# Homework Help: Operating Point

1. Mar 3, 2016

### Marcin H

1. The problem statement, all variables and given/known data
So I have to plot 2 sets of IV curves for my lab and find the operating points of where they cross. Problem is, they don't cross!!! I'm not sure if this is the right place to ask, but my professor isn't responding to my emails so here it is.
This is what the circuit looks like for this part of the lab:

2. Relevant equations
V=IR

3. The attempt at a solution
Here are the measurements

And here is where I get stuck:

And here is what I got. The blue line is the thevenin IV line. the Red is the motor IV. It says in the last table to use data from step 5 and this is what it gives me. I'm not sure where I went wrong. I know this is a lot, but I don't know where else to get help this late :/

2. Mar 3, 2016

### haruspex

How did you calculate the motor current in 4? I make it much higher.

3. Mar 3, 2016

### Marcin H

I just took the motor voltage and divided it by 51ohms. I=V/r

4. Mar 4, 2016

### haruspex

I thought the 51 Ohms was the resistor in series with the motor, not the motor itself.

5. Mar 4, 2016

### Marcin H

It is in series with the motor, but we weren't asked to measure the resistance of the motor itself. I'm not sure if there is a way to calculate it.

6. Mar 4, 2016

### haruspex

You cannot just take any any old resistance value you find lying around and divide it into some handy voltage to find a current. There has to be a certain physical relationship between the resistor and the voltage difference. What relationship?

7. Mar 4, 2016

### Marcin H

I'm not sure what the relationship is. Vs=Vm+Vp. Vs=Vm+IpRp. Vs-Vm=IpRp. Ip=(Vs-Vm)/Rp. Is that right? by KVL? Doin that I get 86mA

8. Mar 4, 2016

### haruspex

I said a physical relationship. V=IR where R is the resistance of a resistor in a circuit, V is the voltage (where?) and I is the current (where?)

9. Mar 4, 2016

### CWatters

+1

A motor doesn't really behave like a resistor so you can't directly calculate the current using the motor voltage. The trick is to realise that the motor and resistor are in series so the same current goes through both. Work out the current in the resistor as haruspex suggests.

10. Mar 4, 2016

### Marcin H

So do I just use the battery voltage (6.186V) and the different resistor for each part 4,5,6. So motor current is the same as I=6.186V/51ohms=3-9.76? This doesn't look right. operating point should be in the middle somewhere i think

11. Mar 4, 2016

### Staff: Mentor

For each resistor you have a current value and a motor voltage value. Plot those (current,voltage) points. They are points on the motor's operating curve.

12. Mar 4, 2016

### CWatters

Unfortunately the OP calculated the motor currents incorrectly.

13. Mar 4, 2016

### CWatters

As I said earlier... The motor current is the same as the resistor current because they are in series.

You can calculate the resistor current using I = V/R where V is the voltage across the resistor (not the battery voltage).

This is what Haruspex was hinting at in #8.

14. Mar 5, 2016

### Marcin H

Yeah I fixed that, but it turns out I get my operating point by plotting my Vth and Ith IV curve over my Motor IV curve from part 1 of the lab. I thought I was supposed to have a seperate graph with 2 lines intersecting, but that was not the right idea. Anyway, this then gave me my operating point(where the thevenin EQ line intersects the turn ON voltage line). My professor said this was correct. I'm not sure what this tells me though. What does the operating point of a motor tell you? My initial thought was that it would tell you it's turn on voltage, but I don't think that right.
Picture of my graph:

15. Mar 5, 2016

### haruspex

I may have this all wrong, but as I understand it, given a specific source and a specific load, an operating point is a state (power level, say) at which the system will equilibrate.
In your experiments, you found the operating point for each of a set of circuits consisting of the battery, the motor, and some resistor placed in series. My guess, then, is that the operating point you are trying to determine is for the case when there is no extra resistance in series.

16. Mar 5, 2016

### Marcin H

What do you mean by equilibrate? The motor will reach a certain speed and not go any faster?

17. Mar 5, 2016

### haruspex

Yes.

18. Mar 5, 2016

### Marcin H

So once it reaches this state, the motor will run at that speed no matter how much current/voltage you pump into it? I thought the motor would keep going faster and faster until it just burns out or breaks.

19. Mar 5, 2016

### haruspex

No, I said the operating point is a characteristic of the whole circuit, not just the motor. If you vary anything, e.g. put in a higher voltage battery, or reduce the inline resistor, the operating point will change. As I understand it, what you really want to know is what the operating point will be with the given battery and given motor, but no extra inline resistance. But if you set that circuit up you will not know what current is going through the motor. So, instead, you experiment with some nonzero inline resistors. In each case, you can determine the current through the motor. From these data you can extrapolate to the case you are interested in.

But please bear in mind I am just guessing based on the information you have provided and what I have found online. I have no prior experience of the topic.

20. Mar 5, 2016

### Marcin H

Hmm, that makes sense. Thank you for the explanation and help!

21. Mar 6, 2016

### CWatters

Normally you are faced with a source and a load that you want to connect together and predict how the combination will behave. One way to do this is to plot two lines..

1) Look at the source and calculate the open circuit voltage and short circuit current. That gives you two points you can join with a straight line.

2) Look at the load and plot voltage vs current for the load.

Where these two lines cross is the operating point for the combination.