JAMB Physics 2013 Paper

Concept:The power rating of an electrical device is the product of voltage and current.Use the relationship $P = IV$ to determine the maximum safe current.Explanation:The device is rated at $P = 2000\ \mathrm{W}$ and $V = 250\ \mathrm{V}$.Rearrange the power formula to make current the subject:$I = \dfrac{P}{V}$.Substitute the given values into the equation:$I = \dfrac{2000\ \mathrm{W}}{250\ \mathrm{V}}$.Simplify the fraction by dividing $2000$ by $250$:$I = 8\ \mathrm{A}$.This calculated current corresponds to the correct option.Answer:The maximum current the device can take is $8\ \mathrm{A}$.