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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 fi...
Answer
#2: Post edited
by
Olin Lathrop
·
2024-09-26T11:30:10Z (about 2 months ago)
- 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 sine on the capacitor, assuming ideal components.
- 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.
#1: Initial revision
by
Olin Lathrop
·
2024-09-11T13:16:32Z (2 months ago)
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 sine on the capacitor, assuming ideal components.