pokemon23
  • pokemon23
Anyone know how to solve compound circuits ? involving with the math
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Maybe give an example of what you are looking for?
pokemon23
  • pokemon23
Hold on let me get my binder
pokemon23
  • pokemon23
|dw:1327207626861:dw|

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pokemon23
  • pokemon23
like basically i know we have to combine like terms
pokemon23
  • pokemon23
so final circuit will be
anonymous
  • anonymous
Shouldn't it just be \[\frac1R=\frac{1}{R_1+R_2}+\frac1{R_3+R_4}+\frac1{R_5+R_6}\]
pokemon23
  • pokemon23
exactly my point
pokemon23
  • pokemon23
But I was wondering I took my exam on Friday and I was struggling on find the RT of this circuit
pokemon23
  • pokemon23
I remember the numbers
pokemon23
  • pokemon23
|dw:1327208223868:dw|
pokemon23
  • pokemon23
The first thing I did was combine like term terms
anonymous
  • anonymous
Mhm, giving you three parallel resistors of 36, 8 and 39 respectively (I believe those are your numbers, correct me if I'm wrong. Then you should get \[R_T=\frac1{\frac1{36}+\frac18+\frac1{39}}=\frac{936}{167}\]
pokemon23
  • pokemon23
Basically i just couldn't figure out R3+r4
pokemon23
  • pokemon23
and r5+r6
pokemon23
  • pokemon23
for r1+r2 I got 9ohms
pokemon23
  • pokemon23
|dw:1327208561762:dw|
pokemon23
  • pokemon23
basically I'm in 9th grade
pokemon23
  • pokemon23
right now where understanding the basics
anonymous
  • anonymous
Okay, I believe you are confusing some of the laws of resistors. If two resistors are in sequence, then their total resistance is the sum of the two. Imagine it as if you're running through an obstacle course. The more obstacles (ie resistors) the harder it is (ie the more resistance to you finishing). On the other hand resistors in parallel use the inverse law, namely that if resistors A and B are in parellel, then the total resistance (R) is 1/R=1/A+1/B. Since current is just a flow of electrons and electrons want to minimize resistance, this law basically is the result of more electrons choosing the resistor with lower resistance. However, they can't ALL go through the smaller one, otherwise it would cause a jam of sorts, causing them to instead follow this law.
anonymous
  • anonymous
Thus, what you must do is first deal with the three sets of sequential resistors. Namely, find R_1+R_2=18+18=36, R_3+R_4=4+4=8, and R_5+R_6=13+23=36.
pokemon23
  • pokemon23
omg -_-'
anonymous
  • anonymous
Now, for all intents and purposes, you just have three resistors of resistance 36, 8, and 36 (I accidentally said 39 earlier, so ignore that).
pokemon23
  • pokemon23
but why do you add 9+9
pokemon23
  • pokemon23
i know we had to do this |dw:1327208987134:dw|
anonymous
  • anonymous
Your final answer should come from the equation \[\frac1R=\frac1{R_1+R_2}+\frac1{R_3+R_4}+\frac1{R_5+R_6}\] so yes, you must do that several times. Plugging in those values will give you your answer.
pokemon23
  • pokemon23
but why do you 9+9 and not use the parallel formula..?
anonymous
  • anonymous
Oh sorry, I thought those were 4's, not 9's.
pokemon23
  • pokemon23
its alright not your fought
pokemon23
  • pokemon23
fault*
anonymous
  • anonymous
So plugging those in, you get \[\frac1R=\frac1{36}+\frac1{18}+\frac1{36}=\frac19\] So your total resistance is 9 ohms :)
pokemon23
  • pokemon23
ok that what i got :D
pokemon23
  • pokemon23
now for r3+r4
pokemon23
  • pokemon23
|dw:1327209449039:dw|
anonymous
  • anonymous
pokemon, 9 is the answer for the total resistance of the ENTIRE diagram. Once you solve for that 9, you are done.
pokemon23
  • pokemon23
you serious
pokemon23
  • pokemon23
my friend told me nine as well -_-
pokemon23
  • pokemon23
I should had listen to him
pokemon23
  • pokemon23
nooooooooooooooo thanks again imperialist
anonymous
  • anonymous
Haha, no problem pokemon. Glad to help :)

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