# Capacitor, Resistance and Inductor in series, help!

Discussion in 'Homework Help' started by ZeroTorrent, Apr 5, 2012.

1. ### ZeroTorrent Thread Starter New Member

Mar 28, 2012
10
0
As shown in the attachment I want to find an expression for U_out in terms of U_in and it's impedances and as an addition I need find the inductance L. I pretty much need someone to say where I go wrong in my calculation.

I set the node where U_out, C and R comes together and call it node e for now.
Also now I choose to mash up L and R to one impedance (to save time):

Z_LR = Z_L + R

Using currents going in and out of node e I get:

(U_in - e) / Z_LR = e/Z_C

U_in/Z_LR = e(1/Z_LR + 1/Z_C), where e = U_out.

This, however is wrong! I can use the voltage divider to get the correct result (I guess) but I refuse(!!!). The voltage divider formula has zero intuition connected to it when I try to use it, that is why I'll rather use the node method as it (should) get me what I want and understand what i'm doing at the same time.

Where am I doing wrong?

File size:
5.5 KB
Views:
12

Feb 17, 2009
3,991
1,115
3. ### ZeroTorrent Thread Starter New Member

Mar 28, 2012
10
0
hmm, i'd have thought they were wrong.

Okay, so assuming it's correct I would in the end get:

U_out = U_in(Z_L*Z_C*R)/((Z_L+R)*(Z_C*Z_L+R)) ?

Is this really correct? I'm asking because this is actually a part of a much larger question where I know U_out and U_in and the frequency need to find the L without knowing R. I tried solving for L but it's impossible without knowing R.

My teacher solves this using a transfer function that can be written as G(jω) = Uout/Uin = (1/jωC)/[(1/jωC) + R + jωL]. Frankly I don't know why or how he came to do that in order to solve this problem.

4. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,991
1,115

Check you maths, the correct answer look like this

Uin/Z_LR = e *(1/Z_LR + 1/Z_C)

e *(1/Z_LR + 1/Z_C) = Uin/Z_LR

e = ( Uin/Z_LR ) / (1/Z_LR + 1/Z_C)

e = Uin/ ( Z_LR * ( (1/Z_LR + 1/Z_C))

e = Uin/ ( Z_LR * ( (1/Z_LR + 1/Z_C))

e = Uin / ( (Z_LR * Z_C )/ Z_C )

e = Uin * Z_C /( Z_LR + Z_C)

Post full question

5. ### jtrent New Member

Mar 11, 2012
26
4
Looking at it as a voltage divider the way you said your instructor does using the formula G(jω) = Uout/Uin = (1/jωC)/[(1/jωC) + R + jωL], It would be helpful if we knew a phase angle associated with the Uout. Do you have that information. Without that information I don't see how you are going to solve this. Can you go ahead and give us all the pieces of the puzzle that you have. You said that you know Uout, Uin, C and the frequency. Tell us what these values are and everything else that you know. Then maybe we can be of help.