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pokemon23
 5 years ago
Anyone know how to solve compound circuits ? involving with the math
pokemon23
 5 years ago
Anyone know how to solve compound circuits ? involving with the math

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Maybe give an example of what you are looking for?

pokemon23
 5 years ago
Best ResponseYou've already chosen the best response.0Hold on let me get my binder

pokemon23
 5 years ago
Best ResponseYou've already chosen the best response.0dw:1327207626861:dw

pokemon23
 5 years ago
Best ResponseYou've already chosen the best response.0like basically i know we have to combine like terms

pokemon23
 5 years ago
Best ResponseYou've already chosen the best response.0so final circuit will be

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Shouldn't it just be \[\frac1R=\frac{1}{R_1+R_2}+\frac1{R_3+R_4}+\frac1{R_5+R_6}\]

pokemon23
 5 years ago
Best ResponseYou've already chosen the best response.0But I was wondering I took my exam on Friday and I was struggling on find the RT of this circuit

pokemon23
 5 years ago
Best ResponseYou've already chosen the best response.0I remember the numbers

pokemon23
 5 years ago
Best ResponseYou've already chosen the best response.0dw:1327208223868:dw

pokemon23
 5 years ago
Best ResponseYou've already chosen the best response.0The first thing I did was combine like term terms

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Mhm, 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
 5 years ago
Best ResponseYou've already chosen the best response.0Basically i just couldn't figure out R3+r4

pokemon23
 5 years ago
Best ResponseYou've already chosen the best response.0for r1+r2 I got 9ohms

pokemon23
 5 years ago
Best ResponseYou've already chosen the best response.0dw:1327208561762:dw

pokemon23
 5 years ago
Best ResponseYou've already chosen the best response.0basically I'm in 9th grade

pokemon23
 5 years ago
Best ResponseYou've already chosen the best response.0right now where understanding the basics

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay, 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
 5 years ago
Best ResponseYou've already chosen the best response.0Thus, 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Now, 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
 5 years ago
Best ResponseYou've already chosen the best response.0but why do you add 9+9

pokemon23
 5 years ago
Best ResponseYou've already chosen the best response.0i know we had to do this dw:1327208987134:dw

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Your 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
 5 years ago
Best ResponseYou've already chosen the best response.0but why do you 9+9 and not use the parallel formula..?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh sorry, I thought those were 4's, not 9's.

pokemon23
 5 years ago
Best ResponseYou've already chosen the best response.0its alright not your fought

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So plugging those in, you get \[\frac1R=\frac1{36}+\frac1{18}+\frac1{36}=\frac19\] So your total resistance is 9 ohms :)

pokemon23
 5 years ago
Best ResponseYou've already chosen the best response.0ok that what i got :D

pokemon23
 5 years ago
Best ResponseYou've already chosen the best response.0dw:1327209449039:dw

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0pokemon, 9 is the answer for the total resistance of the ENTIRE diagram. Once you solve for that 9, you are done.

pokemon23
 5 years ago
Best ResponseYou've already chosen the best response.0my friend told me nine as well _

pokemon23
 5 years ago
Best ResponseYou've already chosen the best response.0I should had listen to him

pokemon23
 5 years ago
Best ResponseYou've already chosen the best response.0nooooooooooooooo thanks again imperialist

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Haha, no problem pokemon. Glad to help :)
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