Impedance of a circuit

1. Aug 28, 2015

Jimbob999

1. The problem statement, all variables and given/known data
The impedance of the circuit shown is:

A. 41.1 Ω

B. 100 Ω

C. 173 Ω

D. 187 Ω

E. 241 Ω

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

3. 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?

2. Aug 28, 2015

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 ?

3. Aug 28, 2015

Jimbob999

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?

4. Aug 28, 2015

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.

5. Aug 28, 2015

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?

6. Aug 28, 2015

BvU

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

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