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KH6OWL
01-04-2009, 03:08 AM
Element 5 is hurting my head.


There are 8 polar coordinate questions and I don't understand them at all. Can anyone help me understand how to come up with the answer?

Thanks


E5C01
In polar coordinates, what is the impedance of a network consisting of a 100-ohm-reactance inductor in series with a 100-ohm resistor
The answer is: 141 ohms at an angle of 45 degrees

NA0AA
01-04-2009, 03:15 AM
This is the trangle question - in this question, note that the answer is the square root of two...So it's a right triangle of 45-90-45 degrees.

I don't have a lot of hints but to study the idea - it's got to do with a vector diagram - one number is out, one up or down, the answer is the hypontenuse and the included angle.

Honestly, I gun-decked these questions cause I did not understand 'em either. I need to re-read the book on that section.

K7UF
01-04-2009, 03:33 AM
These are easy points.
They can all be approached the same way.


E5C01 (B)
In polar coordinates, what is the impedance of a network consisting of a 100-ohm-reactance inductor in series with a 100-ohm resistor?
A. 121 ohms at an angle of 35 degrees
B. 141 ohms at an angle of 45 degrees
C. 161 ohms at an angle of 55 degrees
D. 181 ohms at an angle of 65 degrees

Think about this graphically.
We are going to plot a point on a graph.

Resistance goes along the X axis. Its always positive.
Resistance in this case is 100 ohms.

Next we have the reactive component. Its an inductor. It has a reactance of 100 ohms (positive reactance).

So we go up from the 100 X position, to the 100 position on the Y axis.

If X=Y the line from the graph origin to our point is going to be 45 degrees.

Which of the answers has an angle of 45 degrees?

That one was just too easy. Lets look at a harder one:


E5B13 (D) was E5D11
What is the phase angle between the voltage across and the current through a series RLC circuit if XC is 250 ohms, R is 1 kilohm, and XL is 500 ohms?
A. 81.47 degrees with the voltage lagging the current
B. 81.47 degrees with the voltage leading the current
C. 14.04 degrees with the voltage lagging the current
D. 14.04 degrees with the voltage leading the current

Again, lets start by plotting the resistance on the X axis at 1,000 (1k ohm).

We have XL = 500 and XC = 250.
We know that these are opposites, so we can subtract the smaller from the larger, and we are left with 250. Since the larger was L, the 250 we are left with is going to be inductive (L).

Now, we will be plotting a point at x=1000, y=250.

If x=y the angle is 45 degrees. In this case y is less than x, so we know that the angle has to be less than 45 degrees.

That rules out answers A and B because the angle is too great.
Now, we decided that the reactance was inductive (since we had more inductance than capacitance).

What do we know about current and voltage in inductive circuits?

Think you might know whether its C or D now?

AB3EQ
01-04-2009, 03:57 AM
... There are 8 polar coordinate questions and I don't understand them at all. Can anyone help me understand how to come up with the answer?

I agree that one of the best ways to get a good understanding of this is to focus on the graphics that derive from the equations. But I also strongly suggest that you read something other than the license manuals on the subject. The ARRL Handbook has some useful material on this, as will any decent basic to intermediate Electronics and RF theory book. You can even find some useful articles on the subject on Wikipedia.

One thing is certain, however: you're not the first to get headaches from imaginary numbers and polar coordinates!

Good luck,

-Danny.

AI6DX
01-04-2009, 04:25 AM
Yup, study study study your brains out with those polar coordinates. All to just enjoy the hobbie at the "top" class (extra) of license.

...and after the Extra exam, you'll never stink'n see or NEED to know anything about polar coordinates again. :D

I know, there's purists out there that probably claim to have a NEED to use polar coordinate formulas day in and day out... blah blah blah.....whatever. :p

(I'm probably gonna get "fried" now for the statements made above.) :rolleyes::rolleyes:

KA0GKT
01-04-2009, 06:02 AM
As for remembering leading and lagging, the mneonic ELI the ICE man can be a big help {E is ahead of I in an L (inductor) I is ahead of E in a C (Capacitor)}

You can also use simple trig to calculate the angles. Most simple scientific calculators have Sin, Cosin and Tangent functions along with ArcSin, ArcCos and ArcTan. It is a good idea to actually learn how to key the math into your TI-30.

The tangent (TAN) of the angle is equal to the Opposite (in this case Reactance) over the Adjacent (In this case, the resistance), so you divide the reactance by the resistance then find the ArcTangent (ARCTAN) of that product which equals the angle in question.

Since this is a right angle, if you need to find the impedance of this circuit, Pythagoren's Theorem comes into play (X squared plus Y squared equals Z squared. Let X = the reactance and Y= the resistance. Z will equal the impedance.

