# Finding values of LRC circuit when frequency is given

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?

• 500uF is wrong for C. – Andy aka Apr 18 '13 at 12:31
• @Andyaka can you explain more? why? – Sean87 Apr 18 '13 at 12:32
• It's a homework question so I'm pointing out your first mistake and not solving the whole thing for you – Andy aka Apr 18 '13 at 12:35
• @Andyaka no its not homework! I am too old for that lol – Sean87 Apr 18 '13 at 12:36
• Given your starting point, you can simplify 2*pi*318Hz. That might make the error in C easier to spot. – Brian Drummond Apr 18 '13 at 13:32

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

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

• Thanks a lot...can you tell me if they asked for current for the capacitor, would it be 4mV/W0C ? – Sean87 Apr 18 '13 at 22:56
• It would be $4mV\cdot \omega_0C$ because the capacitor's impedance is $1/(\omega_0C)$ and current = voltage/impedance. – Matt L. Apr 19 '13 at 7:16