Hellllooo!

So, I just had a query wrt this quest:

I found out v(t) .

Now we know p(t) = v(t) * i(t)

I just had one query: HOW DO WE CONVERT p(t) into phasor domain? Basically if p(t) was equal to 1385 cos(377t+30) i could easily convert it into polar. But there is a 1200+ term before. .How do we account that when converting into polar form. Please help guys!!

 

Slide 2: http://hermes.eee.nott.ac.uk/teaching/cal/h51elc/elc0001.html

 

Hey. .yes, I am aware of the basics of how to convert a time domain to phasor. I will try to explain.
Let us assume that p(t)=1385cos(377t+30). . So, in phasor form it will be simply 1385/sqroot2 @ angle 30.
But what now there is a 1200+ term before this. How to take this into account is my question.
Thank you for your time

 

1200 is a offset
http://www.wolframalpha.com/input/?i=1200+++1385cos(377t+30)

 

Yep, 1200 is dc since it does not have any ac parts. So the ac part is going to oscillate around the dc part.

 

Ok.. I have tried one more method.. I have converted 1385/sqrt2 @ angle 30 into Rectangular Form. I got 848.133+j489.67. I added 1200 to this and got 2048.133+j489.67 & then converted this into polar: 2105 @ angle 13.45

is this correct?

 

guys, please confirm

 

guys, please confirm

Hi Lisa,
Something is peculiar about the question itself.
It is well known that the instantaneous power in an AC circuit can be described by an equation which is effectively what you have in the question - an AC component riding on an average value. The average value is what we regard as the average power.

However the AC component is normally twice the frequency of the system source frequency. So a 60 Hz system would have a 120 Hz component riding on the the DC or average power value.
In your question the system frequency appears to be 60 Hz and the AC component in p(t) is also stated as 60 Hz [377 radians/sec]. That's rather curious.

Irrespective of the latter issue one finds the instantaneous power as you show in your first post.

\(p(t)=v(t)\times i(t)\)

You have

\(v(t)=V_m sin(\omega t)\)

and

\(i(t)=I_m sin(\omega t + \theta)\)

hence

\(p(t)=v(t)\times i(t)=\frac{V_m I_m cos(\theta)}{2} - \frac{V_m I_m}{2}cos(2 \omega t+ \theta)\)

Also one can use the cosine forms which give a slightly different result which is perhaps more relevant to the problem statement

\(v(t)=V_m cos(\omega t)\)

and

\(i(t)=I_m cos(\omega t + \theta)\)

hence

\(p(t)= v(t)\times i(t)=\frac{V_m I_m cos(\theta)}{2} + \frac{V_m I_m}{2}cos(2 \omega t+ \theta)\)

Hopefully, given the above equation and the expression you were given for p(t) in the question and the known source voltage, you can take it from here ....

Notwithstanding my misgivings about the angular frequency discrepancy in the expression for p(t) shown in the question.