Equivalent Resistance of a Ciruit

[GeorgeCostanz]

Homework Statement

Find the equivalent resistance of the circuit as shown in the diagram below; where, R1 = 2 Ω, R2 = 1 Ω, R3 = 2 Ω, R4 = 1 Ω, R5 = 4 Ω, R6 = 4 Ω, and R7 = 4 Ω.

http://i.imgur.com/OsAs2.gif

Homework Equations

The Attempt at a Solution

R3 and R4 are in parallel, found R of those 2 then added that to R2 (in series)

R6 and R5 in series, added those together

this is where i get lost

I added R1 + R7 + (R6+R5) in parallel.

then added that to the 1st R-eq i found - in parallel

wrong.

i'm clearly confused about the relationships

thanks

 

[PeterO]

Homework Statement

Find the equivalent resistance of the circuit as shown in the diagram below; where, R1 = 2 Ω, R2 = 1 Ω, R3 = 2 Ω, R4 = 1 Ω, R5 = 4 Ω, R6 = 4 Ω, and R7 = 4 Ω.

http://i.imgur.com/OsAs2.gif

Homework Equations

The Attempt at a Solution

R3 and R4 are in parallel, found R of those 2 then added that to R2 (in series)

R6 and R5 in series, added those together

this is where i get lost

I added R1 + R7 + (R6+R5) in parallel.

then added that to the 1st R-eq i found - in parallel

wrong.

i'm clearly confused about the relationships

thanks

The R 5&6 combination [in series with each other] is in parallel to the R2,3,4 combination you established. Then that whole combination is in series with R 1&7.

 

[PeterO]

Homework Statement

Find the equivalent resistance of the circuit as shown in the diagram below; where, R1 = 2 Ω, R2 = 1 Ω, R3 = 2 Ω, R4 = 1 Ω, R5 = 4 Ω, R6 = 4 Ω, and R7 = 4 Ω.

http://i.imgur.com/OsAs2.gif

Homework Equations

The Attempt at a Solution

R3 and R4 are in parallel, found R of those 2 then added that to R2 (in series)

R6 and R5 in series, added those together

this is where i get lost

I added R1 + R7 + (R6+R5) in parallel.

then added that to the 1st R-eq i found - in parallel

wrong.

i'm clearly confused about the relationships

thanks

It sometimes helps to draw the resistor set up in a straigh line, rather than 3 sides of a square.

This would start with R1 , then divide to two branches, with R5&R6 on the bottom, and R2, along with a parallel R3&R4 on the top, the the branches re-joining to get to R7.

 

[GeorgeCostanz]

hmm, i never thought to look at it that way.
thanks guys