It is indeed possible that you will never need to know about polar coordinates again. Many Amateurs never bother to ever build a single thing preferring to throw money (or slugs from a .30-06 :D) at a problem, but if a person has any interest in how and why things work the way they do, knowing the math rather than memorizing the answers to a multiple guess question is a good idea. When it gets down to brass tacks, nobody stink'n needs to know anything...perhaps, in the end that is better for some, squelch any curiosity and be blissfully ignorant.

KA7O
01-04-2009, 06:12 AM
Yup, study study study your brains out with those polar coordinates. All to just enjoy the hobbie at the "top" class (extra) of license.

...and after the Extra exam, you'll never stink'n see or NEED to know anything about polar coordinates again. :D

I know, there's purists out there that probably claim to have a NEED to use polar coordinate formulas day in and day out... blah blah blah.....whatever. :p

(I'm probably gonna get "fried" now for the statements made above.) :rolleyes::rolleyes:

Ya know - if ya substitute "Morse Code" for "polar Coordinates" in what you said there, that's the same argument many made not so long ago.....

::running, ducking and joining you under the table to hide from the flames. Hey, I brought chips!::

KA0GKT
01-04-2009, 06:18 AM
Ya know - if ya substitute "Morse Code" for "polar Coordinates" in what you said there, that's the same argument many made not so long ago.....

::running, ducking and joining you under the table to hide from the flames::

No argument there. I am beginning to believe the gloom and doomers who predicted that the next target would be to dumb down the written test.

OBTW, I would think that a little extra heat would be welcome in central Wyoming this time of year :D

KA8RAW
01-04-2009, 06:43 AM
These were giving me headaches too, I have finally just about memorized the formulas and to help, I draw a graph. Then all of a sudden, they toss this one in>
In polar coordinates, what is the impedance of a circuit of 100 -j100 ohms impedance?
() 141 ohms at an angle of -45 degrees
() 100 ohms at an angle of 45 degrees
() 141 ohms at an angle of 45 degrees
() 100 ohms at an angle of -45 degrees

Now, I am lost with this one!


EDIT, Ok I have it figured out now, those people that made these test questions, really know how to throw in a curve ball now and then.

K7UF
01-04-2009, 04:49 PM
These were giving me headaches too, I have finally just about memorized the formulas and to help, I draw a graph. Then all of a sudden, they toss this one in>
In polar coordinates, what is the impedance of a circuit of 100 -j100 ohms impedance?
() 141 ohms at an angle of -45 degrees
() 100 ohms at an angle of 45 degrees
() 141 ohms at an angle of 45 degrees
() 100 ohms at an angle of -45 degrees

Now, I am lost with this one!


EDIT, Ok I have it figured out now, those people that made these test questions, really know how to throw in a curve ball now and then.

+j = Inductance
-j = Capacitance

This is actually easier, since you are told that the reactance goes above or below the zero axis.

We know its below, we know x=y so its -45 degrees, we know that the hypotenuse of a triangle is going to be longer than the other two sides.

K7UF
01-04-2009, 06:30 PM
Lets try to make this easy, because really, it is.
These questions should take you no longer than 30 seconds to answer without calculator or even drawing paper.

Its all based around this:

http://vogon.net/plot.png

This is a plot for Resistance equal to 400 ohms and reactance equal to 400 ohms. The reactance is inductive (+j).

When R = X the result is 45 degrees. It is also larger in magnitude and either R or X.

If X is less than R the angle decreases below 45 degrees.
If X is larger than R the angle increases above 45 degrees.

It should be obvious from the basic geometry that you did at school, or just by looking, that the result always has to be larger than either X or R.

When we are dealing with capacitive reactance (-j), the reactance line goes negative (downwards) and so does the result line. The same principles apply, R=X angle is -45 degrees, R>X angle is less than -45 degrees, R<X angle is greater than -45 degrees.

Just think of this graph and map the values you are given onto it, and the angle and its sign should be obvious (at least if its more or less than 45 degrees).

When we have both capacitive (-j) and inductive (+j) components, just subtract the smaller from the larger, and the type left is the same as the larger value.

e.g. If we have inductive reactance of 100 ohms and capacitive inductance of 50 ohms, we subtract 50 from 100 and are left with 50 ohms of inductive reactance -- that is what we plot on the graph.

So on the above graph, if we have the original R=400 ohms and now have an inductive reactance of 50 ohms, the vertical component (inductive reactance) is going to be shorter ... so the angle will be less than 45 degrees.

That 45 degrees is magic for these questions. They are all set so that the answers are more or less than 45 degrees and all the wrong answers are always the opposite (or opposite sign).

