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Image a simple RLC circuit as below:

enter image description here

The source Vs can supply an AC current 40A max., if L1 is absent, the current flows through C1 can achieve 40A. And with the adjustable inductor L1, I wonder if the current flows through C1 can beyond 40A (by adjusting the inductor)?

Maybe this is simple, but now I just can't figure it out!!!

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    \$\begingroup\$ I guess your question boils down to: can the current through C1 (=Ic measured by M2) be greater than the total current, I, supplied by the source and measured by M1? By KCL, I = Ic+ IL, but Ic and IL differ in phase by 180 degrees therefore Ic (and IL) can be significantly greater than I. In the extreme, if XL = Xc, the total current is zero, and the current through C (and L) is of magnitude V/Xc \$\endgroup\$ – Chu May 22 '15 at 9:44
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For a current excitation, with a parallel tuned LC circuit in resonance the current flowing in the capacitor is theoretically infinite for perfect components. The same is true of the inductor theoretically.

Reality isn't perfect and usually the inevitable dc resistance of the inductor ensures the current might be a few mA to several tens of amps.

What is true is this; for decent high Q components the current flowing in both L and C can be many, many times higher than the current supplied by the power source.

For a voltage source the current supplied to the LC circuit may be theoretically zero at resonance whilst the current in C or L is determined by the voltage supplied and the component reactance

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  • \$\begingroup\$ Do you need further calrification of this answer? \$\endgroup\$ – Andy aka Jun 23 '17 at 14:52

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