Discussion in 'Homework Help' started by peter_morley, Apr 9, 2011.

1. ### peter_morley Thread Starter Member

Mar 12, 2011
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The problem is in attachments...I've tried this problem a couple of times but I'm not getting it right. I understand the imaginary part of I needs to be zero so I set the imaginary part to zero and solve. But I'm still getting the wrong answer. The answers for the variable capacitor are 0.04*10^-6F or 0.16*10^-6F. please help thank ya

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2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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Your approach is correct - it's just that there's obviously a problem in the math.

I would try to reduce everything to a simple symbolic formulation before substituting any actual values.

$Z=j\omega L+\frac{(\frac{1}{j\omega C})R}{R+\frac{1}{j\omega C}}$

which after some manipulation reduces to

$Z=\frac{1+\frac{j\omega L}{R(1-\omega^2LC)}}{1+j\omega RC}$

So one then equates the two complex parts in the numerator & denominator

$\frac{\omega L}{R(1-\omega^2 LC)}=\omega RC$

Finally giving after some re-arranging

$\omega^2R^2LC^2-R^2C+L=0$

which has the roots

$C=\frac{1}{2\omega^2L}\stackrel{+}{-} \frac{1}{2\omega}\sqrt{\frac{1}{\omega^2 L^2}-\frac{4}{R^2}}$

This will give you the two possible values of .16uF or .04uF

3. ### peter_morley Thread Starter Member

Mar 12, 2011
179
0
which after some manipulation reduces to

$image=http://forum.allaboutcircuits.com/mimetex.cgi?Z=%5Cfrac%7B1+%5Cfrac%7Bj%5Comega%20L%7D%7BR%281-%5Comega%5E2LC%29%7D%7D%7B1+j%5Comega%20RC%7D&hash=bf8acede6424dae78efe1c89514cd4bc$

So one then equates the two complex parts in the numerator & denominator

$image=http://forum.allaboutcircuits.com/mimetex.cgi?%5Cfrac%7B%5Comega%20L%7D%7BR%281-%5Comega%5E2%20LC%29%7D=%5Comega%20RC&hash=c45bb038eebc80d0d18ef7470b687c2c$

In this part do you just reduce to the second equation by not taking anything out or do you separate out the parts of the equation that have a j coefficient to solve for the imaginary part equaling 0. Can you do this if say the equation was x^2j+5xj +20x+5 and now reduce to this x^2j+5xj=0 and solve for whatever x gives a zero result? Ive been doing multiple problems and i'm always screwing it up some how.

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
784
I have actually mistakenly left out part of the equation for Z, when factoring out.

It should be

$Z=R(1-\omega^2 LC)[\frac{1+\frac{j\omega L}{R(1-\omega^2LC)}}{1+j\omega RC}]$

not just
$image=http://forum.allaboutcircuits.com/mimetex.cgi?Z=%5Cfrac%7B1+%5Cfrac%7Bj%5Comega%20L%7D%7BR%281-%5Comega%5E2LC%29%7D%7D%7B1+j%5Comega%20RC%7D&hash=bf8acede6424dae78efe1c89514cd4bc$

But the rest of the solution remains the same ...... still using the term in the square braces

Because I've manipulated the terms to have real part equal to unity in both the numerator & denominator, then it only remains to have the complex parts equal for there to be no phase shift in the output voltage. That's why I needed to only equate the terms in 'j'.

5. ### peter_morley Thread Starter Member

Mar 12, 2011
179
0
"Because I've manipulated the terms to have real part equal to unity in both the numerator & denominator"

I have no idea what real part equal to unity in both the numerator and denominator. Does this mean real = imaginary and then solve or real = imaginary numerator and real = imaginary denominator?

6. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
784
OK - I think I see what I've written might be confusing. I should be more careful with my terminology and distinguish real and imaginary parts of the complex numbers involved.

Looking at the term

$\frac{1+\frac{j\omega L}{R(1-\omega^2LC)}}{1+j\omega RC}$

The numerator is

$1+\frac{j\omega L}{R(1-\omega^2LC)$

The denominator is

$1+j\omega RC$

The numerator real part is 1.
The denominator real part is 1.

The numerator imaginary part is

$\frac{j\omega L}{R(1-\omega^2LC)$

The denominator imaginary part

$j\omega RC$

It is the imaginary parts that I have made equal for the condition of zero phase shift.

Why is this OK to do? In reality I have equated the two complex numbers for the numerator and denominator. But since the real parts (both equal to 1) immediately drop out I then just equate the imaginary parts as follows....

$1+\frac{j\omega L}{R(1-\omega^2LC)}=1+j\omega RC$

or (after subtracting 1 from both sides)

$\frac{j\omega L}{R(1-\omega^2LC)}=j\omega RC$

& (after dividing both sides by j)

$\frac{\omega L}{R(1-\omega^2LC)}=\omega RC$

Hope that makes more sense....