
#37
Jul811, 10:22 AM

PF Gold
P: 2,551

P2 = 2 x 5.211^2 = 54.31 [W] Pt = Power dissipated in resistor 1 +Power dissipated in resistor 2 + Power dissipated in resistor 3? In which case, I was aware of that! And I did get the correct result, yes? 



#38
Jul811, 10:36 AM

HW Helper
P: 6,189

We got a total power of 258.65 W. Subtract 199.46 W from your previous result. Subtract another 54.31 W that you just got. That leaves.... 4.88 W for the 4 ohm resistor between A and B. Yes! That's it! You got it! 



#39
Jul811, 10:48 AM

PF Gold
P: 2,551

w00000000000000000t
You rock! I'll probably do the other one tomorrow^^ You're incredible, ILS!!! This is pretty fun for me, I don't know about you You're a life saver. I'm pretty stressed about term B in electronics, it's my last chance at "redemption" for this course. 



#41
Jul811, 03:44 PM

Sci Advisor
HW Helper
P: 1,327

You did a nice job !! (!עבודה טובה)
There are typically several approaches to the same question. Some take fewer steps than others. Here is the method I was suggesting you try. (It may help save you some time on an exam) You found currents for the initial question to be I_{1}=1.105, I_{2}=5.211A and I_{3}= I_{0} = 6.316A Next was find V_{AB} and P_{AB} (with respect to R_{1}). V= IR ; V_{AB} = I_{1}R_{1} = (1.105)(4) = 4.42V P = IV; P_{AB} = I_{1}V_{AB} = (1.105)(4.42) = 4.88W 



#42
Jul811, 08:27 PM

PF Gold
P: 2,551

Duly noted Quabache
Next time I will certainly be wiser I will try me hands on another exercise this morning (it's 4:30 AM here. I'm an early riser!) 



#43
Jul911, 12:52 AM

PF Gold
P: 2,551




#44
Jul911, 01:17 AM

HW Helper
P: 6,189

Hmm, it seems you've been progressing.....
But wait! You did not get all the smurfs! Which smurf did you miss? 



#45
Jul911, 01:19 AM

PF Gold
P: 2,551

Yea I know finding the P in each of them...but com'on you just apply a formula it's soooooooooooooo easy! Lemme skip that pretty please?




#46
Jul911, 01:22 AM

PF Gold
P: 2,551

How come they get to pick for me the direction of I? Shouldn't I be the one determining it? 



#47
Jul911, 01:24 AM

HW Helper
P: 6,189

Not that. You made a mistake.
Perhaps I should have said that you dropped a smurf? 



#48
Jul911, 01:28 AM

HW Helper
P: 6,189

Look at the drawing in your first post in this thread. You did not mark the directions of the currents (and as a consequence you made a mistake with it). So I thought I'd better mark them for you! (Just kidding, I just picked the first exercise that fitted your description. ) The real reason would be that the people who made the exercise would want the same answer from all students, so it's easier for them to check the answers. This means naming the currents and preselecting the directions. 



#50
Jul911, 01:55 AM

PF Gold
P: 2,551





#51
Jul911, 02:03 AM

HW Helper
P: 6,189





#52
Jul911, 02:15 AM

PF Gold
P: 2,551

*comes back* Oh right there's an exercise to solve. But if I didn't have....if I didn't....I'd so storm off right now! As if! 



#53
Jul911, 02:18 AM

PF Gold
P: 2,551

Hmm.... would it be reasonable to ask an easier parallel voltage sources problem before I try this one? I really want more practice on basic circuits with parallel voltages first...unless you think it would be redundant to me now? Whatever you tell me



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