discharge slope logic (fixed sloppinees question)

Hello, I want to understand the math of the phenomena of capacitor as temporarary means of increasing the current supply.
First I have created a current limited voltage source which give me 45V and current limit of 4A at most.
Without the limitation i would have current and voltage on my 1.5Ohm load as shown below in the NO LIMITATION photo. But then i plug the limited current source (45V 4A) I see both the current and voltage go down very rapidly.
The slope is too high.
Suppuse i want that over my 2u pulse( ON I'll have only 0.5V drop and current also drop by 0.1A.
I=CdV/dt =10uF(0.5V)/(2usec)=2.5A
I dont understand this formula result what is the meaning of 2.5A in my pulse current plot?
I want my slope to be as small as possible.
LTspice file simulation is attached in the link Thanks. https://file.io/pyYz2mG0nCGa Image_alt_text

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discharge slope logic (fixed sloppinees question) [duplicate]

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