hello. The task is simple: i need to find the resistance of the

>>8798387
Draw the corresponding curcuit.

>>8798534
what is that and how do i do that

>>8798587
Fold down to simplify, do PI to T, get 0.8r.

Sum of current entering a node is the sum of current leaving a node.

Draw the direction in which the current flows.

>>8799794
brainlet

>>8798387
symmetries

>>8798387
Saw this yesterday. Can't believe these people are this incompetent. Thought for sure someone would of solved it for you. Sorry, I'm too lazy. That looks like it's take at least 10 minutes.

>>8798387
Most important thing about solving these is recognizing that the equivalent resistance is independent of the voltage. Therefore, you pretend there is a battery connected, draw the current, and then choose an arbitrary value for the voltage so you can more easily solve your system of equations.

Here I drew my current. I labelled each current [math] x_0,x_1, ... x_6 [/math]

Now we choose loops and junctions. In order to solve your system of equations, you will need to make sure your equations are linearly independent. To ensure that, we must choose at least 1 previously unused circuit element in each loop equation, or in the case of the junction rule, at least 1 unused current.

I choose these equations:
[math]
v - rx_0 - rx_3 - rx_5 = 0\\
v - rx_0 - rx_3 - rx_6 = 0\\
v - rx_0 - rx_2 - rx_6 = 0\\
v - rx_1 - rx_6 = 0
[/math]
These are for the loops. We see that there are no remaining unused resistors to use the loop rule anymore, yet we only have 4 equations and 7 unknown

File: 1490645603094.png (902KB, 1055x736px) Image search: [Google]

902KB, 1055x736px

>>8800577
we now use the junction rule. I chose these equations:
[math]
x_0 = x_2 + x_3\\
x_1 + x_2 + x_4 = x_6\\
x_3 = x_4 + x_5
[/math]

Now we have 7 equations and 7 unknowns. 7 unknown currents. Your answer must be in terms of r of course, since they don't tell you what r is. Here's the trick: we can simply choose a value for r and a value for our imaginare battery v. I choose to let r = 1 ohm, v = 12 volts. You can choose literally any values for r and v, but these are simple. You can solve this the tedious substitution way, but good fucking luck with that. Fuck that shit. Put your system into augmented matrix form, with constants one side of equation and coefficients of unknowns on other. Solve your 7 x 7 matrix. Or let wolfram do it, as I have here: https://www.wolframalpha.com/input/?i=row+reduce+%7B%7B1,+0,+0,+1,+0,+1,0,12%7D,%7B1,0,0,1,0,0,1,12%7D,%7B1,0,1,0,0,0,1,12%7D,%7B0,1,0,0,0,0,1,12%7D,%7B1,0,-1,-1,0,0,0,0%7D,%7B0,1,1,0,1,0,-1,0%7D,%7B0,0,0,1,-1,-1,0,0%7D%7D

>>8800579
Almost forgot to mention that this problem is only 7 unknown currents due to symmetry. If you don't see that, you are welcome to use the same method but with 12 unknowns and get the same answer.

We see that:
[math]
I_{total} = x_1 + 2x_0\\
x_0 = \frac{48}{11}\\
x_1 = \frac{72}{11}\\
2x_0 + x_1 = 132/168\\
I_{total} = 168/11
[/math]

Now recognize that V = IR, where we let V = 12 volts and r = 1 ohm:
[math]
v_{battery} = I_{total}R_{equiv}\\
12 = \frac{168R_{equiv}}{11}\\
R_{equiv} = \frac{132}{168}
[/math]

Now we just see that R equivalent is 132/168 = 132r/168

Which is the answer. In decimal form it's approx: 0.785r

>>8800590
Typo:
[math]
I_{total} = 2x_0 + x_1 = 132/168
[/math]

Use Y-Delta transforms and it'll probably be trivial.

>>8799809
no u

File: rnets.png (21KB, 321x413px) Image search: [Google]

21KB, 321x413px

>>8800593
yup
fold, convert, result

>>8800590
where did u find x0 from? (not op)

>>8800735
I set up 7 equations with 7 unknown and then solved it in matrix form. Follow the wolfram alpha link to see the reduced row echelon form of that matrix.

>>8800694
how are you proceeding from the last image?

>>8800735

Actually I see I made a mistake here: >>8800577

The 2nd equation should be
[math]
v - rx_0 - rx_3 - rx_4 - rx_6 = 0
[/math]

Here it is corrected in wolfram: https://www.wolframalpha.com/input/?i=row+reduce+%7B%7B1,+0,+0,+1,+0,+1,0,12%7D,%7B1,0,0,1,1,0,1,12%7D,%7B1,0,1,0,0,0,1,12%7D,%7B0,1,0,0,0,0,1,12%7D,%7B1,0,-1,-1,0,0,0,0%7D,%7B0,1,1,0,1,0,-1,0%7D,%7B0,0,0,1,-1,-1,0,0%7D%7D

And we see now that:
[math]
x_0 = 9/2\\
x_1 = 6\\
12 = I_{total}R_{equiv} = (2x_0 + x_1)R_{equiv}\\
12 = 15R_{equiv}\\
R_{equiv} = \frac{4}{5} \Omega
[/math]

And since we let r = 1 ohm, it follows
[math]
R_{equiv} = \frac{4r}{5} \Omega
[/math]

>>8800761
Another symmetry: no pd across P, no current. You calculate upper branch parallel lower branch (last line).