Check this please? i(t)=C.(dv/dt), v(t)=A.sin(wt+@), find...

[markbrock30]

Hi,

Given

i(t) = C.(dv/dt)
v(t) = A.sin(wt + @)

To find an equation for i(t) I did the following:

v(t) = A.sin(wt).cos(@) + cos(wt).sin(@)

dv/dt = A.w.@sin(wt).cos(@) + w.@.cos(wt).sin(@)

= w@(A.sin(wt).cos(@) + cos(wt).sin(@))

Therefore:

i(t) = C.w.@(Asin(wt).cos(@) + cos(wt).sin(@))


Is this right?

Can anyone suggest a better solution?


Cheers,

Mark.

 

[royhaas]

Re: Can someone check this please?

The derivative of the sine is the cosine. Why expand?

 

[markbrock30]

Re: Can someone check this please?

I needed it in that form so I could prove the next question easier (prove I = j.w.C.V, where V = v(t) in polar form).

I'm not entirely sure i've done this right though - hence me posting it here for opinions .


Any suggestions are welcome.


Cheers,

Mark.