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I'm so confused on how to find the equivalent resistance of this circuit. Initially, I calculated it to be R_eq = (1/(42+21)+1/63+1/(84+105))^-1 = 27 Ohms, but I think that's wrong.

I think that the 84 Ohm and 105 Ohm resistors are in series, but I don't know about the others.

enter image description here

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    \$\begingroup\$ You can tell by inspection that 27 ohms is wrong. There's a 21 ohm resistor going directly from A to B, so the equivalent resistance won't ever be more than 21 ohms. \$\endgroup\$ – Pete Becker Feb 21 '16 at 20:19
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I think that the 84 Ohm and 105 Ohm resistors are in series

This is correct. They are in series, and you can combine them as such.

Once you have combined them, they will be in parallel with the 63Ω resistor, and you can combine again. Keep doing so, and you will find that the circuit reduces to a single equivalent resistor.

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  • \$\begingroup\$ Thank you! I think I have it now. The equivalent resistance should be 17 Ohms I believe. \$\endgroup\$ – user88062 Feb 21 '16 at 19:42
  • \$\begingroup\$ Yup, that looks correct. \$\endgroup\$ – uint128_t Feb 21 '16 at 19:45
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When in doubt, re-draw the circuit.

schematic

simulate this circuit – Schematic created using CircuitLab

Hopefully the solution is obvious now.

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Redrawing the original schematic for clarity and showing the work, in 3. Rx is the the sum of R4 and R5 in parallel with R3, and in 4. Rt is the sum of R2 and Rx in parallel with R1.

enter image description here

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    \$\begingroup\$ The simplifications are shown, as well as the work and the right answer, so why the downvotes? \$\endgroup\$ – EM Fields Feb 21 '16 at 22:00
  • \$\begingroup\$ The downvotes are for giving a full solution to a homework problem. The other full solution answer is getting downvoted too, while the two answers that merely show a strategy without actually doing the work are getting upvoted. \$\endgroup\$ – Olin Lathrop Feb 21 '16 at 23:33
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    \$\begingroup\$ @OlinLathrop: Well, seeing as how the OP stated: "Thank you! I think I have it now. The equivalent resistance should be 17 Ohms I believe." and the system logged it as: " [the OP] 4 hours ago", the cat was out of the bag an hour before I posted my answer, so either someone's not paying attention or you're wrong, or both. Notice that I posted nothing substantially different from what had already been posted except for the minimization sequence and the use of the formula for two parallel resistances, where \$ Rt = \frac {R1\times R2} {R1+R2}\$, instead of the clumsy reciprocal way. \$\endgroup\$ – EM Fields Feb 22 '16 at 0:58
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I get 17 ohms.

84 and 105 are in series, = 189 ohms. Then, 189 and 63 are in parallel, and that is in series with the 42.

42 + (1/63 + 1/(84 + 105))^-1 = 89.25 ohms,

Then, that total is in parallel with the 21. (1/21 + 1/89.25)^-1 = 17 ohms

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