Understanding Capacitor and Inductor Behavior: A Physically-Based Explanation

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    Capacitor Inductors
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The voltage on a capacitor does not change immediately because it depends on the accumulation of electric charge on its plates, which takes time. The relationship is described by the equation I = C * dV/dt, indicating that if the time interval (dt) is infinitesimally small, the change in voltage (dV/dt) and thus the current (I) can become infinitely high. Similarly, for inductors, the voltage is related to the rate of change of current, expressed as V = L * dI/dt, which also implies that an instantaneous change in current would result in an infinite voltage. When a DC current is suddenly interrupted in an inductor, it can generate a spark as it attempts to maintain the current flow. Understanding these principles is crucial for grasping the behavior of capacitors and inductors in electrical circuits.
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Physically, can someone explain why does not the voltage on the capacitor change immediatly?
and why does not the current in the inductor also do?
:smile:
 
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The voltage on the capacitor is a function of the amount of electric charge on its two plates. It takes time, and work, for the charge to accumulate.
 
As for the capacitor:

I = C * dV/dt

So if dt is infinite small, dV/dt and thus I will become infinite high.

As for the inductor:

V = L * dI/dt

So if dt is infinite small, dI/dt and thus V will become infinite high.
Practically, if you feed an inductor by a DC-current and switch it off, the inductor will immediately produce a (long) spark, thereby keeping up the current.
 
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