Because the derivative of ln(f(x))=f'(x)/f(x), so in this video's case since the integration is with respect to vc and we're trying to integrate it using the previously mentioned rule it means that the derivative of the denominator(With respect to vc) will be on the numerator so 1/(vs-vc) will become -ln(vs-vc) (The ln doesn't have the absolute value brackets because the input will always be positive)
@@ElectronXLab wow, thank you!! What a surprise to get a special video in response to a question. I'm coming back to calculus after a very long gap due to a recently acquired fascination with the maths behind radio signals. It's a bit of a struggle, but videos like this really help 🙂
The formula for capacitance is: C = Q/V, let's rearrange it to Q = CV. We now know what electric potential we need to apply to get our desired amount of charge on a capacitor of known capacitance. Let's say we go incrementally and ask, what is the amount of charge we get for a small increase in applied voltage -> dQ = C*dV We know, that what a circuit does is that it moves these charges en-masse over time, so we divide both sides of the equation by the small change in time it takes our "river" to move it: dQ/dt =C*dV/dt Hmm, incremental movement of charge over time, where have I seen that, hmmm... :D Yes so we just replace dQ/dt with current to get the final formula: I = C*dV/dt (feel free to correct me if I got something wrong, I am just a student and any improvements will be helpful)
Was hard to find people who cared to explain how we arrived at this formula, thank you for doing that, sir.
I don't quite understand why a definite integral is used here. Could someone explain? 3:09
Immensely helpful, thank you.
thank you, helped a lot!
Respect to you sir 🎉
Hi. I don't understand why there is a negative sign at 3:14.
because of the vc in the denomiator as well
Because the derivative of ln(f(x))=f'(x)/f(x), so in this video's case since the integration is with respect to vc and we're trying to integrate it using the previously mentioned rule it means that the derivative of the denominator(With respect to vc) will be on the numerator so 1/(vs-vc) will become -ln(vs-vc) (The ln doesn't have the absolute value brackets because the input will always be positive)
thank you very much sir
Thats well explained
Thank you. Would it be possible to show the integration step in a future video?
@Iwbnwif Yiw You're welcome! I put together a quick video showing the integration step. You can see it here: th-cam.com/video/AU9XJkahd0U/w-d-xo.html
@@ElectronXLab wow, thank you!! What a surprise to get a special video in response to a question. I'm coming back to calculus after a very long gap due to a recently acquired fascination with the maths behind radio signals. It's a bit of a struggle, but videos like this really help 🙂
thank you
Respect sir
skipped a whole section of the workings at 3:12 of the video. didnt show us how to integrate
you should show why i=c dv/dt after all you are teaching .
The formula for capacitance is: C = Q/V, let's rearrange it to Q = CV. We now know what electric potential we need to apply to get our desired amount of charge on a capacitor of known capacitance. Let's say we go incrementally and ask, what is the amount of charge we get for a small increase in applied voltage -> dQ = C*dV
We know, that what a circuit does is that it moves these charges en-masse over time, so we divide both sides of the equation by the small change in time it takes our "river" to move it: dQ/dt =C*dV/dt
Hmm, incremental movement of charge over time, where have I seen that, hmmm... :D
Yes so we just replace dQ/dt with current to get the final formula: I = C*dV/dt
(feel free to correct me if I got something wrong, I am just a student and any improvements will be helpful)