Resistance And Electric Power

In a practice problem A motor rated at 20 A with a voltage of 115V exerts a force of 4900 N over a distance of 10 m in 30 s. Using the formulas $$P=VI$$ and $$P=\frac{Fs}{t}$$, we can see that the motor uses 2300 W of power while the action it does uses 1633.33 W of power. You are then asked to calculate the resistance which causes the remaining power $$P=2300$$ W $$- 1633.33$$ W $$= 666.67$$ W to dissipate.

The book does this using the formula $$P=I^2R$$ which yields 1.67 $$\Omega$$

When I try to do this with the formula $$P=\frac{V^2}{R}$$ I get roughly 19.837 $$\Omega$$

Why does this not yield the same answer?

• What voltage did you use for the second formula?
– nasu
Aug 21 '21 at 14:19

For a direct current motor, (at constant speed), you must write $$V=Ri+E$$ with $$E$$ the back electromotive force associated with the movement of the rotor : $$E=\emptyset\omega$$. So, you can't use $$V=Ri$$
• So neither answer is valid? Because $P=I^2R$ is also derived using $V=RI$ through $P=VI=I^2R$ Aug 21 '21 at 9:17
• Since you know the current passing through the resistance, you must use $P=RI^2$ to calculate the power dissipated in this resistance. Aug 21 '21 at 9:39