Hi, can anyone please explain to me how to solve the diagonal part of this circuit please? Thanks.

 

The way Homework Help works is you have to show your best effort so we can see where you're having difficulties.

 

The way Homework Help works is you have to show your best effort so we can see where you're having difficulties.

But i don't even know how to start solving this because of the crisscross resistors

 

But i don't even know how to start solving this because of the crisscross resistors

We can't help you if you don't know where to start because we don't know how you're expected to solve the problem.

Instructors don't typically ask students questions that they won't know how to answer. Check your notes/book. Talk to your instructor.

 

We can't help you if you don't know where to start because we don't know how you're expected to solve the problem.

Instructors don't typically ask students questions that they won't know how to answer. Check your notes/book. Talk to your instructor.

This isn't from class, i'm just solving random exercises. It says i'm supposed to find the equivalent resistance that would replace this whole circuit between a and b and the only thing i know how to do it is put the right most 2 resistors into a parallel, but i don't know how to solve the x in the middle

 

Also see this https://forum.allaboutcircuits.com/...he-equivalent-resistance.163560/#post-1436911
And post #2

 

Also see this https://forum.allaboutcircuits.com/...he-equivalent-resistance.163560/#post-1436911
And post #2

What does that have to do with my problem?

 

Try redrawing the schematic to eliminate the crossed conductors.

(Think in terms of nodes, not lines.)

 

Hi, can anyone please explain to me how to solve the diagonal part of this circuit please? Thanks.

View attachment 301755

This circuit is drawn to deliberately confuse you. Identify resistors that are in parallel by looking at which resistors share the same pair of nodes.

Try coloring each node with a different color. Any two resistors that are connect to the same two colors are in parallel and can be combined.

If this fails you, then you can always fall back on applying a test voltage source between 'a' and 'b' and then determining what the current is that flows out the positive terminal of the source. The ratio of V to I is the effective resistance.

A good practice to get into is to estimate what you expect the answer to be or, even better, come up with upper and lower bounds that a correct answer must fall between.

In this case, you can immediately see that the resistance has to be at least 15 Ω because of the series resistor connected to 'a'. But you can also see that the maximum the total resistance can be is 21 Ω because the left most 6 Ω resistor in in parallel with everything to the right of it. A bit of looking will reveal that the upper limit can easily be lowered to 17.5 Ω. So, with very little effort, we can say that the answer is within 10% of 16 Ω. Any answer that we get that is not within that range, we know is wrong (or, possibly, that we messed up our estimate somehow). In either case, we should not continue until the discrepancy is resolved.

 

But i don't even know how to start solving this because of the crisscross resistors

Then un-crisscross them.

 

I'll give you a little clue:
It would be tricky if there were a resistor directly below the 1Ω resistor in the b line.

 

diagonal part

(??? which way "diagonal")

It reduces at several (3x) places (you can solve it numerically by iterating the currents and voltage points based on "terminal voltage")

 

Thanks everyone, so would this be correct then?

 

Thanks everyone, so would this be correct then?

That's what I get...

You should be more careful with the way you draw '1'. Sometimes the "tail" is so long that it looks like '^'. In the third configuration, 1.5 looks like 4.5.

You should also be more consistent in your use of connection dots. Using humps with dots pretty much went out of favor 60 years ago.

 

Thanks everyone, so would this be correct then?

Yep! Notice how close the estimate of 16 Ω was.