Finding values of LRC circuit when frequency is given

Question

I am trying to solve the following question. The original question is:

The given impedances of the series LRC circuit are established at an angular frequency of 2000 rad/s.

Calculate the resonance angular frequency from the given circuit and calculate the current I through the inductor at resonance, given the voltage across the capacitor is 4j mV at resonance.

This is my approach:

First, I think that the given impedances are when the frequency is 2000 rad/s , so I tried to find their values when the frequency is 2000 rad/s. Is the approach correct?

Then I remember the formula for finding the critical frequency was as how I wrote it in part 2 of answer. Any ideas if this is correct?

Also, I have no idea how to find the current across L at resonance. What I remember is that in resonance the impedance of C and L are canceling eachother so the given voltage should be the voltage same across the resistor. Can you check please?

Answer 1

There have been a lot of comments but I think it's still valuable to sketch the solution:

The impedances of \$L\$ and \$C\$ are given at an angular frequency \$\omega = 2000\$ rad/s. This means that

$$\omega L = 6\Omega\text{ and } \frac{1}{\omega C}=8\Omega$$ which gives the following values for \$L\$ and \$C\$: $$L=3mH\quad C=62.5\mu F$$

If you don't know the formula for the resonance frequency by heart, it's very easy to derive it (if you know that the impedance must be purely real-valued at resonance):

$$Z = R + j\left (\omega L -\frac{1}{\omega C} \right)$$ The imaginary part of the impedance \$Z\$ disappears for \$\omega L = \frac{1}{\omega C}\$, i.e.

$$\omega_0 = \frac{1}{\sqrt{LC}} = 2309.4\text{ rad/s}$$

Since at resonance the impedance of the capacitor and the inductor are equal, the voltages across them must be equal. So the current through the inductor at resonance must be

$$\frac{4mV}{\omega_0 L} = 0.577\text{ mA}$$