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Calculate the absolute value of the complex charge in the RLC circuit:

$$Q(t)=\frac{V_0e^{i\omega t}}{-\omega^2L+i\omega R+\frac{1}{c}}.$$

Find the frequency where $|Q(t)|$ is maximum.

This question is from Mathematical methods, by Boas. I am not so sure what the question is asking?

If I take absolute value then I will have:

$$\left|Q(t)\right|=\left|\frac{V_0e^{i\omega t}}{-\omega^2L+i\omega R+\frac{1}{C}}\right|= \frac{\left|V_0e^{i\omega t}\right|}{\left|-\omega^2L+i\omega R+\frac{1}{C}\right|}=\frac{V_0}{\left|-\omega^2L+i\omega R+\frac{1}{C}\right|}.$$

Robben
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1 Answers1

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You're on the right track. First, evaluate the $|.|$ in the denominator to get an expression in terms of $R,L,C$, and $\omega$:

$$ \left| \left(\frac{1}{C}-L\omega^2\right)+i\left(R\omega \right)\right| = \sqrt{\left(\frac{1}{C}-L\omega^2\right)^2 + \left(R\omega\right)^2} $$

Now differentiate $|Q(t)|$ w.r.t. $\omega$, set it to $0$ and solve for $\omega$. This will be the frequency at which $|Q(t)|$ is maximum.

PS: If you want to be rigorous, differentiate $|Q(t)|$ again and use the second derivative to show that the solution is indeed a maximum, not a minimum or a saddle point.

MGA
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  • It's just a complex number in the form $x+iy$, and $|x+iy|$ is simply $\sqrt{x^2 + y^2}$. Where are you getting stuck exactly? – MGA Aug 28 '14 at 03:54
  • See if the edit helps? Now we're rid of the $i$ ... what remains is just 'brute force', if that's what was confusing you. – MGA Aug 28 '14 at 03:57
  • None, you're there. You can expand what's inside the $\sqrt{\cdot}$ into a quartic polynomial if you like to make the differentiation easier, but what you have now is the absolute value. – MGA Aug 28 '14 at 04:03
  • Oh, that is easy enough! Thanks for the help! – Robben Aug 28 '14 at 04:04
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    @Robben: One small trick is that maximizing $|Q(t)|$ is the same as maximizing its square, but the latter is easier to differentiate because it doesn't have the square root. – Semiclassical Aug 28 '14 at 04:10
  • @Semiclassical is right, but in any case you'll get rid of the square root anyway when you equate to $0$. – MGA Aug 28 '14 at 04:24
  • Sure, if I could figure out how to create a chat room :) – MGA Aug 28 '14 at 04:32
  • Hm, I do not know how to either, lol. But I will post my solution as a comment then. What I did was just differentiate $|Q(t)|$ by using the chain rule and setting it equal to $0$ and got that $$\sqrt{\frac{1}{LC}-\frac{R^2}{2L^2}}=\omega.$$ Is that correct? – Robben Aug 28 '14 at 04:38
  • @Robben Yup. Well done. Now show it's a maximum! – MGA Aug 28 '14 at 04:47