0
\$\begingroup\$

I have an asignment where I am supposed to express a current:

$$12sin(400t+100^o)$$

as a phasor. I am not shure what that means but I guess that I should express it on the form $$A\angle\theta$$ where $$Acos(\omega t+\theta)$$

How do I do that?

I tried, but looking at the solution, it seems to be that I should divide 12 by $\sqrt{2}$ and I don't know why.

Here is my solution which is wrong:

$$12sin(400t+100^o)=12cos(90^o-400t-100^o)=12cos(-400t-10^o) \Rightarrow 12\angle 10^O$$

I am also curious to know why $\omega$ seems to not affect the answer...

\$\endgroup\$
0
\$\begingroup\$

12 is the maximum (peak) voltage - you need to convert to the RMS value and that's why you need to divide by 2^0.5.

The omega * t part is the frequency of the wave. You are only really considering one cycle for the phase angle (since an angular change of 2*pi brings us back to 0 again)

|improve this answer|||||
\$\endgroup\$
0
\$\begingroup\$

Your answer is correct. The amplitude is 12 and the phase is 10°. Why "the solution" gives the RMS value rather than just the amplitude, whoever wrote the solution must answer (unless the assignment asks for it).

The angular frequency ω doesn't affect the answer simply because it stands right beside t, and t is zero at the time the phasor refers to.

You could visualize phase as longitude. 0° is defined to be in London, and then the longitude of any other place can be given, without mentioning the rate of rotation of the earth. There is nothing really special about either London or t = 0, but as points of reference they are convenient enough.

|improve this answer|||||
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.