# EERM 27-13 Example 27.2



## jdd18vm (Jul 28, 2007)

A Resistor 10 Ohms in // with an Inductor of 4 Ohms.

Can someone explain how to arrive at 3.714 Ohms? And /or the Calculator operations on the Casio fx 115MS

John


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## Dark Knight (Jul 28, 2007)

jdd18vm said:


> A Resistor 10 Ohms in // with an Inductor of 4 Ohms.
> Can someone explain how to arrive at 3.714 Ohms? And /or the Calculator operations on the Casio fx 115MS
> 
> John


Stay here. I am going for my stuff upstairs.


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## Dark Knight (Jul 28, 2007)

Don't let the suckers confuse you.

What they did, since they are using complex numbers, use the inverse of R. That is S or conductance, if I am not wrong. S in parallel are added as Rs in series. It is just a convinience. They are giving you the impedance of the inductor so don't fall in the trap of trying yo convert it again using j(L)omega.

You can add the impedances in parallel but it is going to be messy. Take my word for that. I did it and did not like it. Changing to S and then back to R is the best way.

Once you have the value of S in terms of a real and imaginary components just invert that to get magnitude and angle.

Now let us go to the Casio. Mine has a function where you can select how you want your results no matter the way you enter the numbers. Press "Shifht" and then "Set Up". Press the "Down Arrow" at the wheel. Select 3 and then select "1" or "2".

As far as I know my 115-ES and your 115-MS are pretty similar in that aspect. Let me know if that works.


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## jdd18vm (Jul 28, 2007)

Thanks Luis,

I think I understand what they are doing, I just cant get the get the 3.714 Ohms. I keep getting 1.379

Doesnt appear i have a "setup" key, would it be something else. I glanced through the instructions, they didnt mention a setup feature?

John


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## mudpuppy (Jul 28, 2007)

There is also a shortcut for two impedances in parallel:

Zeq = (Z1*Z2)/(Z1+Z2)

This is equivalent to (1/Z1+1/Z2)-1 or (Y1+Y2)-1

The MS may be a little different to set up than the ES. First you should make sure you are in complex mode by pressing Mode then 2. The screen will say CMPLX at the top when you're in complex mode. To set the complex display mode on mine, you press Mode six times, press 1, then scroll to the right once and select either (1) rectangular (a+bi) or (2) polar (r/_theta). I use rectangular because I got confused with polar when I entered 1 - 2 and got the result 1 (at 180 degrees) when I wasn't expecting to have to deal with complex numbers.

To get the result to this example (using the first method I mentioned above), type in (10 X 4i) [divide] (10 + 4i) [equals]. If you are in polar mode it will output 3.714 on the screen. If you are in rectangular mode you will have to convert it to polar by pressing [shift] [plus] [equals]. You can get the imaginary part or the angle by pressing [shift] [equals]. Note that when in complex mode, you can get i by simply pressing the [ENG] key (no shift required).

You could also enter the impedances in polar form (10*4/_90)/(10+4/_90).

Clear as mud?


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## benbo (Jul 28, 2007)

Do you understand how they get the -j0.25 + 0.10 Siemens ?

Basically by rationalizing the first fraction (mult top and bottom by -j4)

I don't know about the calculator, but just get the magnitude of the admittance, and take the reciprocal =

|Z| = 1 / |Y|

|Y| = squareroot of ((.25^2) + (.1^2)) = squareroot of (.0625+ .01) = squareroot of (.0725) = .269

|Z| = 1 / .269 = 3.714

Now, if you get the phase of the admittance it will be the opposite (negative) of the phase for the impedance because they are reciprocals.

I don't know an easy way to do it with a calculator, but there are calculators that can do all sorts of complex math, I just didn't need one because I did the eCC module, not the power.


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## mudpuppy (Jul 28, 2007)

Whoops, error in my post above; I will correct it.

1.379 is the real (resistive) part of the answer. You need to put it in polar form to get the magnitude of 3.714.


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## Dark Knight (Jul 28, 2007)

Try this link my friend. Did not look too deep into it but might help you.

Casio FX-115-MS

Sorry I can't help more but do not have the 115-MS. I wonder why this motherflowers at NCEES banned my 115-ES.


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## mudpuppy (Jul 28, 2007)

Anyone else feel like we're descending on these technical questions like hungry vultures? 

My respect Benbo! I never learned the efficient methods for converting rectangular &lt;--&gt; polar by hand. I doubt I could have finished the power module in four hours without my calculator--all the square roots and and tangents and dividing by j tend to trip me up. Not that it can't be done; I'm just not adept enough to do it correctly and quickly every time.

And John, just a word of unsolicited advice; make sure you spend a lot of time practicing with your Casio before the exam. It works a lot differently than my TI-85 and it took me a while to get comfortable with it--but now that I've learned to use it, I'm quite happy with it. If you have any questions on how to use it, I'm sure many of us here would be happy to share our tips.


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## Dark Knight (Jul 28, 2007)

Luis said:


> As far as I know my 115-ES and your 115-MS are pretty similar in that aspect. Let me know if that works.


I was really wrong. They are not alike in the way they handle complex numbers operations.


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## benbo (Jul 28, 2007)

mudpuppy said:


> Anyone else feel like we're descending on these technical questions like hungry vultures?


Yes, I sort of miss working problems and studying. Of course it's a lot easier and more fun when the pressure's off. And i also always mess up the complex arithmetic. In this case I knew the answer so I wasn't going to post unless I could get it. At first, I even forgot about the complex part entirely. That's why I didn't take power. It was my worst class in school.


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## jdd18vm (Jul 28, 2007)

Thanks for all your help guys. And I hope you are truly Hungry vultures, I'm sure I can keep you all well fed until Oct.

Mud, that way helps for the Calculator stuff, and Benbo thanks for the reasoning. I had done the longhand before but forgotten it once someone told me how to convert it with the Calculator.

to get it to work on the calc input SQRT ((.25)^2 + (.1)^2) X‾¹ (invert)

I think I actually got it. :thankyou:

Luis, I came across that site in a search a while back thanks for reminding me its very helpful.

John


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## Art (Jul 28, 2007)

Z1 = 10

Z2 = 4j

simple impedence in parallel

Z = Z1 Z2/(Z1 + Z2)

Z = (10 x 4 ang 90)/(10 + 4j)

10 + 4j = 10.77 ang 21.8

Z = 40/10.77 ang (90-21.8) = 3.714 ang 68.2


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## Dark Knight (Jul 29, 2007)

Try this in the Casio:

(10)(4i)/(10+4i)

That should do the trick.

Just have to find the set up to get the results in polar form


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## jdd18vm (Jul 29, 2007)

Worked like a charm Luis. Thanks Art. I think I got it.

John


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## grover (Jul 30, 2007)

mudpuppy said:


> Anyone else feel like we're descending on these technical questions like hungry vultures?


What, looking up cable sizes in the NEC isn't the kind of "engineering" you busted your ass in college for 4+ years to do? :sniff:


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