Basic equivalent resistance problem

AI Thread Summary
The discussion centers on calculating the equivalent resistance Rab in a circuit with multiple resistors. The user outlines their method, first combining the bottom 2Ω resistors in series to find a total of 4Ω. They then calculate the equivalent resistance for the top right 12Ω and 4Ω resistors in parallel, resulting in 3Ω, which is further combined with a 6Ω resistor in parallel to yield 2Ω. The final steps involve combining two sets of resistors, leading to a total equivalent resistance of 5Ω for Rab. The calculations are confirmed to be accurate, with a slight variation in approach yielding the same result.
qpham26
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Homework Statement


Find Rab
https://sphotos-a.xx.fbcdn.net/hphotos-ash4/199915_496321627054024_1130826860_n.jpg

Homework Equations



Req = R1R2 / R1+ R2

The Attempt at a Solution


bottom 2Ω resistors in series ---> Req1 = 4Ω

top right 12 and 4 Ω in parallel ---> Req2 = 3Ω

this 3Ω and the left top 6Ω are in parallel ---> Req3 = 2Ω

2 resistors of 4Ω bottom left are in parallel -->Req4 = 2Ω

Req4 and Req3 are in series --->Req5 = 4Ω

Req5 and Req1 are in parallel ----> Req6 = 2Ω

Rab = 3Ω + 2Ω = 5Ω

please let me know if I have gotten the correct answer.
Thanks for your time.
 
Last edited by a moderator:
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I also make it 5 Ohms slightly different way..

[{(6//12//4) + (4//4)} // {2+2}] + 3

[{2 + 2} // {4}] + 3

[4//4] +3

2 + 3 = 5
 

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