Simple AC circuit with L and parasitic R

[anthonuc]

Homework Statement

[PLAIN]http://sphotos.ak.fbcdn.net/hphotos-ak-snc4/hs894.snc4/72731_10150110340044992_712009991_7710053_709779_n.jpg

The circuit has a non-ideal inductor with parasitic R. v₀(t) goes to coswt as the limit of w goes to zero.

The output signal has a +15 degree phase shift at 50kHz.

Determine L

Homework Equations

arctan(z1)-arctan(z2)=arctan[(z1-z2)/(1+z1z2)]

V₀(t)= X_a * |H| * cos(wt+angle H)

H = a+bj
angle H = arctan(b/a)

The Attempt at a Solution

I change the impedances to 50 for the resistor and jwL for the inductor. Then using voltage division I get Vo = Vi(jwL/(jwL+50)) where (jwL/(jwL+50)) = H

However when trying to get the angle of H i get arctan(wL/0)-arctan(wL/50) and the arctan identity formula doesn't seem to help as the first value just goes to pi/2. Can I set the angle to 15 degrees and use w = 2pi*f where f =50khz to solve for L?

Trying this I get L = 5.9e-4 Henrys. Does this seem right? Feels low.

I don't know what to make of the first hint either with the limit of w going to zero.

Thanks if you read/help, much appreciated.

 

[Phrak]

I don't know what all those equations are for. Impedances in series add. Z_total = Z_R + Z_L.

Use the complex equivalent of V=IR: V=IZ, or V(t)=I(t)Z_total

V(t) = 2cos(wt)

Now you can find the current through the series circuit using V(t)=I(t)Z_total. This is also the current through the inductor.

Next, apply V=IZ again to find the voltage across the inductor. This time V=V_inductor and Z=iwL. V_inductor will have a different phase from the voltage source. This phase difference is what you are after.

 

[anthonuc]

Hmm. Well in class we were taught to use sinusoidal steady states to solve these problems which is where the relevant equations came from.

Using your method I get I=I_L=(jwL+50)/(2costwt)

Then solving for V_L in V=IZ I get

V_L = (-w²L²+50jwL) / 2coswt

However I'm not sure what to do with this. This has a different phase from the voltage source?

 

[vk6kro]

[PLAIN]http://dl.dropbox.com/u/4222062/L-R.PNG

You are asked to work out the inductance, so you can work it out from the above diagram.

 

[anthonuc]

Thanks for the help but I've managed to solve it correctly.

You use sinusoidal steady state to find the parasitic R in the inductor. The R attached in the circuit is not the parasitic R. It's just a regular resistor. There's also some unknown parasitic resistance in the non-ideal inductor.

Parasitic R = R_p

You use Vo = 2coswt*(jwl + R_p) / (jwl + R_p + 50).

Set this equation equal to coswt as w goes to zero and you'll get R_p = 50.

Knowing this value you can plug it back into this equation:

Vo = 2coswt*(jwl + R_p) / (jwl + R_p + 50)

and use H = (jwl + R_p) / (jwl + R_p + 50)

the angle of H = 15 degrees when f = 50Khz where w=2pi*f

the angle of H = arctan(50/(100000*pi*L) - arctan(100/100000*pi*L) = 15 degrees

Then you can use the identity

arctan(z1)-arctan(z2)=arctan[(z1-z2)/(1+z1z2)]

and get a solvable equation for L. You get a quadratic polynomial but it should be easy enough to solve in the end.

I ended up with L = 85mH

Actually I still have to find out what the correct answer will be since my quadratic actually gives me negative roots.

Hope this can be help for future references at least! I appreciate your guys' help.