Question

The starter motor of a car draws a current \(I = 300\) A from the battery of voltage 12 V. If the car starts only after 2 minutes, what is the energy drawn from the battery?

• A. 3 kJ
• B. 30 kJ
• C. 7.2 kJ
• D. 432 kJ

Hint:

Time must be in seconds when using SI units.

Answer 1

The Correct Option is D

To find the energy drawn from the battery, we need to use the formula for electrical energy, which is given by:

\(E = V \cdot I \cdot t\)

where:

• \(V\) is the voltage across the battery (in volts)
• \(I\) is the current drawn (in amperes)
• \(t\) is the time for which the current is drawn (in seconds)

Let's plug in the given values:

• \(V = 12 \, \text{V}\)
• \(I = 300 \, \text{A}\)
• \(t = 2 \, \text{minutes} = 2 \times 60 = 120 \, \text{seconds}\)

Substitute these values into the equation:

\(E = 12 \times 300 \times 120\)

Calculate the result:

\(E = 3600 \times 120 = 432\,000 \, \text{Joules}\)

Since 1 kJ = 1000 J, we convert the energy into kilojoules:

\(E = \frac{432\,000}{1000} = 432 \, \text{kJ}\)

Therefore, the energy drawn from the battery is 432 kJ. This matches the correct answer option: 432 kJ.