How to solve this network of resistors for equivalent resistance?

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The discussion focuses on solving a network of resistors for equivalent resistance, specifically identifying parallel resistors. Participants note that segments PA and PB, as well as QA and QB, are in parallel. By solving these, they determine that each side of segment AB has a resistance of 2 ohms in series, resulting in an equivalent resistance of 4 ohms. The conversation emphasizes the importance of recognizing parallel connections and the potential impact of circuit configuration on resistance calculations. Overall, the participants successfully clarify the resistor arrangement and arrive at the correct equivalent resistance.
Harsharma17
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Homework Statement
The question asks to find the equivalent resistance for this group of resistors in a network.
Relevant Equations
I used Ohm's law and parallel and series combination to solve it, but don't know where to start. I thought about using Kirchoff's laws, but still I can't find the answer.
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Did you spot any resistors that are in parallel?
 
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Straight segment AB looks like the letter I. Would anything change if you replaced it with more straight segments that look like the letter H and tied the two resistors on the left to the left side of the H and likewise for the right side?

Remember, the convention is that, in a schematic circuit, any point on a straight line segment is at the same potential as any other point as long as there are no circuit elements, e.g. resistors, batteries, etc., in between.
 
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kuruman said:
Straight segment AB looks like the letter I. Would anything change if you replaced it with more straight segments that look like the letter H and tied the two resistors on the left to the left side of the H and likewise for the right side?

Remember, the convention is that, in a schematic circuit, any point on a straight line segment is at the same potential as any other point as long as there are no circuit elements, e.g. resistors, batteries, etc., in between.
Thanks for the reply!
Going through your suggestion, I think PA and PB are in parallel, and QA and QB are also in parallel. Solving them would lead to 2 ohm on each side of AB in series. And the equivalent resistance would be 4 ohm.
I got the answer! Thanks for the help.
 
gneill said:
Did you spot any resistors that are in parallel?
Yes, now I able to separate them in parallel circuits.
Thanks for the reply!
 
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