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What occurs when $$\omega = \omega_r$$ The circuit frequency equals its natural frequency? So what. And why is it important?

JYelton
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40pro
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  • At resonance, the reactance, or the "imaginary" part of the impedance, is zero. For capacitors and inductors, which are ideally lossless, this means you have zero reactive power loss on these components when operating in the resonant frequency. – Shabab Oct 20 '13 at 21:31

3 Answers3

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A series RLC circuit has an impedance of: -

\$R + j\omega L +\dfrac{1}{j\omega C}\$

Knowing that "j" in the denominator is the same as "-j" in the numerator means there is a frequency when the inductive and capacitive reactances totally cancel and this means that the only component left is resistance, R.

This occurs when \$j\omega L = -\dfrac{1}{j\omega C}\$

Or \$\omega^2 = \dfrac{1}{LC}\$ or \$\omega = \dfrac{1}{\sqrt{LC}}\$

So what?

It's a really big deal when wanting to filter just one small band of frequencies whilst disregarding the rest; in simple terms, at resonance, a series RLC circuit lets through a small band of frequencies and progressively attenuates frequencies that are not in the "pass-band".

And why is it important?

It's a big deal for radio receivers and transmitters so it's really very, very important.

If you analyse a parallel RLC circuit it has the same formulas and, depending on the circuit configuration, it can do pretty much the same as a series RLC circuit. The "R" in a parallel circuit is a different way of expressing losses and it functions inversely to the "R" in a series RLC circuit.

Resonance can do bad things too - if not controlled it can produce voltages that cause semiconductor devices to breakdown - voltage regulators is a device that springs to mind.

Have you seen the video of that suspension bridge rocking backwards and forwards and then collapsing - that was uncontrolled resonance and mechanically it has virtually the same formula - it is a second order filter.

Andy aka
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    Re the Tocoma Narrows bridge failure: there's some debate as to whether the famous bridge collapse is due to uncontrolled resonance. If I understand the counter-arguments correctly, the wind added energy to the system and the "poles moved into the RHP" so the bridge became an oscillator: http://www.ketchum.org/billah/Billah-Scanlan.pdf‎ – Alfred Centauri Oct 20 '13 at 22:02
  • @AlfredCentauri - that's the one; couldn't think of its name. One side issue - is that why you get small, strange mechanisms attached to the wires on a lot of suspension bridges - are these to do with damping resonance or is it my imagination and the beer? – Andy aka Oct 20 '13 at 22:05
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First, a preface.

When a sinusoidal source drives a pure resistive load, the source delivers power to the resistance, the resistance never delivers power to the source.

However, for a pure reactive load, the source alternately delivers and then receives power back from the load, the reactive load stores energy from the source half of the time and then "gives it back" during the other half.

When the load is an RLC circuit, at most frequencies, the load is part resistive and part reactive (either inductive or capacitive depending on whether one is driving with a frequency that is above or below the resonance frequency respectively).

So, to answer your question: "What happens to RLC circuits at resonance frequency, conceptually"

When the RLC circuit is driven at the resonance frequency, it appears as a pure resistance so that the load never returns power back to the source, all the power is delivered to the load.

This is important, especially in RF circuits, and in other areas too.

Alfred Centauri
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Conceptualy, you get minimum impedanance and therefore maximum current

privera
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