If a square wave voltage waveform is applied across a pure capacitor or inductor of a given capacitance/inductance, how will the resulting current waveform look? How should I calculate this using the theoretical formulas of Ic and IL?
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\$\begingroup\$ Please flag this as homework. What are your initial thoughts about this? How does a real-world capacitor/inductor behave and why? What is the difference between a real-world and idealised device in each case and how do you think that will change the behaviour as it tends towards an idealised device? \$\endgroup\$– mhaselupCommented Oct 15, 2020 at 21:16
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\$\begingroup\$ RLC series and parallel responses are well documented for various Q’s. Falstad sim shows the best \$\endgroup\$– D.A.S.Commented Oct 15, 2020 at 21:24
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1 Answer
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Obvious homework.
I'll talk about the capacitor only. "In theory" the capacitor will charge "instantly" and you'll have an infinitesimally narrow yet infinitely tall pulse of current. This is because there's no "R" in the circuit (as you have defined "ideal" components"). A capacitor charging time is often called the "RC Time constant".
https://en.wikipedia.org/wiki/RC_time_constant
It's given by the equation: t=RC
Because "R" is zero, and "C" is non-zero, then "t" must also be zero i.e. it takes no time to charge the cap.
In the real world, there would be some resistance in your power supply ("output impedance") that will serve to limit the current even if you have no discrete "R" in your circuit. To actually calculate anything, you need to know that "R" (or just tell your professor it happens "instantly" as described above)
Inductors are the exact opposite, but I leave that to you to think about. The current in an LR circuit can be found on this page: https://www.electronics-tutorials.ws/inductor/lr-circuits.html
What happens when "R=0" in that equation??
