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Analysis of LC circuit using intuition

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This problem investigates the behavior of the voltage across the capacitor in response to a step input: -

Image_alt_text

I am asked to find the voltage Vc(t) without the use of differential equations or simulation. This is difficult but here is my attempt.

If the capacitor is much smaller than the inductors (say L1 = L2 = 1 mH and C1 = 1 pF) then C1's impedance is much higher than the inductors', and is effectively an open circuit. Hence, all the current through L1 equals the current through L2. If L1 and L2 have the same value then Vc becomes equal to 1/2 Vin.

If the components are of the same size (L1 = L2 = 1 mH and C1 = 1 mF) then it gets harder. Let's say 1 A passes through L1 and 0.8 A goes through L2 and 0.2 A goes through C1. Both L2 and C1 get energized but by currents of different magnitude. I suppose L2 and C1 seek to arrive at some kind of energy equilibrium which would involve the voltage Vc having decaying oscillations. I'm not sure about this explanation. Can someone help me out?

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2 answers

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A quick intuitive way of looking at this circuit is that the voltage source, L1, and L2 can be thought of as a lower voltage source with a single lower inductance in series. This is the same as finding the Thevenin equivalent if L1 and L2 were just resistors. Put another way, L1 and L2 are in parallel AC-wise.

Now you have a simple L-C circuit. When a voltage step is applied, the result is an infinite-length sine on the capacitor, assuming ideal components.

Another wrinkle is that the DC current thru the two inductors increases indefinitely, which obviously can't happen with real world components.

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Not infinite (2 comments)
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without the use of differential equations or simulation

Then use Laplace transforms to derive the transfer function. Multiply it by 1/s to get the Laplace result when a step is applied then, use inverse Laplace tables (and/or partial fractions) to derive the transient response in the time domain.

These methods (as are all analyses like this) based on differential equations but, you would not be explicitly using them.

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