Equivalent Resistance of a Ciruit

[Kajayacht]

1. Find the equivalent resistance of the circuit as shown in the diagram below; where R1 = 1 Ω, R2 = 2 Ω, R3 = 2 Ω, R4 = 4 Ω, R5 = 1 Ω, R6 = 1 Ω, and R7 = 2 Ω.
[URL]http://img21.imageshack.us/i/prob06v2.gif/[/URL]

http://img21.imageshack.us/i/prob06v2.gif/

Homework Equations

R(parallel) 1/R(total) = 1/R1 + 1/R2
R(series) R(total) = R1 + R2

The Attempt at a Solution

I know I can do this one, its easy. But the arrangement of the resistors is throwing me off

I think I can take R3 and R4 in series, then take R5 in parallel with R3,4, take that in series with R2.

Then R7 and R6 are in series, and R7,6 is in parallel with R1. R1,7,6 in series with R2,5,3,4

Is this right?

 

[doublea500]

not 100% on this but here's my help. id do r3,4 in a series, then Rt of that with r5 in a parallel. then the rt3,4,5 with R2 in a series. r7 and r6 are a series so just add those. then i believe you can just do the 3 final resistances all in one parallel circuit.

 

[jhae2.718]

I know I can do this one, its easy. But the arrangement of the resistors is throwing me off

I think I can take R3 and R4 in series, then take R5 in parallel with R3,4, take that in series with R2.

Then R7 and R6 are in series, and R7,6 is in parallel with R1.

This is correct.

R1,7,6 in series with R2,5,3,4

Is this right?

This is not correct.

 

[alacey11]

After you have added together R7 and R6, you should have 3 resistors in parallel:
1. R1
2. R6 + R7
3. The resistor you got from merging R2, R3, R4 and R5

Use the parallel resistor formula on all three to get R Total

...the arrangement of the resistors is throwing me off

When I used to have this problem, I found that redrawing the circuit makes it easier to see which resistors are in parallel :)

 

[doublea500]

is what i wrote correct?

 

[Kajayacht]

Thanks guys, I got it now. R1,7,6 should be in parallel with R2,5,3,4 not series.