Yep -- thanks for pointing out the typo. I'll edit the post.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.For the avoidance of confusion: "changing the parallel capacitance" should read "changing the parallel resistance"
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.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?
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