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Hi how do you solve this problem? I am confused.

schematic

simulate this circuit – Schematic created using CircuitLab

Find \$R_{AB}\$, \$R_{CD}\$, \$R_{EF}\$, \$R_{AE}\$

I came up with this: \$R_{A-B}\$ = 1 \$\Omega\$ , \$R_{C-D}\$ = 4 \$\Omega\$ , \$R_{E-F}\$ = 7 \$\Omega\$ , \$R_{A-E}\$ = 2 + 5 = 7 \$\Omega\$

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  • \$\begingroup\$ +1 for posting a schematic with your question, even if the first version was rudimentary. Most questions come without one and that makes them a lot harder to answer. \$\endgroup\$ – Ricardo Aug 1 '14 at 23:57
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Have you covered equivalent resistances? Essentially, it is a way to combine resistors in series or parallel until you get a single resistance, working your way from out to in.

Let's say you want to find \$R_{AB}\$.

First, your have a series combination \$R_5\$, \$R_6\$, and \$R_7\$. $$ R_{5,6,7} = R_5 + R_6 + R_7 = 18 \Omega $$

schematic

simulate this circuit – Schematic created using CircuitLab

Now you have a parallel combination between \$R_4\$ and \$R_{5,6,7}\$. $$ R_{4,5,6,7} = \frac{R_4 \cdot R_{5,6,7}}{R_4 + R_{5,6,7}} = 3.27 \Omega $$

schematic

simulate this circuit

You are almost done! Now, since you are looking for the equivalent resistance between A and B, you have a series combination for \$R_2\$, \$R_3\$ and \$R_{4,5,6,7}\$: $$ R_{2,3,4,5,6,7} = R_2 + R_3 + R_{4,5,6,7} = 8.27 \Omega $$

schematic

simulate this circuit

Last Step!

You have a parallel combination between \$R_1\$ and \$R_{2,3,4,5,6,7}\$: $$ R_{AB} = \frac{R_1 \cdot R_{2,3,4,5,6,7}}{R_1 + R_{2,3,4,5,6,7}} = 0.892\Omega $$

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    \$\begingroup\$ you saved me so much time, have my upvote! I'm just editing your answer to include some nice formatting, have a look at it after to learn how to use mathjax. \$\endgroup\$ – Vladimir Cravero Aug 1 '14 at 18:36
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    \$\begingroup\$ Ahh, thank you. I was trying to figure out how to clean the equations up. Thanks for the edit! And you're welcome. \$\endgroup\$ – Ornusashas Aug 1 '14 at 18:46
  • \$\begingroup\$ @Ornusashas to understand how to write great answers just check out some of the frequent ones. That's what I did at least... You might also learn some electronics (not sure if you need to...) \$\endgroup\$ – Vladimir Cravero Aug 1 '14 at 21:27
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$$ R_{ab} = (((5 + 7 + 6) || 4) + 2 + 3) || 1 $$

Combine the resistors in series and parallel until you end up with a single resistance between the two nodes and you have your answer.

I believe for \$R_{ae}\$ one would need to employ a wye-delta transform.

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