How Does Changing Frequency Affect Current in a Series LC Circuit?

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SUMMARY

The discussion focuses on the behavior of current in a series LC circuit as the frequency of an AC generator is altered. It establishes that the impedance (Z) of the circuit is defined by the equation Z = |ωL - 1/ωC|, where ω represents angular frequency, L is inductance, and C is capacitance. As frequency increases, if the inductive reactance (ωL) exceeds capacitive reactance (1/ωC), the impedance increases, leading to a decrease in current. Conversely, if capacitive reactance exceeds inductive reactance, the current increases. The correct answer to the posed question is that the current behavior cannot be definitively determined without additional information.

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Homework Statement


A capacitor and inductor are connected in series to an AC generator. As the generator’s frequency is increased, the current in the capacitor

A. increases

B: decreases

C: does not change

D: can't answer this question due to no enough information.


Homework Equations


The impedance in the LC circuit is Z = │ωL- 1/ωC│
Current I = V/Z


The Attempt at a Solution


If ωL > 1/ωC, then then ω increases, Z also increases, and current will decrease.
If ωL < 1/ωC, then then ω increases, Z decreases, and current will increase.
So the answer to this question is D.

Is the above answer right?

 
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E: goes up then goes down.
Not a fair question if the right answer is not included.

A series tune circuit becomes low impedance at resonance, so the current goes up when the frequency equals the resonant frequency. Above or below that frequency, the current is less than at resonance.
 

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