what is the power dissipated in 2 ohm resistor when input current is

i(t)=2+((2√2)cos(2t+45°))+2sin2t

please help?

 

This is the Homework Help, not Homeword Done For You forum. So you need to show whatever work you have done in your best attempt to date to solve the problem. If nothing else, describe what you think the approach needs to involve, even if you don't see how to bring it all together just yet.

 

calculated rms value of current and multiplied with 2 ohm.

p=(√6^2)*2= 12w
but answer is wrong!!!

 

Okay, so let's take a step back.

How did you come up with that as the RMS current?

 

first simplified above equation to 2+2cos2t
by adding rms value of each term, I got √6

 

You summed real power vs imaginary power directly.

That method is incorrect, which is why Power Factor correction is important.

 

You summed real power vs imaginary power directly.

That method is incorrect, which is why Power Factor correction is important.

thanks for your reply,
then how can I proceed?

 

first simplified above equation to 2+2cos2t
by adding rms value of each term, I got √6

Is the correct answer 20W?

 

Is the correct answer 20W?

correct answer is 3W ,

 

thanks for your reply,
then how can I proceed?

Do you know what imaginary/reactive power is?

Is that topic something your course has covered?

 

Do you know what imaginary/reactive power is?

Is that topic something your course has covered?

yes , I know it.
till now I did not get expected help, so please close the thread.

 

Could you do us, and others with a question similar to yours, a favor and post your work to get the correct answer?

 

correct answer is 3W ,

Yeah, I came up with that doing it in my head and almost as soon as I posted that I realized that I had screwed it up because one of the cross terms doesn't go away. I get 12W, the same as you. I got this by performing the square on the three terms directly.

I've confirmed that you combined the two trig functions properly and I, too, get that the current is

i(t) = 2A+2A*cos(2t)

I even plotted it in the time domain in Excel and found the average power numerically and got 6W.

So I think that your answer is correct. I don't know how they are getting 3W.

Actually, I think I do. If this were the voltage across the resistor, and not the current through it, then you would have 3W.

 

You summed real power vs imaginary power directly.

That method is incorrect, which is why Power Factor correction is important.

I'm not following how you came to the conclusion that he did this.

 

I'm not following how you came to the conclusion that he did this.

Math error on my part.

 

Math error on my part.

It seems to be going around.

The OP's the only one that seems to have their math act together on this one!

 

Letting Mathematica do the hard work, I get:

]

 

Yep. I think the author just screwed up and divided the square of his function by 2 instead of multiplying it by 2. The reason the mistake didn't get caught is because the author can't be bothered to track their units, so they forfeited the ability to spot the problem.