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[itex] u(t) = \begin{Bmatrix}

1, & t \geq 0 \\ 0, & t<0 \end{Bmatrix}[/itex]

and lets have a simple circuit. Solo capacitor, connected to a DC voltage U0, a switch S exists.

For purposes of this problem, I can mark the voltage across the capacitor as Vc(t)

Vc(t)=u(t)*U0

Current through the capacitor is

[itex] i(t)=C \frac{du_{c}(t)}{dt}=CU_{0} \frac{du(t)}{dt}=CU_{0}\delta(t)[/itex]

As we know that the derivative of the Heaviside step function is Dirac Impulse function.

From this we can conclude that there is a initial spike in current when you turn the switch on.

Does this mean, that at first, very first moment you turn the switch on, capacitor acts as short circuit?

Second part:

When I saw graphs of currents of capacitors in RC circuits connected to a DC voltage, I saw that the graph starts at V/R and falls of with exponential law.

But how can current be immediately U/R, isn't there a transient state, due to RC self-inductance, to V/R, then exponential fall?

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# Capacitor Dirac impulse

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