
#1
Jan813, 10:48 AM

P: 273

Once again, my textbook seems to do a poor job of explaining another concept just like I commented earlier here. Anyways, here is the relevant part of the textbook: http://www.mediafire.com/view/?cw5t2wr9vqqprnv
To calculate p_{4} in example 1.7, voltage of 8V is multiplied by a current of 0.2I. Firstly, why is the current negative? Secondly, why do they substitute a value of 5A for I. Where did they get this value? I tried to do practice problem 1.7, but I don't seem to get the right answer for p_{3} when I use a voltage of 3V and a current of 0.6(3)A. 



#2
Jan813, 10:59 AM

P: 390

Because the dependent current source is supplying the power. So current flows from its positive to its negative. But when you calculate power of a branch you take the current value (which is to be multiplied) same as what will be entering the branch from positive.
They defined the dependent power that way. What ever will be the I (which is going through p1 brance) will be multiplied by 0.2 and that will be the current coming out of dependent source. 



#3
Jan913, 03:54 PM

P: 273





#4
Jan1013, 03:06 AM

P: 5,462

Electric Circuit Analysis: Calculating power absorbed/supplied by each element
Well it says on your diagram I = 5 amps.
Incidentally, I cannot acces your link on my ofice computer, so others may well have the same problem. It is better to post the doc or image here at PF, and also to properly reference the book as others may have it. You will also likely receive more attention that way. 



#5
Jan1013, 05:46 AM

P: 273





#6
Jan1013, 08:41 AM

P: 390

The value of the second current source is not constant, but changes when the current through p2 branch changes (and the proportionality constant is 5). 



#7
Jan1313, 08:55 PM

P: 273




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