2
\$\begingroup\$

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:

enter image description here

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?

\$\endgroup\$
  • 1
    \$\begingroup\$ 500uF is wrong for C. \$\endgroup\$ – Andy aka Apr 18 '13 at 12:31
  • \$\begingroup\$ @Andyaka can you explain more? why? \$\endgroup\$ – Sean87 Apr 18 '13 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 Apr 18 '13 at 12:35
  • \$\begingroup\$ @Andyaka no its not homework! I am too old for that lol \$\endgroup\$ – Sean87 Apr 18 '13 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\$ – Brian Drummond Apr 18 '13 at 13:32
2
\$\begingroup\$

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

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

\$\endgroup\$
  • \$\begingroup\$ Thanks a lot...can you tell me if they asked for current for the capacitor, would it be 4mV/W0C ? \$\endgroup\$ – Sean87 Apr 18 '13 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. Apr 19 '13 at 7:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.