# First Order Circuits

1. Jul 3, 2008

### asi123

Ok, lets say I have a first order circuit with i_s = Acos(wt+phase) (like in the pic).
I'm having problem with the private solution, lets say I pick my private solution like this:
i_p = Bcos(wt + phase) = Ccos(wt) + Dsin(wt) (right?)
After I place this i_p in the equation I come up with this:
f[C,D]*cos(wt) + g[C,D]*sin(wt) = Acos(wt+phase) (right?)
Can I say
f[C,D]*cos(wt) + g[C,D]*sin(wt) = Acos(wt)?
I mean, How do I come up with C and D?

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2. Jul 3, 2008

### Defennder

I'm afraid I'm not understanding your question. What is i_p, f[C,D} and g[C,D]?

3. Jul 4, 2008

### CEL

You must develop $$Acos(\omega t + \phi) = A_1cos(\omega t) + A_2sin(\omega t)$$

Now, $$f[C,D] = A_1$$ and $$g[C.D] = A_2$$