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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?

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  • $\begingroup$ What voltage did you use for the second formula? $\endgroup$
    – nasu
    Aug 21 '21 at 14:19
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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 $V=Ri$

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  • $\begingroup$ So neither answer is valid? Because $P=I^2R$ is also derived using $V=RI$ through $P=VI=I^2R$ $\endgroup$
    – John Doe
    Aug 21 '21 at 9:17
  • $\begingroup$ Since you know the current passing through the resistance, you must use $P=RI^2$ to calculate the power dissipated in this resistance. $\endgroup$ Aug 21 '21 at 9:39
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The voltage might be applied across more components than just the resistor in question.

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