I'm new to electronics, still learning, and I need some help understanding something. I've got an aerator that runs on two 1.5V D batteries. I'm trying to figure a way I can wire this to work on my 12 volt boat battery and do away with the D cells which are always dying at inconvenient times. Anyway, I connected the motor to a 3.45 volt power supply and read the power draw as 0.2 amps. Just for kicks I read the resistance of the motor and it reads 1.9 ohms. But according to ohm's law, at that voltage with a 0.2 amp draw the resistance should be 17.25 ohms. On the other side, a motor with 1.9 ohms resistance should draw 1.81 amps! And the 1.9 ohms * 0.2 amps is .38 volts! These numbers aren't even close. What's going on? Am I not applying the formula correctly? Isn't it R = V/I, I = V/R and V = I*R? This should be about as simple and basic as you can get, but I'm lost. Evidently I'm not learning as well as I thought.
BTW I checked all the readings with 2 different digital meters and got identical readings.

 

You're neglecting meter resistance when measuring current. Classical novice mistake.

Try using paragraphs in your posts; correct grammar makes them easier to read.

 

A running motor has much higher resistance than a stalled one.

 

Your calculations are correct as far as they go. The problem is you don't understand how a motor operates.

The measured resistance of the motor is just the parasitic resistance of the windings.
That is what determines the current during startup, before the rotor starts turning, which is generally much higher than the running current.

When running, most of the voltage drop across the motor is due to the back EMF from the spinning rotor, which is the part of the electrical energy used that is doing the work.
Any power dissipated in the motor resistance is just wasted and contributes to the motor inefficiency.

So your resistance measurements can roughly determine what the startup motor current is, but cannot be used to predict the motor running current, which is determined by the motor load and efficiency.

Make sense?

Note that if you want to operate the motor from 12V, the best way is likely to use a regulator, such as the common LM317, to generate 3V from the 12V battery.
That way the motor can use whatever current it needs for the high current required during startup, and as determined by the load when running.

 

You cannot use resistance measurement with a motor. A motor is an inductive element. As the motor starts to spin the back EMF opposes the supply voltage and the current is reduced.

Ohm's Law does not work in this case.

(as crutschow says. He beat me to it).

 

Use voltage regulator to change 12V to 3V. Something like AMS1117 3.3V regulator.

 

The inductances of the motor windings are alternately switched. The current through the inductance increases smoothly and during normal operation of the motor does not reach a maximum. But when the motor starts up, a larger current develops (until the rotor rotates). If you brake the motor (stop), then the current will be equal to the maximum, determined by Ohm's law.

 

I'm new to electronics, still learning, and I need some help understanding something. I've got an aerator that runs on two 1.5V D batteries. I'm trying to figure a way I can wire this to work on my 12 volt boat battery and do away with the D cells which are always dying at inconvenient times. Anyway, I connected the motor to a 3.45 volt power supply and read the power draw as 0.2 amps. Just for kicks I read the resistance of the motor and it reads 1.9 ohms. But according to ohm's law, at that voltage with a 0.2 amp draw the resistance should be 17.25 ohms. On the other side, a motor with 1.9 ohms resistance should draw 1.81 amps! And the 1.9 ohms * 0.2 amps is .38 volts! These numbers aren't even close. What's going on? Am I not applying the formula correctly? Isn't it R = V/I, I = V/R and V = I*R? This should be about as simple and basic as you can get, but I'm lost. Evidently I'm not learning as well as I thought.
BTW I checked all the readings with 2 different digital meters and got identical readings.

Hi,

Crutschow explained it pretty well, i'll just add a tiny bit more.

Resistors are linear and follow Ohm's Law, but motors are in a class we can call "non linear loads" and do not follow Ohm's Law directly. So simply put, there are at least two classes of electrical devices, where some follow Ohm's Law and some just dont. If you want to measure a device with an Ohm Meter, you have to know if it follows Ohm's Law. If it does not, then you cant use an Ohm Meter, and most likely more extensive tests will be required.

The wire used for the armature follows Ohm's Law for the most part, but once it starts turning we see other things happen, like it acts partly like a self generator which creates a voltage that opposes the applied voltage and that makes the motor look like it is running on a lower voltage and so we see lower current while it runs.

There are a host of other devices out there that also do not follow OHm's Law directly. Another good example is the silicon diode. For devices like this the apparent resistance changes with the DC operating point.

As the complexity of the device increases, the less likely it is to follow Ohm's Law. So unfortunately not everything follows Ohm's Law.

 

Eventually, readings do not match the law.

 

You can observe a similar discrepancy with a lightbulb. The filament resistance increases as it gets hot. The cold resistance you measure with your meter would suggest a much higher current draw and power usage.

Despite the discrepancy, the filament is always obeying Ohms law whether hot or cold.

As noted, a motor also obeys ohms law but is not strictly an ohmic device like the lightbulb.

 

You can observe a similar discrepancy with a lightbulb. The filament resistance increases as it gets hot. The cold resistance you measure with your meter would suggest a much higher current draw and power usage.

Despite the discrepancy, the filament is always obeying Ohms law whether hot or cold.

As noted, a motor also obeys ohms law but is not strictly an ohmic device like the lightbulb.

Please do not obfuscate the TS and other readers lacking the relevant knowledge.
I believe that you made this statement before.

Ohm's Law states that the current through a resistor is directly proportional to the voltage across the resistor and inversely proporational to the resistance.

Mathematically, we can express this as:

I = V/R

If the resistance R is not constant, this is a non-linear function and hence Ohm's Law does not apply.