KA8RAW
01-04-2009, 08:50 PM
K7UF, Thanks for that graph. The way I was drawing them is where zero inductance and resistance was in the lower right hand corner. Your example makes it easier.
One of the problems I originally had was the following formula: X = XL - XC

If my study material had said that X = XL - XC or X = XC - XL and explained that it can't go negative, I would have caught on earlier than I did.

I had assumed that you always subtracted XC from XL, even if it would result in a negative number.

After working with these, and making some sample problems of my own, polar coordinates will no longer be any more trouble for me.

K7UF
01-04-2009, 09:29 PM
Glad it clicked for you.

These questions can all be solved by inspection and a bit of thought.
No need for the calculator.

In the real world, you don't get multiple choice answers, so you will need a calculator, but the calculations are all basic trigonometry, the stuff you had difficulty staying awake for in school ;)

K0CMH
01-04-2009, 09:59 PM
Now that you understand polar coordinates, I will tell you how I answered the questions on the test.

I knew the math that would calculate the numbers, however, that was to much agravation for me.

I would use the scratch paper they gave me for the test and draw my own XY graph by hand. Then I would plot the points, like K7UF did. I would just eyball the angle and look for the answer with the angle that looked correct.

As an example, the angle would look to be about 30 degrees. I would find and answer that had 32 degrees. All the other answers were way off.

Sometimes there would be two answers with the same angle. Then you can estimate the length of the line making the angle and look for that in the answer. Say the line looked about 7 units long. There was an answer with 32 degrees and 15, and an answer with 32 degrees and 6.5. You know the one I chose.

I didn't feel I was "cheating", since I knew the math to calculate the length of the one side of the trinagle, but I just didn't want to take the time to run all those calculations, when the "eye" did the job.

K7UF
01-04-2009, 10:24 PM
Now that you understand polar coordinates, I will tell you how I answered the questions on the test.

I knew the math that would calculate the numbers, however, that was to much agravation for me.

I would use the scratch paper they gave me for the test and draw my own XY graph by hand. Then I would plot the points, like K7UF did. I would just eyball the angle and look for the answer with the angle that looked correct.

As an example, the angle would look to be about 30 degrees. I would find and answer that had 32 degrees. All the other answers were way off.

Sometimes there would be two answers with the same angle. Then you can estimate the length of the line making the angle and look for that in the answer. Say the line looked about 7 units long. There was an answer with 32 degrees and 15, and an answer with 32 degrees and 6.5. You know the one I chose.

I didn't feel I was "cheating", since I knew the math to calculate the length of the one side of the trinagle, but I just didn't want to take the time to run all those calculations, when the "eye" did the job.

You know, I don't consider this cheating at all.
To do this, you have to understand the basic principles you are working with. Learning a formula to blindly use with your calculator is much more "cheating" IMHO.

K8ERV
01-05-2009, 12:10 AM
I'll bet Eric is an expert on Polar coordinates---

TOM K8ERV Montrose Colo

KD0DKI
01-05-2009, 07:00 PM
I didn't sleep in Trig class but I did sit there with the open mouth stupid look on my face most of the time. I'm looking at general class test in March, but it will be a while before I even look at Extra. I think I'll take a trig class before the Extra exam just so I can relive feeling stupid.

Glad I work with software applications that do business math :)

I must have been in the Wizard of OZ as the scare crow.

AE1PT
01-06-2009, 12:41 AM
Let's make it even easier. Memorize what you can--there will only be one or two of these questions. And if you miss every one it will not cause you to fail the exam unless everything else is a mystery as well...

Fuggedaboudit and move on!

K7UF
01-06-2009, 02:24 AM
Let's make it even easier. Memorize what you can--there will only be one or two of these questions. And if you miss every one it will not cause you to fail the exam unless everything else is a mystery as well...

Fuggedaboudit and move on!

Its actually easier to understand and answer these -- save the memory cells for the licensing questions which you either know or don't and if you don't here is no way to work out the answer.

Once you get the hang of the triangle stuff these are just 10 second, free points.

AC0FP
01-06-2009, 03:54 AM
Element 5 is hurting my head.


There are 8 polar coordinate questions and I don't understand them at all. Can anyone help me understand how to come up with the answer?

Thanks


E5C01
In polar coordinates, what is the impedance of a network consisting of a 100-ohm-reactance inductor in series with a 100-ohm resistor
The answer is: 141 ohms at an angle of 45 degrees

I probably got the polar coordinate questions right on my test because they asked the simple 45 degree ones! ;) I am used to using rectangular coordinates but I know the formulas to convert rectangular to polar and I took my calculator! :eek:

The questions I missed, were probably the band allocations for certain classes! :p I figured if I ever got a radio I could just get a chart to tell me where I could operate, after all you don't have to score 100% :cool:

73,

Frank :)

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