Solve RLC Circuit Problem - Small & Large Frequencies

AI Thread Summary
When analyzing an RLC circuit, at very low frequencies, the inductor behaves like a short circuit while the capacitor acts like an open circuit, resulting in most current flowing through the inductor. Conversely, at very high frequencies, the inductor behaves like an open circuit and the capacitor like a short circuit, causing most current to flow through the capacitor. The resonant frequency is determined by the formula wo=1/sqrt(LC). Understanding these behaviors allows for accurate predictions of current flow in the circuit. This knowledge is crucial for solving RLC circuit problems effectively.
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RLC Circuit problem. Please Help!

Homework Statement



In the figure, what is the current supplied by the emf when the frequency is very small?

What is the current supplied by the emf when the frequency is very large?
knight_Figure_35_52.jpg

Homework Equations

I am not sure but:

wo=1/sqrt(LC)

Vr=IR

Vl=IXl

Vc=IXc
 
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When the frequency is very small. XL = 2*pi*f*L is nearly zero and XC = 1/2*pi*f*C is near open circuit, and most of the current will flow through inductor branch
When the frequency is very high. XL = 2*pi*f*L very high and XC = 1/2*pi*f*C is near zero and most of the current will flow through capacitor branch.
 
For high enough frequencies you can replace the inductor with an open circuit, and replace the capacitor with a short circuit. For low enough frequencies you can replace the inductor with a short circuit and the capacitor with an open circuit.
 
Thank You!

I am pretty sure I will be able to get it now.
 

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