Question

Circuits - @whpalmer4 @yakeyglee - If I know power for resistor two and the circuit voltage, how do I solve for resistance? Let me get a drawing and what I have so far...

Answer 1

Power is related to voltage drop across a resistor and the resistance of that resistor by \(P = V^2/R\) so \(R = V^2/P\).

Answer 2

P2=V2*I therefore P2=(R2/(8)R2)I

Answer 3

You can also use \[P=VI\]\[V = IR\]\[P=I*IR = I^2R\] to find the resistance from the power because you know \[I=\frac{V}{8R_2}\]

Answer 4

I'm sorry @whpalmer4... I still can't get it. \[P=V_{2}I\] but when we get down to the I=V/8R2 i cant find I because i don't know what r2 is?

Answer 5

I'm so confused

Answer 6

Okay, we know that \[V=IR\]so\[I=\frac{50}{4R_2+R_2+3R_2} = \frac{25}{4R_2}\]We also know that \[P = I^2R_2 = 160 \text{mW}\]Combine the two equations and you should be able to solve for \(R_2\)

Answer 7

\[.16=\left(\begin{matrix}50 \over\ 8R_{2}\end{matrix}\right)^2 \times R_{2}\]\]

\[.16=\left(\begin{matrix}2500 \over\ (64R_{2})^2\end{matrix}\right) \times R_{2}\]

\[.16=\left(\begin{matrix}2500 \over\ 64R_{2}\end{matrix}\right)\]

Answer 8

= .004096 ?

Answer 9

\[(\frac{25}{4R_2})^2R_2=160\text{mW}\]\[\frac{625}{16R_2}=160\text{mW}\]\[R_2=\frac{625}{16*160*10^{-3}} \approx 244.141 \Omega\]\[I=\frac{25}{4*244.141}=25.6\text{mA}\]\[V=I*(R_1+R_2+R_3) =25.6*10^{-3}*(8*244.141) = 50\]

Answer 10

Wowzers i messed that up quite a bit
Let me take in your math so i can actually use this effectively

Answer 11

You mistyped the denominator in the second equation, but didn't actually heed what it said, so got the right 3rd equation.
\[0.16=\frac{2500}{64R_2}\]\[0.16*64*R_2=2500\]\[R_2=\frac{2500}{0.16*64}=\frac{2500}{10.24} = 244.140625\]Maybe you just pushed the buttons wrong on the calculator? Or your slide rule needs lubricating :-)

Answer 12

That is what i was just about to say, when i plugged it into the calc it was a miss type from the start... thank you for clearing this up. looks like this was the hardest problem from the lot, i was able to finish most of them in the time it took working on these two problems with you. Thank you very much @whpalmer4 !!!

Answer 13

You're welcome, of course! What class is this for?

Answer 14

Ha, embarrassing enough, this is circuits I, which is just a simple 1140 series class...

Answer 15

What's the textbook you are using?

Answer 16

http://books.google.com/books/about/Circuit_Analysis.html?id=5UiGuAAACAAJ

Answer 17

Are you in agreement that the text provides "a thorough, engaging introduction to the theory"? :-)

Answer 18

I think they write that in the review of every circuit theory textbook ever written. I'm not sure it has ever been true :-)

Answer 19

Wellllllll it seemed like that until i started working on this homework, plus it didnt help that the prof. didnt prepare us for this kind of work... even though it is algebra.

Answer 20

Haha, no it has not, unfortunately.

Answer 21

Well, it could be worse, in physics textbooks they always want you to compute the resistance of grids of interconnected resistors stretching off to infinity and beyond :-)

Answer 22

Damn, well that gives me a little peek into my future :( haha. I really appreciate the help though!

Answer 23

But, a lot of this is just a matter of seeing some good examples and working a bunch of problems until you get the hang of it.

Answer 24

Thats what I am really hoping for

Answer 25

Well, shoot me a message if you get stuck on one and no one seems to be helping. Hopefully the reason for that won't be that the problem is impossibly difficult :-)

Answer 26

Thank you sooo much, I'd hate to bug you but you really know what you're doing!

Answer 27

Don't sweat it. You won't occupy any more of my time than I'm willing to spend here. As long as whoever is using my time is learning something, I'm happy.

Answer 28

And it turns out that explaining things to someone else is a good way to reinforce your own understanding.

Answer 29

That is so awesome of you. I try and help out others in math when I can, and when I know how to get it right haha, but lately its been working on this and calcII for hours and hours on end

Answer 30

It does help quite a bit, that is why I enjoy helping others if I get the chance.

Answer 31

Well, time for me to turn in, it's almost 2:30 AM here on the west coast.

Answer 32

Sitting here at 3:30 AM almost done though. Thank you so much for the help again!