What Happens When ω²LC=1 in an RLC Circuit?

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When ω²LC=1 in an RLC circuit, the inductive and capacitive reactances become equal and opposite, resulting in a net reactance of zero. This condition indicates resonance, allowing for maximum current oscillation at the resonant frequency without the need for additional voltage. The impedance at this point can appear as negative, which suggests a theoretical scenario that doesn't account for real-world losses. In practical terms, while the circuit can support large currents, it cannot operate losslessly due to inherent resistances. Understanding this resonance condition is crucial for analyzing RLC circuit behavior.
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Homework Statement


Consider a circuit with a capacitor C in series an inductor of inductance L. Explain what happens when ω2LC=1, without calculations, using your knowledge and intuition.

Homework Equations


1/Z = 1/impedance = 1/(iωL) +1/(iωC)
i is pure imaginary.

The Attempt at a Solution


Well, unfortunately I have no intuition. I don't know what the quantity ω2LC represents, but it appears in the equation when you rearrange to find Z. So when I worked out Z, it was -1/(iωC+iωL) when the set condition is met. Which is a bit weird, because -ve resistance makes no sense...?
I've obviously gone wrong somewhere!
 
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Yes, a capacitor has impedance $$1\over j\omega C$$
Look here for some ideas
 
So when ω2LC=1 then the inductive and capacitive reactances are equal. The current is oscillating at the resonant frequency of the circuit?
 
That's right. Nothing wrong with your intuition.
Reactances are equal and opposite. So together they are 'zero'. ##\bf V = Z I## (bold face to mark them as complex): there can be huge currents shooting back and forth with 'no' voltage needed to whip them up. In reality you can't have a lossless system, so characteristics of an RLC circuit sneak in.
 

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