# Solve Xp

Discussion in 'Math' started by KCHARROIS, Oct 21, 2012.

1. ### KCHARROIS Thread Starter Member

Jun 29, 2012
292
1
Hello,

I'm currently reading a book called rf circuit design by chris bowick and theres an example in the book to design a simple parallel circuit. The part that i dont understand is when they say substitute the equation:

Rp=(85)Xp in the equation: 10 = [Rp(500)/Rp +500] / Xp to solve Xp.

Thanks

Jun 29, 2012
292
1

3. ### KCHARROIS Thread Starter Member

Jun 29, 2012
292
1
Heres the page to the whole problem.

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4. ### WBahn Moderator

Mar 31, 2012
17,762
4,800
First off, be sure that you express things with the proper order of operations:

10 = [Rp(500)/Rp +500] / Xp

is the same as

10 = [(Rp(500)/Rp) +500] / Xp

Which is not what you want. You meant:

10 = [Rp(500)/(Rp +500)] / Xp

Second, track your units (even though the author is sloppy)

10 = [Rp(500Ω)/(Rp +500Ω)] / Xp

Now you just do what it says. Everywhere that you see Rp in the above expression, you replace it with (85)Xp. Once you have done that, you have a single equation with only one unknown, namely Xp, and so you can solve for Xp.

5. ### KCHARROIS Thread Starter Member

Jun 29, 2012
292
1
I did do that and my answer turns out to be 7.21 and that is not the right answer.

Last edited: Oct 22, 2012
6. ### KCHARROIS Thread Starter Member

Jun 29, 2012
292
1
I did get the answer but by cross multiplying both formulas.

7. ### THE_RB AAC Fanatic!

Feb 11, 2008
5,435
1,305
Xp = (Vista-1)
Vista = ****
therefore Xp = SHIS (in base 26)

Sorry, i was bored.

8. ### WBahn Moderator

Mar 31, 2012
17,762
4,800
Then please show us your work, step by step if possible, so that we can spot where you are going wrong.

9. ### WBahn Moderator

Mar 31, 2012
17,762
4,800
All valid methods should yield the same answer, so you need to figure out what you are doing wrong with the other way. It's probably just a simple algebra error, but it might be one that you have a habit of making, so if you don't find it now it will continue to haunt you.

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10. ### KCHARROIS Thread Starter Member

Jun 29, 2012
292
1
10 = [(85Xp * 500)/(85Xp + 500)]/Xp
10 = (42500Xp/585Xp)/Xp
Xp = 72.64/10
Xp = 7.264 Ohms

Voila

11. ### WBahn Moderator

Mar 31, 2012
17,762
4,800
Not so viola.

85Xp + 500 is NOT 585Xp. It is 85Xp + 500.

4a+3a is 7a because you can factor out the 'a' to get (4+3)a. This you can add the 4 and the 3. But if it is 4a+3b, you can't do that. You also can't do it if you have 4a+3.

This is a VERY fundamental misconception that will cause you no end of difficulty unless you get it under your belt.

You probably know that I always harp on units and this is an example where tracking units can catch this.

If 85Xp and 500 can be added, then they need to have the same units. In this case, the 500 has units of Ω, which should be carried throughout the work:

10 = [(85Xp * 500Ω)/(85Xp + 500Ω)]/Xp

Since Xp also have units of Ω, that means that the 85 is dimensionless, which is obvious in the above equation because there are no units attached to it, but this is only obvious if you consistently track units throughout your work. You therefore can't add 85 to 500Ω, because you can only add things that have the same units.

Now, while tracking units will catch a huge number of your errors, it can't catch them all. For instance, if the Xp had been dimensionless (which, of course, means that it would have represented something other than reactance), then the 85 would have been 85Ω, which would have been dimensionally consistent with the 500Ω, but you still couldn't add them for the fundamental algebra reasons given above. Bottom line, if the units work out, you have a good change, but not a guarantee, that the answer is right. If the units don't work out, you are guaranteed that the answer is wrong (baring coincidental collisions such as 2+2 and 2*2 yielding the same numerical result (although the units would be messed up in this case, telling you that you had done something wrong).

12. ### KCHARROIS Thread Starter Member

Jun 29, 2012
292
1
Ok so now,

10 = [(85Xp * 500Ω)/(85Xp + 500Ω)]/Xp
10 = [85(500Ω^2)/85(501Ω)]/Xp
Xp = (250000/501)/10
Xp = 49.9

I'm terrible...

13. ### WBahn Moderator

Mar 31, 2012
17,762
4,800
How are you getting from your first line to your second line?

I think you have some serious weaknesses in your algebra skills that you need to overcome. That's fine, as long as you recognize them and overcome them.

Remember that the core of algebraic manipulation is to do the same thing to both sides of an equation in order to maintain equality. Until you skills come up to speed, don't try to do too much in one step.

$
10 = \frac{\left( \frac{85Xp * 500\Omega}{85Xp + 500\Omega} \right) }{Xp}
$

Multiply both sides by Xp

$
10Xp = \frac{ \left( \frac{85Xp * 500\Omega}{85Xp + 500\Omega} \right) } {Xp}Xp
\
10Xp = \frac{85Xp * 500\Omega}{85Xp + 500\Omega}
$

Now multiply both sides by (85Xp+500Ω)

$
10Xp(85Xp+500\Omega) = \left( \frac{85Xp * 500\Omega}{85Xp + 500\Omega} \right) (85Xp+500\Omega)
\
10Xp(85Xp+500\Omega) =85Xp * 500\Omega
$

Now divide both sides by Xp

$
10(85Xp+500\Omega) =85 * 500\Omega
$

Distribute the 10 on the left hand side and perform the multiplication on the right

$
850Xp+5000\Omega = 42500\Omega
$

Subtract 5000Ω from each side

$
850Xp = 37500\Omega
$

Divide both sides by 850

$
Xp = \frac{37500\Omega}{850}
\
Xp = 44.1\Omega
$

You can verify that this is the right answer by simply plugging it back into the original equation.

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14. ### KCHARROIS Thread Starter Member

Jun 29, 2012
292
1
Wow...

I do have some work to do then, thanks again Wbahn. Its funny how I'm capable of going to a math class and do integration and then a question such as this causes me so much frustration.

Thanks

15. ### WBahn Moderator

Mar 31, 2012
17,762
4,800
Actually, not as strange as you might think. My high school calc teacher once said that he could teach calculus to a third grader and they would understand the concepts, but that the algebra would kill them. Now, clearly he was exaggerating for effect, but I have shown the basic ideas of both differential and integral calculus to numerous seventh and eighth grader and they did seem to catch them quite fully; but, to be sure, they lacked the algebra skills to really run with it.