Power in voltage/current sources, passive sign conventionby caesius Tags: convention, passive, power, sign, sources, voltage or current 

#1
Mar409, 02:40 AM

P: 26

1. The problem statement, all variables and given/known data
Determine which of the five sources in Fig. (attached) are being charged (absorbing positive power), and show that the algebraic sum of the five absorbed power values is zero. I've labelled how I'm referring to the sources in red 2. Relevant equations p = vi, passive sign convention 3. The attempt at a solution source one: p = (2v)(2A) = 4W generated (not PSC) source two: p = (8V)(2A) = 16W generated (not PSC) source three: p = (2V)(4A) = 8W generated (not PSC) > 8W absorbed source four: p = (10V)(5A) = 50W generated (not PSC) source five: p = (10V)(3A) = 30W generated (not PSC) > 30W absorbed  So I know I'm wrong because the question wants to know which ONE is absorbing power. And the sums don't add up in the second part. I'm clueless, been looking at this problem for a long time now, what have I missed? 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution 



#2
Mar409, 04:13 AM

P: 220

Edit: caesius, ignore my posts. mplayer seems to have gotten it right...




#3
Mar409, 05:37 AM

P: 154

You have the wrong voltage for Source 3.
Source 5 seems fine to me unless I'm missing something. The rest are correct, just correct source 3 and then: (power generated) = (power absorbed) (power generated + power absorbed) = 0 W If the problem is asking for just one power absorbing element, then I'm not sure what to tell you...I'm getting 2 elements absorbing, and the numbers add up to zero. Hopefully that can help, though. 



#4
Mar409, 05:43 AM

P: 220

Power in voltage/current sources, passive sign convention
Maybe, I should shut my mouth now. My method seems to be wrong...




#5
Mar409, 05:56 AM

P: 154

Source 2 seems OK by me. Positive current flowing from the negative terminal to positive terminal on the voltage source. That generating power P = (2A)(8V) = 16W.
I don't think there's a problem with Source 5, it looks like the current and voltage parameters are explicitly defined there. You may be seeing a problem I'm not though, what do you think? 



#6
Mar409, 05:59 AM

P: 154

All currents flowing into that top node are summing to zero so that's good:
2A  4A + 5A  3A = 0A I don't think there are any other effects on Source 5. 



#7
Mar409, 06:09 AM

P: 154

P = (4A)(10V) = 40W = 40W absorbed Source 5 is a current source of (3A) moving from negative to positive terminals through the element. This satisfies the passive sign convention. The voltage across that device is given, 10V. P = (3A)(10V) = 30W = 30W absorbed Did that help? 



#8
Mar409, 06:14 AM

P: 220





#9
Mar409, 06:19 AM

P: 154

Source 3 is in paralell with 10 V Source 4. Source 5 has a 10V across the current souce becasue it was labeled that way. 



#10
Mar409, 06:31 AM

P: 220

I guess the same methods don't apply with supplies and resistors. 



#11
Mar409, 07:16 AM

Mentor
P: 11,986





#12
Mar409, 08:06 AM

P: 220

Okay, just to let you know, I had a total brain freeze there. Obviously you can't use the superposition method here, since all the elements are supplies. You just need to make sure each loop has [tex] \sum V=0[/tex].




#13
Mar409, 12:58 PM

P: 26





#14
Mar409, 01:10 PM

P: 154

For example: In Source 3's case, my assumption was the source was generating power. The convention for power generation is current flow from  to + terminal. I calculated power by P = (4A)(10V) = 40W. The answer came out negative, so my assumption of power generation was incorrect. Therefore, the source is absorbing. I hope that's clear and am not confusing you more. 



#15
Mar409, 01:17 PM

P: 26

Thanks, that makes sense. Another question though, looking at source three, it looks to me like there are two voltages over it, 2V from the very left hand side and 10V from the voltage source. Why do you pick 10V over the 2V?
As far as I understood everything in parallel has the same voltage drop, how can there be two DIFFERENT voltage drops I guess is what I'm asking... Thanks again 



#16
Mar409, 03:43 PM

P: 154

Source 3 is not in parallel with Source 1. But, notice that Source 3 is in parallel with the combination of Source 1 and Source 2. If Source 1 and Source 2 were combined, it would have a potential difference of 10 V, just like Source 3 and Source 4. Does that make more sense?



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