/sci/ - Science & Math

[Anonymous No.4765620]

/sci/, I need you to explain the following:
<span class="math">U(t)=U_0\cdot \sin(\omega t)[/spoiler]
<span class="math">U=Q/C[/spoiler]
<span class="math">U(t)=\frac{Q}{C}\cdot \sin(\omega t)[/spoiler]

How come, that this is what comes out?
It looks like the current was the derivative of the voltage:
<span class="math">\frac{dQ}{dt}=I=\frac{Q}{C}\cdot\omega\cdot \cos(\omega t)[/spoiler]

[Anonymous No.4765632]

>>4765620
I'm trying to figure out the current of a circuit with a capacitor and AC.
If you didn't know.

[Anonymous No.4765633]

>>4765620
There was a very basic differential equation for U(t). Solving it (with the proper inital conditions) gives the shit you have.

What part is troubling you?

[Anonymous No.4765649]

>>4765633
can you do it step by step?

[Anonymous No.4765674]

>>4765649
it troubles me, that if U_0 is Q/C,
<span class="math">U(t)=/frac{Q}{C}/cdot /sin (/omega t)[/spoiler]
Taking the derivative takes you to
<span class="math">dU/dt=\frac{Q}{C}\cdot\omega\cdot \cos(\omega t)[/spoiler]
which would be equal to the correct current. That doesn't make sense.

[Anonymous No.4765714]

I looked it up and found, that this holds (pic).
Why is so?

[Anonymous No.4765723]

>>4765674
1) Solve your diff eq for Q(t) to get Q(t) = f(U)

2) Solve Q(t) = f(U) for U to get U(t)

It is very very very very simple.

Are you a fucking retard (engineer)?