You have to recognize the formula for an LC circuit and its resonance frequency.

[Jimbob999]

Homework Statement

The impedance of the circuit shown is:

A. 41.1 Ω

B. 100 Ω

C. 173 Ω

D. 187 Ω

E. 241 Ω

Homework Equations

Z= sqrt (R^2 + [Xinduct - Xcapac]^2)
Wd = 1/sqrt(L * C)
Xinduct = WdL
Xcapac = 1/WdC

The Attempt at a Solution

I think my issues come with the 2nd to 4th equations.
The answer I get is 100 ohms, which i get as wrong, I follow the below logic
Wd = 1/sqrt (0.0002*0.5)
Wd=100
Xinduct = 0.02
Xcapac = 0.02
thus Z = R which is 100 ohms.

Where have I gone wrong here?

 

[BvU]

Hi Jimbo,

You want to work with the complex impedance, because there is a phase involved.
So $Z_L = j\omega L$ and $Z_C = {1\over j\omega C}$.

Are you comfortable with that approach ?

 

[Jimbob999]

Hi Jimbo,

You want to work with the complex impedance, because there is a phase involved.
So $Z_L = j\omega L$ and $Z_C = {1\over j\omega C}$.

Are you comfortable with that approach ?

Is Jw here the same as the angular frequency wd?

The textbook chapter that this refers to says nothing about complex impedance as opposed to regular impedance, so I am not sure I get what you mean?

 

[BvU]

$\omega$ is the angular frequency allright, but not the LC resonance frequency $\omega_r$ (*)
It is the driving frequency in the circuit under consideration, so in this case $50 \times 2\pi = 100 \pi$ rad/s.


(*) your wd, for which $\ |Z_{LC}| = 0\$ from $$\ j\omega L + {1\over j\omega C} = j\omega L\; \left ( 1 + {1\over j^2\omega^2 LC }\right ) = j\omega L\; \left ( 1 - {1\over \omega^2 LC }\right )\ = 0 $$ if $\omega^2 LC = 1$, a frequency of $100/2\pi$ Hz.

 

[Jimbob999]

I have a feeling this picture tells me all I need?

w = 2pi *50
Xc = 15.92
XL = 157.08

Thus Z = 173, thus C.

Is that correct reasoning?

 

[BvU]

Yes. But now I am afraid you have no idea where the $X_L - X_C$ comes from ...