Resistance in a series / parallel?

AI Thread Summary
The discussion revolves around calculating the equivalent resistance between points A and B using given resistor values. Initially, the user incorrectly identifies the configuration of resistors R3, R4, and R5 as parallel, leading to confusion in calculations. After several iterations and corrections, the user determines that the equivalent resistance of R3, R4, and R5 is in series with R6, and then combines these with R1 and R2 in parallel. Ultimately, the correct equivalent resistance is found to be approximately 1.56 ohms after resolving calculation errors. The thread highlights the importance of accurately applying series and parallel resistor formulas in circuit analysis.
dolpho
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Homework Statement



Find the equivalent resistance between points A and B shown in the figure(Figure 1) . Consider R1= 2.2 , R2= 1.0 , R3 = 3.6 , R4= 5.0 , R5= 4.0 , and R6= 7.6 .

http://i.imgur.com/EBJmSvk.png

Homework Equations



For resistors in parallel, 1 / Req = 1/r1 + 1/r2 + 1/r3...etc
for resistors in series, Req= R1 + R2 + R3...etc

The Attempt at a Solution



So I think R3, R4 and R5 are parallel

1/3.6 + 1/5 + 1/4 = .72

and then I think we can combine R6, R1 and R2 into one since they are parallel also.

1 / 2.2 + 1/1 + 1/7.6 = 1.58

Now add up .72 and 1.58 since they are in series? = 2.3

Would appreciate any help, this is pretty confusing to me lol
 
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dolpho said:
So I think R3, R4 and R5 are in a series.

Check this again.
 
Pranav-Arora said:
Check this again.

Ahhh I'm sorry I did a typo. I meant parallel
 
dolpho said:
Ahhh I'm sorry I did a typo. I meant parallel

Did you get the correct answer now?
 
Pranav-Arora said:
Did you get the correct answer now?

nope :( I'm not sure what I'm doing wrong
 
dolpho said:
So I think R3, R4 and R5 are parallel

1/3.6 + 1/5 + 1/4 = .72

0.72 is reciprocal of equivalent resistance of R3, R4 and R5. The equivalent of these three resistors is in series with R6.
 
Pranav-Arora said:
0.72 is reciprocal of equivalent resistance of R3, R4 and R5. The equivalent of these three resistors is in series with R6.


Ok I was just thinking that, so its actually.

.72 = 1 / Req

Req = 1.39?

Then we add that to R6 in a series? so

1.39 + 7.6 = 8.99

Now we add R6 to R2 and R1 which are parallel.

So... 1/2.2 + 1 + 1/8.99 = 1 / Req

= 1.24 = 1 / Req -----> 1 / 1.24 = .8?
 
dolpho said:
Ok I was just thinking that, so its actually.

.72 = 1 / Req

Req = 1.39?

Then we add that to R6 in a series? so

1.39 + 7.6 = 8.99

Now we add R6 to R2 and R1 which are parallel.

So... 1/2.2 + 1 + 1/8.99 = 1 / Req

= 1.24 = 1 / Req -----> 1 / 1.24 = .8?

I haven't checked out your calculations but your steps look fine to me.
 
Pranav-Arora said:
I haven't checked out your calculations but your steps look fine to me.

ARG...It's not .8...what the f...heck
 
  • #10
dolpho said:
So... 1/2.2 + 1 + 1/8.99 = 1 / Req

= 1.24 = 1 / Req -----> 1 / 1.24 = .8?

Calculation mistake. 1/Req=1.56
 
  • #11
Pranav-Arora said:
Calculation mistake. 1/Req=1.56

Finally! Got it right, thanks so much for your help!
 
  • #12
dolpho said:
So... 1/2.2 + 1 + 1/8.99 =
= 1.24

Try that step again (btw, 8.97 is more accurate)
 
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