Find total resistance in mixed circuit

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Princess19
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Homework Statement



http://img59.imageshack.us/img59/9321/photo2100.jpg

R1 = R2 = R3 = 5 ohms
R4 = R5 = R6 = 10 ohms
R7 = 20 ohms

Homework Equations



1/Rp = 1/R1 + ... +1/Rn (parallel)
Rs = R1 +...+Rn (series)

The Attempt at a Solution



Here's what I've got so far. I need help this is due tomorrow. :(

R67 = 10 + 20 = 30 ohms

1/R567 = 1/10 + 1/30 = 7.5 ohms

R4567 = 10 + 7.5 = 17.5 ohms

1/R34567 = 1/5 + 1/17.5 = 4 ohms

R234567 = 5 + 4 = 9 ohms

1/R1234567 = 1/5 + 1/9 = 3 ohms


okay, now tell me where I am wrong!

Thanks a bunch!
 
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Princess19 said:

Homework Statement



http://img59.imageshack.us/img59/9321/photo2100.jpg

R1 = R2 = R3 = 5 ohms
R4 = R5 = R6 = 10 ohms
R7 = 20 ohms

Homework Equations



1/Rp = 1/R1 + ... +1/Rn (parallel)
Rs = R1 +...+Rn (series)

The Attempt at a Solution



Here's what I've got so far. I need help this is due tomorrow. :(

R67 = 10 + 20 = 30 ohms

1/R567 = 1/10 + 1/30 = 7.5 ohms

R4567 = 10 + 7.5 = 17.5 ohms

1/R34567 = 1/5 + 1/17.5 = 4 ohms

R234567 = 5 + 4 = 9 ohms

1/R1234567 = 1/5 + 1/9 = 3 ohms


okay, now tell me where I am wrong!

Thanks a bunch!

First, to be accurate,

1/R567 = 1/10 + 1/30 = 7.5 ohms should be 1/R567 = 1/10 + 1/30 = 1/7.5 ohms

Next, this is correct:

R4567 = 10 + 7.5 = 17.5 ohms

But then, R3 is in series with this value, not in parallel...
 
Last edited by a moderator:
berkeman said:
First, to be accurate,

1/R567 = 1/10 + 1/30 = 7.5 ohms should be 1/R567 = 1/10 + 1/30 = 1/7.5 ohms

Next, this is correct:

R4567 = 10 + 7.5 = 17.5 ohms

But then, R3 is in series with this value, not in parallel...

oh, 7.5 ohms is already the reciprocal of the answer i got which was .1333

oh, why is it in series? arent they facing each other?
 
Princess19 said:
oh, 7.5 ohms is already the reciprocal of the answer i got which was .1333

oh, why is it in series? arent they facing each other?

R3 is in series with the 4567 total, and that is in parallel with R2.

In fact, I would have just added the series combination of R4 + R3 + R567 as a single step.

When two resistors are next to each other and connected with shorts at both ends, they are in parallel. When they are connected with a resistor at one end, all 3 of those resistances are in parallel (nose-to-tail).
 
R67 = 10 + 20 = 30 ohms

1/R567 = 1/10 + 1/30 = 7.5 ohms

R4567 = 10 + 7.5 = 17.5 ohms

R34567= 5 + 17.5 = 22.5 ohms

1/R234567 = 1/5 + 1/22.5 = 4 ohms

R1234567 = 5 + 4 = 7 ohms

how about now?
 
Princess19 said:
R67 = 10 + 20 = 30 ohms

1/R567 = 1/10 + 1/30 = 7.5 ohms

R4567 = 10 + 7.5 = 17.5 ohms

R34567= 5 + 17.5 = 22.5 ohms

1/R234567 = 1/5 + 1/22.5 = 4 ohms

R1234567 = 5 + 4 = 7 ohms

how about now?

That looks good, except you keep doing that shortcut with the parallel addition.

1/R567 = 1/10 + 1/30 = 7.5 ohms Needs to be 1/R567 = 1/10 + 1/30 = 1/7.5 ohms

(so R567 = 7.5 Ohms)

And

1/R234567 = 1/5 + 1/22.5 = 4 ohms Needs to be 1/R234567 = 1/5 + 1/22.5 = 1/4 ohms

(so R234567 = 4 Ohms).
 
Thanks for your help! :)