Diodes, incandescent filaments, motors are non-linear devices.

 

The inductances of the motor windings are alternately switched.

But that's not what reduces the motor current when it runs.
It's the back EMF generated by the rotating magnetic field of the rotor and that looks like a resistive load dissipating real power, not an inductive load.
The motor inductance generates a small reactive part of the total load.

 

Your calculations are correct as far as they go. The problem is you don't understand how a motor operates.

The measured resistance of the motor is just the parasitic resistance of the windings.
That is what determines the current during startup, before the rotor starts turning, which is generally much higher than the running current.

When running, most of the voltage drop across the motor is due to the back EMF from the spinning rotor, which is the part of the electrical energy used that is doing the work.
Any power dissipated in the motor resistance is just wasted and contributes to the motor inefficiency.

So your resistance measurements can roughly determine what the startup motor current is, but cannot be used to predict the motor running current, which is determined by the motor load and efficiency.

Make sense?

Note that if you want to operate the motor from 12V, the best way is likely to use a regulator, such as the common LM317, to generate 3V from the 12V battery.
That way the motor can use whatever current it needs for the high current required during startup, and as determined by the load when running.

Your calculations are correct as far as they go. The problem is you don't understand how a motor operates.

The measured resistance of the motor is just the parasitic resistance of the windings.
That is what determines the current during startup, before the rotor starts turning, which is generally much higher than the running current.

When running, most of the voltage drop across the motor is due to the back EMF from the spinning rotor, which is the part of the electrical energy used that is doing the work.
Any power dissipated in the motor resistance is just wasted and contributes to the motor inefficiency.

So your resistance measurements can roughly determine what the startup motor current is, but cannot be used to predict the motor running current, which is determined by the motor load and efficiency.

Make sense?

Note that if you want to operate the motor from 12V, the best way is likely to use a regulator, such as the common LM317, to generate 3V from the 12V battery.
That way the motor can use whatever current it needs for the high current required during startup, and as determined by the load when running.

Thank you crutschow and everybody else that responded. Yes your answer makes sense. In fact as soon as I read it the little light bulb went off and I thought 'oh yeah, non linear loads'. Duh.
The good thing about this is I bet I won't forget to consider linear and non linear loads in the future. Screw ups can be a good teacher.
And thanks for the advice about using a regulator for the 3 volts. I had been reading about voltage dividers using resistors and thought I could divide it once to get 6 volts, then divide it again to get the 3 volts I needed.
I was trying to determine what size resistors would be required to make it work, which is why I was taking the readings.
It may not have even worked, but hey, tinkering like this is a good way to learn.
I appreciate everybody's help

 

I had been reading about voltage dividers using resistors and thought I could divide it once to get 6 volts, then divide it again to get the 3 volts I needed.

Resistive voltage dividers don't work well with any significant load because the resistance of the divider causes the voltage to drop under load.

But note that a resistive divider can drop the voltage by any value as determined by the resistor ratios.
The ratio is not limited to 1/2, so you could go from 12V to 3V with just two resistors (but as stated, that wouldn't work to drive your motor).

 

Ah, I didn't know that. I will try using the regulator mentioned before. Thanks

PS - you should be a teacher, you have a way of explaining things that makes it easy to understand. Sadly, not many teachers have that quality

 

If the resistance R is not constant, this is a non-linear function and hence Ohm's Law does not apply.

Ohms law does indeed apply, that's why it's called a law. Ignorance of what R is does not negate Ohms law. It just make it harder to apply.

 

If you want to see the motor match ohms law, clamp the shaft in a vise and apply voltage. Then it will draw 1.8A. For a little while at least, then it will probably burn up.

 

An assumption of Ohms law is that R is constant with changing current. For a motor (60hz quasi-synchronous) with a constant torque and power load reducing voltage increases current to keep the power equation balanced. While this is happening the lossy ohmic part of the resistance remains unchanged so the increased current causes an increased voltage drop on that element of the impedance with increased power dissipation (sometimes to the point of destruction) inside the motor and wiring instead of the load.

https://www.allaboutcircuits.com/textbook/direct-current/chpt-2/nonlinear-conduction/

Ohm’s Law is not very useful for analyzing the behavior of components like these where resistance varies with voltage and current. Some have even suggested that “Ohm’s Law” should be demoted from the status of a “Law” because it is not universal. It might be more accurate to call the equation (R=E/I) a definition of resistance, befitting of a certain class of materials under a narrow range of conditions.

 

You can observe a similar discrepancy with a lightbulb. The filament resistance increases as it gets hot. The cold resistance you measure with your meter would suggest a much higher current draw and power usage.

Despite the discrepancy, the filament is always obeying Ohms law whether hot or cold.

As noted, a motor also obeys ohms law but is not strictly an ohmic device like the lightbulb.

Neither is an ohmic device.

An ohmic device is one that obeys Ohm's Law (to whatever level of agreement you want to specify) which states that the current through the device is proportional to the voltage across the device. In other words, if you double the voltage, you get double the current while if you cut the voltage in half you get half the current. Neither a motor or a lightbulb even remotely behave this way and hence are, by definition, nonohmic.

 

@wayneh is confused as to what constitutes Ohm's Law.

You can calculate the current I by dividing the voltage by the resistance. That is not Ohm's Law.

Ohm's Law is:

I ∝ V

in other words:

I = k x V
where k is a constant.

I = V/R(V) is not Ohm's Law.

Hence, the formulas

I = V/R
V = I x R

represent Ohm's Law if V or R is constant.