How to Convert Phasor Back to Time Domain for Evaluating Current at t=2s?

[dimsum1]

Homework Statement

Homework Equations

ω=2\pif

The Attempt at a Solution

After converting it from time domain to phasor domain
my answer is I=17.6\angle25mA.

How am I suppose to show the current at time = 2 sec? Do I just plug in 2 and find the sine of it?

 

[berkeman]

Homework Statement

Homework Equations

ω=2\pif

The Attempt at a Solution

After converting it from time domain to phasor domain
my answer is I=17.6\angle25A.

How am I suppose to show the current at time = 2 sec? Do I just plug in 2 and find the sine of it?

That's what I would do...

 

[dimsum1]

So the answer is 6.47mA? Or is it I=17.6∠25mA? I don't understand what the hint is referring to.

 

[berkeman]

So the answer is 6.47mA? Or is it I=17.6∠25mA? I don't understand what the hint is referring to.

That answer looks okay to me. The hint (more of a reminder) is just saying to always be careful to be consistent in the units of the argument to the sin() function. Since they show the 25 with a degrees sign on it, you need to check that the 250t term is also in degrees. By inspection it seems plausible that it is, since radian measurements will generally have a ∏ in them somewhere. If the function were 25sin(3∏t + 25^o), then you would need to convert one or the other into different units, and make sure your calculator was set to those units when calculating the answer.

Make sense?

 

[dimsum1]

That answer looks okay to me.

I typed two answers, 6.47mA and I=17.6∠25mA, which answer are you saying looks okay?

I thought the hint imply that what I was originally doing was wrong (where I originally arrived at I=17.6∠25mA ) because I had to do something with the 2 seconds which I did not do.

 

[berkeman]

I typed two answers, 6.47mA and I=17.6∠25mA, which answer are you saying looks okay?

I thought the hint imply that what I was originally doing was wrong (where I originally arrived at I=17.6∠25mA ) because I had to do something with the 2 seconds which I did not do.

To be honest, I haven't been understanding what you were trying to do with the phasor. Your 6.47mA answer is correct, I believe. 25sin(525^o) = 6.47

 

[dimsum1]

Hmm, okay, it just seems to easy to be a question, this question feels kind of ambiguous because I don't think that was what the professor wants. Thanks for the input.

 

[aralbrec]

okay? I typed two answers, 6.47mA

I don't like how that question is worded. Just be sure that 250t is in degrees and not radians.

I thought the hint imply that what I was originally doing was wrong (where I originally arrived at I=17.6∠25mA ) because I had to do something with the 2 seconds which I did not do.

The phasor has nothing to do with this question.

A phasor is another representation of a sinusoid where the time-varying part (e^jwt) has been factored out. It's as if you wrote down the sin equation in another form and it won't tell you what value the waveform has at a specific time (t=2) unless you convert it back to sinusoidal form, plug t=2 into it and take the real part.

However, to be clear, you should mark the 25 in the angle part as 25 degrees because without units it could be understood to be in radians.