Finding the EMF of a motor

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Homework Statement



A motor as an internal resistance of 0.5 ohms. It draws 20 amperes at full load from a 120-V supply.

a) What is the emf of the motor?

b) What is the load resistance?

Homework Equations



Potential difference between the terminals = emf - Ir (r = internal resistance)
(P.D. = emf - Ir)

V = IR

The Attempt at a Solution



Part a:

P.D. = emf - Ir
120 = emf - (20)(0.5)
120 = emf - 10
emf = 130 volts

I'm not sure I understand what the 120-V supply is referring to. I assumed it means the potential difference between the terminals when the current is 20 amperes?

My hesitation is that it says "120-V supply" so I was thinking that maybe 120-V is the emf, and in that case the potential difference would be 120 - 10 = 110 Volts? This would also change the answer in part b.

Part b:

Here I assume if we have a circuit with a 120-V source and a 0.5 ohms internal resistance, the load resistance is a second resistance (R) on the circuit.

So I apply:

V = IR (where R is the load resistance)
R = V / I

R = 120 / 20 = 6 ohms
 

Answers and Replies

  • #2
haruspex
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Potential difference between the terminals = emf - Ir (r = internal resistance)
That equation relates emf of a battery to the potential difference across its terminals. The question concerns a motor, not a battery.
It is a bit unusual to describe a motor as having an emf. I assume they mean "back emf". You can think of this as the potential difference related to useful work done by the motor. How do you think internal resistance would adjust that to arrive at the potential difference across its terminals?
 
  • #3
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Hi, thanks for your response. I'm not sure I understand the difference. I searched but I haven't found any reference to the words "back emf" in the course material. I was a bit surprised by the word "motor" because it doesn't show up anywhere in any of the assignments until now nor in the course pack. I looked up definitions of "motor" online but what I found seems to be beyond the level of the course.

The only mention of "emf" in the course material is this:

esbIA05.jpg



The way I approached this problem was just to note variables as I read it, so I set:

Current = 6 amperes
Internal resistance = 0.5 ohms
Voltage = 120 volts

Is this wrong?
 

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  • #4
haruspex
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Hi, thanks for your response. I'm not sure I understand the difference. I searched but I haven't found any reference to the words "back emf" in the course material. I was a bit surprised by the word "motor" because it doesn't show up anywhere in any of the assignments until now nor in the course pack. I looked up definitions of "motor" online but what I found seems to be beyond the level of the course.

The only mention of "emf" in the course material is this:

View attachment 224223


The way I approached this problem was just to note variables as I read it, so I set:

Current = 6 amperes
Internal resistance = 0.5 ohms
Voltage = 120 volts

Is this wrong?
As I posted, the section of text you quote is for a source of power, such as a battery. Here we are dealing with the other side of the picture, the consumer of power.
Did you google "back emf"?
Crudely, you can think of the motor as being two resistors in series, the "internal resistance", which is not useful, and the useful load resistance. It is the second which represents the back emf. (I say crudely because a back emf is not the same as normal electrical resistance. It is the motor behaving as an electrical generator, creating a voltage that opposes the applied one.)
From the perspective of the external circuit, how would those resistances relate to the voltage across the terminals and the current that flows?
 
  • #5
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I googled it but what I found was talking about Lenz's law... is that the right thing? We kind of skipped Lenz's law, except for the equation Ei = (-1) I / t where Ei is the electric field induced by a current.

If I understand your example, the back emf would be the second resistor that is in series with the internal resistance? Does this mean that, basically, part a and part b of the question give the same answer?
 
  • #6
haruspex
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If I understand your example, the back emf would be the second resistor that is in series with the internal resistance?
Yes, except that one is a voltage and the other a resistance. Likewise below. What relates them?
Does this mean that, basically, part a and part b of the question give the same answer?
 
  • #9
haruspex
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Ok thanks I'll print this out and see if I can work through it
Ok, but that first diagram illustrates what I wrote.
Edit: it looks just like the diagram for internal resistance of a battery, but the key diffeence is that the back emf points the other way, opposing the battery voltage.
 
  • #10
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Ok, but that first diagram illustrates what I wrote.
Edit: it looks just like the diagram for internal resistance of a battery, but the key diffeence is that the back emf points the other way, opposing the battery voltage.
Hmm... it says the motor has a voltage drop Vmotor which appears to be equal to the emf.

They the resistance in the armature ra produces a voltage drop Va = Ira. This looks a lot like the emf equation I'm used to.

But then they say the back emf is produced by a rotating armature? I'm not sure I get that - the only time we talked about anything rotating was with regards to magnetism?

I'm trying to wrap my mind about the equations there... I see Vback = emf - Ira...

I'm not sure I'm getting this but then this would mean that in my problem the emf they're talking about is actually the V in the equation? This would make my answer 110 volts instead of 130 I think.

I'm gonna keep reading to see if something clicks in my head
 
  • #12
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Thanks for the help!

It's still not intuitive for me though. I'm gonna try to read through this stuff again. It feels like this is something I need to understand.
 

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