Question 2.5: An electric hotplate has a power rating of 1440 W and a curr......

The quantity to be found is resistance, and the known quantities are power and current. Use the formula shown in Figure 2–21

R = \frac{\mathbf{P}}{\mathbf{I}^{2}}

R = \frac{1440\,\mathrm{W}}{12\,\mathrm{A}\times12\,\mathrm{A}}

R = \frac{1440\,\mathrm{W}}{144\ A}

R = 10 Ω