charge from mAh to Coulombs (C).

The conversion is based on the definition of an Ampere (\(1 \text{ A} = 1 \text{ C/s}\)) and an hour (\(1 \text{ hr} = 3600 \text{ s}\)):

\[ 1 \text{ mAh} = 10^{-3} \text{ A} \times 3600 \text{ s} = 3.6 \text{ C} \]

Step-by-Step Solution:

Step 1: List the given values from the problem statement.

• Voltage, \( V = 4.2 \, \text{V} \)
• Capacity (Charge), \( Q_{rated} = 5800 \, \text{mAh} \)

Step 2: Convert the battery capacity from mAh to the SI unit of charge, Coulombs (C).

First, convert milliampere-hours (mAh) to ampere-hours (Ah):

\[ Q_{rated} = 5800 \, \text{mAh} = 5.8 \, \text{Ah} \]

Next, convert ampere-hours (Ah) to Coulombs (C), knowing that \( 1 \, \text{Ah} = 3600 \, \text{C} \):

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

Step 3: Calculate the total energy stored in the battery.

Using the formula \( E = V \times Q \), we substitute the given voltage and the calculated charge in Coulombs.

\[ E = 4.2 \, \text{V} \times 20880 \, \text{C} \]

Final Computation & Result:

Performing the final multiplication:

\[ E = 87696 \, \text{J} \]

The energy can also be expressed in kilojoules (kJ):

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

The total energy stored in the battery when fully charged is 87696 J or approximately 87.7 kJ.