# help please!



## frecoder78 (Jul 21, 2009)

I have a few questions from the Camera Power Practice Problems Book.

1.) Chapter 27, problem 7 and 8:

Problem 7: Why does the solution manual tell you to use "rms values"? The reference book shows to solve the power triangle differently.

Problem 8: So what does it mean when Q=0? Does this mean S=P?

2.) Chapter 28, Problem 7: Where does the equation Z=sqrt*((sumR)^2+(sumXL - sumXC)^2) come from? I didnt see it at all in Camera's reference book.

3.) Chapter 29, problem 9(B): Any reason why the solution to the current equation i(t) reduces to 0.625 + (.2926...)?

Any help on these problems would be greatly appreciated.

Thanks!!


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## benbo (Jul 21, 2009)

I used to have this book but can't find it. If you could post a pdf maybe me or somebody else could help.


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## RIP - VTEnviro (Jul 22, 2009)

The original EB.com lady's man is back!

What's up Frecoder?


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## frecoder78 (Jul 22, 2009)

VTEnviro said:


> The original EB.com lady's man is back!
> What's up Frecoder?


haha what's up VTEnviro? Yes I'm back but putting my extra-curricular activities aside to study for the PE. How you been?


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## dzdave00 (Jul 23, 2009)

frecoder78 said:


> I have a few questions from the Camera Power Practice Problems Book.
> 1.) Chapter 27, problem 7 and 8:
> 
> Problem 7: Why does the solution manual tell you to use "rms values"? The reference book shows to solve the power triangle differently.
> ...


Hello, I do not have the book with me right now, but I will give it my best:

1.) Chapter 27, problem 7: My experience with problems where there is a question on whether to use peak values or RMS values is if they want peak values, they will specifically ask for peak values. Otherwise I always use RMS values.

chapter 27, problem 8: Apparent power is made up of real (watts) and reactive (VARS) power. You are correct that when Q=0, then S=P, which is unity power factor.

2) The equation you posted is the rectangular formula for impedance (Z). The equation is like pythagoreans theorem due to the phase angles of resistance and reactance. Impedance is comprised of resistance ® and reactance (X). Reactance is comprised of inductive and capacative reactance, which are 180 degrees apart in terms of their phase angles. Hence, when you want to find the total reactance, you take the difference between inductance reactance and capacative reactance.

3) I do remember this problem somewhat, but I thought the solution was pretty clear. Y0ou need to figure out whether the circuit is underdamped or overdamped, or (one other option I think, can't remember) and use the apropriate equation. Is there any part of the problem you are having trouble understanding?

Hope that helped, Dave :sharkattack:


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## frecoder78 (Jul 27, 2009)

dzdave00 said:


> Hello, I do not have the book with me right now, but I will give it my best:
> 1.) Chapter 27, problem 7: My experience with problems where there is a question on whether to use peak values or RMS values is if they want peak values, they will specifically ask for peak values. Otherwise I always use RMS values.
> 
> chapter 27, problem 8: Apparent power is made up of real (watts) and reactive (VARS) power. You are correct that when Q=0, then S=P, which is unity power factor.
> ...


Thanks so much Dave! For problems #3 I understand that the equation being used is underdamped. I'm just having trouble figuring out how the Camera book reached the solution. Check out the PDF of the solution and see if you can help me further. I trully appreciate the help! :gora:


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## benbo (Jul 27, 2009)

For Chapter 29 number 9, I can't really tell what the circuit diagram is. But it looks just like arithmetic. I(0) = .2926 A from part a of the problem.

V(50)/R = 50V/80ohm

Then they rearranged and factored out the exponential.

Which step exactly is it that you don't understand? Is it something beyond this? I'm just not sure I understand the question.

PS - sometimes it is difficult to see the algebra they are trying to do when they solve these problems, because it doesn't seem to make sense. So if that is it, remember that on the exam you will generally be looking for a number, so you can always stop the algebraic manipulation once you have something you can plug into.


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## dzdave00 (Jul 28, 2009)

frecoder78 said:


> Thanks so much Dave! For problems #3 I understand that the equation being used is underdamped. I'm just having trouble figuring out how the Camera book reached the solution. Check out the PDF of the solution and see if you can help me further. I trully appreciate the help! :gora:


Oh, this is a different problem than the one I was thinking of, my apologies. Part A of this problem is solving for the (steady state condition) current flowing in the circuit with the switch in position A.

The proper equation to use is: I(t) = Io*e(-t/T) + (Vbatt/R)*(1-e(-t/T)

where T=L/R, and Io = initial current

Since there is no initial current flowing (Io=0), the first term is equal to zero, so you are left with the equation shown in the solution.

Part B is asking for the total current flowing in the circuit when the switch is flipped to position B.

You use the same overall equation of:

I(t) = Io*e(-t/T) + (Vbatt/R)*(1-e(-t/T)

only this time you do have initial current flowing from when the switch was in position A (Io=.2926

So now your equation will look like this, with T= L/R = .15/80 and 1/T = 533:

I(t) = .2926*e(-533t) + (50/80)*(1-e(-533t))

I(t) = .2926*e(-533t) + .625 - .625*(e-533t)

I(t) = .625 - .332*(e-533t)

I did not really understand how they broke it down either, so I did it the way I could understand it. :dancingnaughty:


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## frecoder78 (Jul 28, 2009)

dzdave00 said:


> Oh, this is a different problem than the one I was thinking of, my apologies. Part A of this problem is solving for the (steady state condition) current flowing in the circuit with the switch in position A.
> The proper equation to use is: I(t) = Io*e(-t/T) + (Vbatt/R)*(1-e(-t/T)
> 
> where T=L/R, and Io = initial current
> ...


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