An electric motor connected to a $\mathbf{120}\mathit{V}$, $\mathbf{60}\mathbf{.}\mathbf{0}\mathit{H}\mathit{z}$ac outlet does mechanical work at the rate of$\mathbf{0}\mathbf{.}\mathbf{100}\mathit{h}\mathit{p}\left(1hp=746W\right)$ . (a) If the motor draws an rms current of $\mathbf{0}\mathbf{.}\mathbf{650}\mathit{A}$, what is its effective resistance, relative to power transfer? (b) Is this the same as the resistance of the motor’s coils, as measured with an ohmmeter with the motor disconnected from the outlet?

• Voltage,$V=120V$.
• Frequency of ac outlet,$f=60.0Hz$.
• The magnitude of RMS current$I_{rms}=0.650A$.
• Rate of mechanical work, $P_{mech}=0.100;hp=74.6W$.

Use the formula of relative power transfer to find the effective resistance and to find if the resistance of the motor’s coils is the same as measured with an ohmmeter with the motor disconnected from the outlet.

The formula is as follows:

$P_{mech}=I_{rms}^2{R}_{eff}$

Where, P is power, I is current and Ris resistance.

The relative power transfer is,

$P_{mech}=I_{rms}^2{R}_{eff}$

Rearranging the terms,

$R_{eff}=\frac{P_{mech}}{I_{rms}^2}$

Substituting the values,

$\begin{array}{rcl}{R}_{eff}& =& \frac{\left(74.6W\right)}{{\left(0.65A\right)}^2}\\ R_{eff}& =& 176.56\\ & \approx & 177\Omega \end{array}$

Therefore, the effective resistance is $177\Omega$.

Energy loss across the resistor is$I_{rms}^{2}R$. This tell how much energy is lost due to thermal dissipation. The effective resistance just gives the resistance for power transfer from the electrical to mechanical form.

Thus,the resistance of the motor’s coils is not the same as measured with an ohmmeter with the motor disconnected from the outlet.

Find the effective resistance using the formula of relative power transfer. The resistance of the motor’s coils is not the same as measured with an ohmmeter with the motor disconnected from the outlet.