## pokemon23 4 years ago Anyone know how to solve compound circuits ? involving with the math

1. anonymous

Maybe give an example of what you are looking for?

2. pokemon23

Hold on let me get my binder

3. pokemon23

|dw:1327207626861:dw|

4. pokemon23

like basically i know we have to combine like terms

5. pokemon23

so final circuit will be

6. anonymous

Shouldn't it just be $\frac1R=\frac{1}{R_1+R_2}+\frac1{R_3+R_4}+\frac1{R_5+R_6}$

7. pokemon23

exactly my point

8. pokemon23

But I was wondering I took my exam on Friday and I was struggling on find the RT of this circuit

9. pokemon23

I remember the numbers

10. pokemon23

|dw:1327208223868:dw|

11. pokemon23

The first thing I did was combine like term terms

12. 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 $R_T=\frac1{\frac1{36}+\frac18+\frac1{39}}=\frac{936}{167}$

13. pokemon23

Basically i just couldn't figure out R3+r4

14. pokemon23

and r5+r6

15. pokemon23

for r1+r2 I got 9ohms

16. pokemon23

|dw:1327208561762:dw|

17. pokemon23

basically I'm in 9th grade

18. pokemon23

right now where understanding the basics

19. 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.

20. 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.

21. pokemon23

omg -_-'

22. 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).

23. pokemon23

but why do you add 9+9

24. pokemon23

i know we had to do this |dw:1327208987134:dw|

25. 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.

26. pokemon23

but why do you 9+9 and not use the parallel formula..?

27. anonymous

Oh sorry, I thought those were 4's, not 9's.

28. pokemon23

its alright not your fought

29. pokemon23

fault*

30. anonymous

So plugging those in, you get $\frac1R=\frac1{36}+\frac1{18}+\frac1{36}=\frac19$ So your total resistance is 9 ohms :)

31. pokemon23

ok that what i got :D

32. pokemon23

now for r3+r4

33. pokemon23

|dw:1327209449039:dw|

34. anonymous

pokemon, 9 is the answer for the total resistance of the ENTIRE diagram. Once you solve for that 9, you are done.

35. pokemon23

you serious

36. pokemon23

my friend told me nine as well -_-

37. pokemon23

I should had listen to him

38. pokemon23

nooooooooooooooo thanks again imperialist

39. anonymous

Haha, no problem pokemon. Glad to help :)