Question

What power is supplied to the starter motor of a large truck that draws \(250\;A\) of current from a \(24.0\;V\) battery hook-up?

Short answer

The power supplied to the starter motor by the battery is obtained as: \(P{\rm{ }} = {\rm{ }}6.00{\rm{ }}kW\).

Step-by-step solution

Define Resistance

In an electrical circuit, resistance is a measure of the resistance to current flow. The Greek letter omega\({\rm{(\Omega )}}\)is used to represent resistance in ohms.

Concepts and Principles

If a potential difference\(\Delta V\)is maintained across a circuit element, the power, or the rate at which energy is supplied to the element, is:

\(P{\rm{ }} = {\rm{ }}I\Delta V\)

The given data and the required data

• The current drawn from the battery is:\(I{\rm{ }} = {\rm{ }}250{\rm{ }}A\).
• The potential difference across the battery is: \(\Delta V{\rm{ }} = {\rm{ }}24.0{\rm{ }}V\).

Evaluating power supplied to the starter motor

Power supplied to the starter motor by the battery is found from the above equation as:

\(P{\rm{ }} = {\rm{ }}I\Delta V\)

Entering numerical values and we obtain:

\(\begin{aligned}{}P{\rm{ }} &= {\rm{ }}(250{\rm{ }}A)(24.0{\rm{ }}V)\\ &= {\rm{ }}6.00{\rm{ }} \times {\rm{ }}{10^3}{\rm{ }}W\\& = {\rm{ }}(6.00{\rm{ }} \times {\rm{ }}{10^3}{\rm{ }}W)(\frac{{1{\rm{ }}kW}}{{1000{\rm{ }}W}})\\ &= {\rm{ }}6.00{\rm{ }}kW\end{aligned}\)

Therefore, power supplied to the starter motor by the battery is:\(P{\rm{ }} = {\rm{ }}6.00{\rm{ }}kW\).