Anyone know how to solve compound circuits ? involving with the math

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Anyone know how to solve compound circuits ? involving with the math

Mathematics
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like basically i know we have to combine like terms
so final circuit will be
Shouldn't it just be \[\frac1R=\frac{1}{R_1+R_2}+\frac1{R_3+R_4}+\frac1{R_5+R_6}\]
exactly my point
But I was wondering I took my exam on Friday and I was struggling on find the RT of this circuit
I remember the numbers
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The first thing I did was combine like term terms
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 \[R_T=\frac1{\frac1{36}+\frac18+\frac1{39}}=\frac{936}{167}\]
Basically i just couldn't figure out R3+r4
and r5+r6
for r1+r2 I got 9ohms
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basically I'm in 9th grade
right now where understanding the basics
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.
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.
omg -_-'
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).
but why do you add 9+9
i know we had to do this |dw:1327208987134:dw|
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.
but why do you 9+9 and not use the parallel formula..?
Oh sorry, I thought those were 4's, not 9's.
its alright not your fought
fault*
So plugging those in, you get \[\frac1R=\frac1{36}+\frac1{18}+\frac1{36}=\frac19\] So your total resistance is 9 ohms :)
ok that what i got :D
now for r3+r4
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pokemon, 9 is the answer for the total resistance of the ENTIRE diagram. Once you solve for that 9, you are done.
you serious
my friend told me nine as well -_-
I should had listen to him
nooooooooooooooo thanks again imperialist
Haha, no problem pokemon. Glad to help :)

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