5 years ago
Yes, it should work
Current ratings on power supplies or batteries don't indicate the amount of current that they are going to supply. It is only an indication of the maximum amount of current the power source is capable of supplying.The amount of current is determined by two things, the voltage and the load. For example if your motor has a 10Ω impedance at 12V it will draw 1.2A.
5 years ago
No, for a motor, V=IR+Back emf. The terminal resistance does not tell you what the current will be, except in stall.
And that's why I said impedance and not resistance, you can't measure it with a multimeter but its value is still measured in Ohms.
Why ? Impedance isn't a get-out. Its still a dissipative component, so its resistance. Its a DYNAMIC resistance though, since the mechanical load affects it. .
The resistance is a property of the conductor and it does not change when you apply load to the motor. When load is applied to the motor the Back-EMF decreases. This decrease to the Back-EMF is what causes the current to increase I = (V - BackEMF) / R. You can't say that the resistance changes because the resistance is a property of the conductor, that's why I said impedance.
Electrical impedance is the measure of the opposition that a circuit presents to a current when a voltage is applied.
You are confusing the issue at hand. For one thing, we are not talking about a conductor, but a DC motor system. You measure the resistance of the motor in your example IT WILL NOT tell you what current the motor will pull when its running. My point is still made.
Pick a better example, or give up when you're wrong. The motor's impedance is, to all intents and purposes, resistive, except transiently. Using the word impedance is just wrong here.
I never measured the resistance of the motor in my example and I never said, that if you measure it with an Ohm meter it's going to tell you the amount of current the motor will draw at a given voltage. By measuring the resistance with an Ohm meter you will only get the resistance of the copper wire that forms the windings inside the motor.In my example I was trying to explain the concept of the load and I said "let's say we have a 10Ω load". It doesn't matter from where that 10Ω are coming from and I never said they were measured with an Ohm meter, I only said if you have a 10Ω load at 12V it's going to draw 1.2A.Now about the terminology, why I said impedance instead of resistance. Let's see what Wikipedia says the resistance is:The electrical resistance of an electrical conductor is the opposition to the passage of an electric current through that conductor.You can't use the word resistance here because as you also said "we are not talking about a conductor". And the resistance is a property of a conductor.Now about the impedance:Electrical impedance is the measure of the opposition that a circuit presents to a current when a voltage is applied.The correct term is impedance because it doesn't only refer to conductors but it's a more general term. The impedance is indeed resistive and I never said the opposite. Of course you have resistance too the windings are not made of an ideal conductor, but it's not resistance only. Now let's take a look at the following equation:I = (V - BackEMF) / RWhat happens when the Back-EMF decreases? The answer is obviously that the current is going to increase. But as we see from the equation the R has no reason to change when the Back-EMF changes. That means the resistance never changes when load is applied to the motor the Back-EMF does, and that change to the Back-EMF is going to change the current that the motor will draw.
Anyway I really don't think that this conversation leads anywhere. I was just trying to explain the concept of the load to chemical70 and this whole conversation completely misses the point. The current that a motor draws sure depends on the supplied voltage, the conductor resistants and the Back-EMF, but what I was trying to do is to combine the last two into one thing by just saying impedance. If you take the equation:I = (V - BackEMF) / RAnd you write it as:I = (V - I*Z) / RAnd then solve for I you get:I = V / R+ZWhen I said impedance in the first place I was referring to the sum of R and Z. And since both R and Z are measured in Ohms it is perfectly valid to just say impedance. There is no point making the things more complicated by introducing the concept of the Back-EMF.
And to be more precise, you can actually measure it with a multimeter but not by measuring resistance. If you measure the same time the voltage drop across the motor and the current that goes through it, then you can very easily calculate its impedance by using the Ohms law.
Yes. The motor will pull the current it wants.