Peak current in series LC resonance circuit
In the following series LC circuit , maximum input current (IL1) reaches to 321mA in the simulation which is decided by V1/2 x π x f x L1 (f being resonant frequency).
why total impedance seen by V1 is not addition of inductor and capacitor impedances but only inductor impedance?
In series LC resonance both impedances (XL=XC) cancel each other then how come current does not reach infinity ?
2 answers
You are driving the series L-C with a voltage step. With theoretical ideal components, the impedance of the L-C is indeed 0 at the resonant frequency. However, you're not driving it at the resonant frequency. The step does contain a broad spectrum for a short time, which is why any current flows at all. In the steady state, a fixed voltage (0 Hz) results in 0 average current due to the series capacitor.
With real components that have real losses, a voltage step input would result in the sine current you see initially, but the amplitude of that current would exponentially decay towards 0 over time as a fraction of the energy is lost each cycle.
why current is limited by impedance of inductor only ?
It's not. If the total impedance was only from the inductor, then the current would linearly rise forever.
In the immediate time after the step input, the voltage on the capacitor is still zero. That means all the voltage is across the inductor. For that infinitely brief moment after the step, the current is the same as if only the inductor were there. However, current thru the inductor charges up the cap such that the cap voltage opposes the step input voltage.
As long as the voltage across the inductor is still positive, the current increases. When the cap voltage equals the step voltage, then the voltage across the inductor is zero, and the current stops rising. The current is still flowing, so the voltage on the cap builds up more. This now causes a negative voltage across the inductor, so the current decreases.
The current continues decreasing until it is zero, which happens at the point where the cap is charged to 2x the step voltage. The voltage on the inductor is -1x the step voltage, so the current keeps decreasing. The current continues until the voltage on the cap becomes zero, at which time the voltage on the inductor is +1x the step voltage.
With theoretical ideal components, energy keeps sloshing back and forth between the inductor and capacitor forever. All the voltages and currents are sinusoids.
In the real world, some fraction of the total energy is dissipated due to resistive ad other losses in the components. That causes the energy to decay towards zero exponentially over time.
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maximum input current (IL1) reaches to 321mA in the simulation which is decided by V1/2 x π x f x L1 (f being resonant frequency).
Given that both capacitor and inductor have equal magnitude impedance at resonance, you can also see that the same formula applies for the capacitive reactance. So, what is your point exactly? You found a formula that worked and stopped right there without delving more deeply? Maximum current also occurs when the capacitor voltage also equals the input step voltage (and inductor voltage is zero). All these things are related and a little bit more complicated than just simple observations.
why total impedance seen by V1 is not addition of inductor and capacitor impedances but only inductor impedance?
Impedance is a steady state AC parameter and it's much more mathematically complicated when circuits are driven with a transient voltage step change. You need to use Laplace transforms for example.
In series LC resonance both impedances (XL=XC) cancel each other then how come current does not reach infinity ?
Because your input voltage is a transient step change and not a continuous sinewave at the resonant frequency.
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