Can someone please give me insight on how to solve for the total resistance of this circuit? Thank you EDIT: Image link fixed

Right click the image, press "view image", copy the url, paste it in your thread, add [noparse] [/noparse] code around it. Spoiler: the image

There is one way but it's gonna take a lot of time. Find every possible route from point A of the circuit to point B (through every resistor possible). List them down by adding the resistor values in each route and making combinations (e.g. 1+2). Then try to categorize them by the similarity of the routes. Find the ones that are most similar between them (up to the last resistors) and use the parallel resistor addition for the resistors that differ between them. After you run out of those, move to the previous resistor position and do the same until you reach the first very first resistors for a final addition. For example: 1+2 1+15+4+5 1+15+5 Take the last two and find a parallel total resistance for (5) and (4+5). Once you do, add it to (15). Continue to find more possible routes with (1) to apply this. Then when you reach the final addition with (1), you are done for all possible routes that start with (1). Do so for every starting resistor and then add them all as parallel resistors again for the final. I warn you though, there are a LOT of routes.

I haven't touched it yet, but you can try to use Millman formula for each intersection. Since Wiki does not give you a comprehensible formula, I will take an example for point G: G = (A/3+B/15+H/4+F/7)/(1/3+1/15+1/4+1/7) With A, B, F, G, H are potential of respective points. There will be 10 equations with 11 variables, but the most difficult ones are the 5 inner points. Try to translate them all to 5 outer points then you should end up with A and C. In case of A and C there is an input current and an output current. Just include that current in the numerator of the formula: a positive current for input current (current that heads toward the point) or a negative current for output one (current that heads outward from the point). The equation is solved when you find out the voltage of AC and the current (input or output one, it is still the same value). It is better to consider the potential of C as zero to simplify your equations, so you should be left with A and current i at the end. The total resistance is A/i, obviously. If you want I can try testing it with real circuit

Yeah. Are the lines inwards just series resistances, or are they all connected to the other internal lines?

I'm sure that all nodes are connected, if not, there is no use of using series resistances other than confusing you.

The solutions given are extremely convoluted and more difficult than it really should be. Use Y-delta transformations to eliminate nodes and simplify it step by step. The wiki (surprisingly) has a good explanation on how it works. https://en.wikipedia.org/wiki/Y-%CE%94_transform Be sure to use the equations given to make sure you're putting in the correct resistances. EDIT: I drew a picture. You can see this creates 3 points where you can use series and combine, which will in turn create a few parallel connections. If you need to keep repeating the process and going back and forth. EDIT2: Assuming I hooked this up properly on this crappy site it seems you're getting a voltage of 21.6 when you're putting in 10 amps. So the equivalent resistance is around 2.16 ohms.

I'm assuming they are connected because it would be kinda simple otherwise. And it would also clarify that in the instructions of the problem. That seems like a convenient simplification of the problem, but the main issue I have with it is that you have to simplify some intersections that are multifaceted. It really depends on how you break the circuit down. I haven't actually calculated all the sums though, so it could be insignificant in the end.

Its unfortunate but I'm pretty sure that is the simplest way to do it. Since everything in this circuit is so interconnected you can't really simplify it much (without the trick I've shown). You can get some parallel calculations but you end up with the same exact problem. I tried myself doing the the 4 resistor parallels on both the right and left side and it only made me face the same exact brick wall. You would turn 4 resistors into 1, but the way the wires were connected still caused a problem, so you were never really going anywhere.