Convert V: 10< 90 Degrees + 66 - j10V at 10k Rads/s

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csmith23
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Question: Convert V = 10< 90 degrees + 66 - j(10 V) at angular frequency = 10k rads/s.

I am stuck here 10(cos(90)+ j(sin(90)) + 66 - j(10)

which would then be: 0 + j + 66 - j(10)
 
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Convert to what?

csmith23 said:
Question: Convert V

Do you have the original problem wording?
Convert to what?
1) V(t)=V(0)sin(wt+p) where t=time and V(0), w, p are real?
2) V(t)=V(0)Cos(wt+p) where t=time and V(0), w, p are real?
3) Other?
 
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csmith23 said:
V(a)cos(\omegat+\phi)
V(t)=V(0)cos($\omega$t+$\phi$)
You are very close,
what is the angle represented by j + 66 - j(10)? That is $\phi$.

Can you find $\omega$ from the given frequency?
Can you find V(0); it is the magnitude of j + 66 - j(10)?
 
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actually I am already given \omega, that is what angular frequency is. Although just re reading my initial post, I can spot my problem. I made an algebraic error:

10(cos(90)+ j(sin(90)) + 66 - j(10)

corrected: 10(0) + j(10) + 66 - j(10)

which just simplifies to 66, while the imaginary cancel out

Final answer: 66cos(10^4t)​

Thanks for your help!