For the avoidance of confusion: "changing the parallel capacitance" should read "changing the parallel resistance"

Yep -- thanks for pointing out the typo. I'll edit the post.

 

WBahn thank you for the time constant formula you provided me with. I used it and in parallel (like not In a parallel circuit, but in parallel to what I was doing) in parallel to that I solved it again but this time using Thevenin's theorem. It was the same! . But I only need to clarify one last thing. And that is if I am to find how much voltage the capacitor will generate for 1time constant, where the formula for doing this is 0,63 x V, this Voltage V , is it the maximum voltage of the voltage source -10v or is it the maximum voltage that the capacitor will acquire - 6,6v?

 

I'm not WBahn and I hope he doesnt mind me jumping in, but it would be 0,63 of the thevenins voltage that you calculated (6.6)

Usefull stuff !

 

WBahn thank you for the time constant formula you provided me with. I used it and in parallel (like not In a parallel circuit, but in parallel to what I was doing) in parallel to that I solved it again but this time using Thevenin's theorem. It was the same! . But I only need to clarify one last thing. And that is if I am to find how much voltage the capacitor will generate for 1time constant, where the formula for doing this is 0,63 x V, this Voltage V , is it the maximum voltage of the voltage source -10v or is it the maximum voltage that the capacitor will acquire - 6,6v?

In the general case the capacitor will start with some voltage, Vini, and it will end with some voltage, Vfin. It's possible that neither of those are zero. The voltage difference, ΔVc = (Vfin-Vini) is how much the voltage across the capacitor will change by and it will change by 63.2% of ΔVc (toward Vfin from Vini) in one time constant.

In any one time constant's worth of time, the voltage will change from where it was at the beginning of the time period to 63.2% of the way toward it's final voltage. So at the end of one time constant it still has 36.8% of the way to do. After another time constant passes it will be 63.2% of this 36.7% of the way there, or an additional 23.3% of the way on top of the 63.2% it already was for a total of 86.5% of the way after two time constants.

 

noweare, I appreciate everyone's help in this.

 

I've made some calclations on time constant with zener and a transistor in parallel, id like to ask if they are true. Should I post the picture here or should I make another thread?. I mean I dunno, it could be considered spamming hah

 

I'd recommend a different thread. Posting multiple problems in the same thread causes lots of confusion as people respond to different problems without making it clear which one they are talking about (usually because they don't know that there's more than one).