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:

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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?

  • 1
    \$\begingroup\$ 500uF is wrong for C. \$\endgroup\$
    – Andy aka
    Commented Apr 18, 2013 at 12:31
  • \$\begingroup\$ @Andyaka can you explain more? why? \$\endgroup\$
    – Dumbo
    Commented Apr 18, 2013 at 12:32
  • \$\begingroup\$ It's a homework question so I'm pointing out your first mistake and not solving the whole thing for you \$\endgroup\$
    – Andy aka
    Commented Apr 18, 2013 at 12:35
  • \$\begingroup\$ @Andyaka no its not homework! I am too old for that lol \$\endgroup\$
    – Dumbo
    Commented Apr 18, 2013 at 12:36
  • \$\begingroup\$ Given your starting point, you can simplify 2*pi*318Hz. That might make the error in C easier to spot. \$\endgroup\$
    – user16324
    Commented Apr 18, 2013 at 13:32

1 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

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

  • \$\begingroup\$ Thanks a lot...can you tell me if they asked for current for the capacitor, would it be 4mV/W0C ? \$\endgroup\$
    – Dumbo
    Commented Apr 18, 2013 at 22:56
  • 1
    \$\begingroup\$ It would be \$4mV\cdot \omega_0C\$ because the capacitor's impedance is \$1/(\omega_0C)\$ and current = voltage/impedance. \$\endgroup\$
    – Matt L.
    Commented Apr 19, 2013 at 7:16

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