Resistance of an electric motor

[pc2-brazil]

Homework Statement

An electric motor connected to a 120 V, 60 Hz supply does work at a rate of 0.1 hp (1 hp = 746 W). If it consumes an rms (root mean square) current of 650 mA, what is its resistance, in terms of power transmission? Would this result be the same thing as the resistance of its coils, measured by an ohmmeter, with the motor disconnected from the generator?

Homework Equations

The Attempt at a Solution

The average power delivered by the power supply to the motor is $P_{av} = i_{rms}^2R$. Since $P=(0.1)(746)=74.6\ \mathrm{W}$, and $i_{rms} = 0.65\ \mathrm{A}$, then $R = \frac{74.6}{0.65^2} = 176.6\ \Omega$ (this is the correct value at the back of the book).
My doubt regards the second question. I imagine that, since the resistance R that is used in the power formula of an RLC circuit is simply the resistance of the circuit, the value of the resistance should remain the same, even if it was measured by an ohmmeter with the motor stopped (it's just the impedance that would be different). However, the answer to this says that: "This would not be the same as the direct current resistance of the coils of a stopped motor, because there would be no inductive effects.".

Is my reasoning correct?

Thank you in advance.

 

[Andrew Mason]

The Attempt at a Solution

The average power delivered by the power supply to the motor is $P_{av} = i_{rms}^2R$. Since $P=(0.1)(746)=74.6\ \mathrm{W}$, and $i_{rms} = 0.65\ \mathrm{A}$, then $R = \frac{74.6}{0.65^2} = 176.6\ \Omega$ (this is the correct value at the back of the book).

Resistance is what you would measure with the ohmmeter across the input terminals of the motor with the motor stopped. If I understand the problem correctly, we can calculate that from the information provided.

Here, the power input is P = V_rmsI_rms. That is 120 x .65 =78 Watts. The power output is 74.6 W. Efficiency = 74.6/78 = 95.6%

Use that efficiency to determine the percentage of energy that is lost through resistance (ie. the I^2R loss). You can determine R from that.

AM

 

[CWatters]

I don't really like the book answer.

You calculated R=176.6 from the power lost but that assumes all the power lost is in the winding resistance. It's not. There are other sources of loss in a motor such as friction in the bearings, air drag on the rotor, eddy current losses in ferrite parts etc. These also account for some of the loss. So the real resistive losses would be lower. So when you just measure R with a meter it will be different.

 

[NoPoke]

The book answer is okay since the question is asking what the motor looks like to the power transmission network. An equivalent load "black-box"approach.

The book answer to the second part is fine too as long as you realize that "inductive" effects is covering all the interesting behaviour of a real motor. For example a spinning motor behaves like a generator and produces back-emf . So very crudely inside the black box is the resistance from the copper winding in series with a variable voltage power supply whose voltage is given by how fast the motor is spinning. The I-V characteristic of a spinning motor does not look like the resistance in an R L C circuit.

I share CWatters concern that the book answer may mislead you into thinking that this 177 ohm resistance is a physical resistor.

 

[Nickie]

Yes, your reasoning is correct. The resistance measured by an ohmmeter when the motor is disconnected from the generator would not be the same as the resistance used in the power formula. This is because the resistance measured by the ohmmeter is only the direct current resistance of the coils, while the resistance used in the power formula takes into account the inductive effects of the motor. When the motor is running, the inductive effects contribute to the overall impedance of the motor and affect the amount of current that flows through it. Therefore, the resistance used in the power formula is not just the direct current resistance measured by the ohmmeter.