# Please tell solution of this resistance problem

Three resistors are connected in form of circle. Resistance between X and Y is 2 ohm. Resistance between Y and Z is 3 ohm and resistance between X and Z is 5 ohm. Find resistances R1, R2, R3.

2 лайка

When we find the resistance between the X and Y nodes, then the equivalent circuit will look like this:

Knowing the formulas of parallel connection we get the systems:

R_{xy} = \frac{R_{1}\cdot(R_{2}+R_{3})}{R_{1}+R_{2}+R_{3}} = 2 \text{Ohm}\\
R_{xz} = \frac{R_{2}\cdot(R_{1}+R_{3})}{R_{1}+R_{2}+R_{3}} = 5\text{Ohm}
R_{yz} = \frac{R_{3}\cdot(R_{2}+R_{1})}{R_{1}+R_{2}+R_{3}} = 3\text{Ohm}
7 лайков

Here will be a system of three equations:
First:

R(x-y) =R1*(R2+R3) /R1+R2+R3

Second:

R(y-z) =R3*(R1+R2) /R1+R2+R3

And the third:

R(x-z) =R2*(R1+R3) /R1+R2+R3

After that, you just decide this system.
If you don’t understand where these equations come from, just imagine that at points (x or y or z) you are connected to the network: two resistors are connected in series and the same two resistors are connected in parallel to another resistor.

P. S. Thanks for my Google translater

4 лайка

Yes, you right! Good work!

2 лайка

my answers are coming out as R1 = R2 = R3 = 0

2 лайка

There are 2 other set of answers as well.

3 лайка

These are wrong values

1 лайк

yeah there will be 3 values because equations are 3

2 лайка

я же тебе кинул ссылку на удобную статью про использование теха на аске, а ты проигнорировал) эх

3 лайка

Эх, тебе не понять…

Мне лень…

1 лайк

i cross checked the answers are 2,3 and infinite… so I think I have done some calculation error… can you find the answer what you are getting?

2 лайка

Sure, just a minute

2 лайка

What answer did you get?

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I tend to consider that there is a typo in the condition of this problem, namely a typo related to the values of equivalent resistances, because taking the third equation from the second, and then equating the result to the first, it turns out that either R_{1} or R_{3} is zero, which of course is incorrect.
Anyway, all such problems are solved in this way

3 лайка

oh ok.

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It’s harder than I thought((

Me and the photomath can’t solve it(

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Ouch
Swap the second and first pictures

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yeah i plugged it into wolfram and got some weird results involving square roots as well.

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The only thing I can help is throw you this funny joke:

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Note that we can turn this system into Y-formed 3 pole curcuit with R_x, R_y, R_z resistances all connected to the center. So @Amir’s equations can be turned into

R_{xy} = R_x+R_y, \quad R_{xz}=R_x+R_z, \quad R_{yz} = R_y + R_z,

or

R_x = \frac{R_{xy}+R_{xz}-R_{yz}}{2}, \quad R_y = \frac{R_{xy}+R_{yz}-R_{xz}}{2}, \quad R_z=\frac{R_{xz}+R_{yz}-R_{xy}}{2}.

It’s much easier to solve this system that way. So I get (using R'=R_xR_y+R_yR_z+R_xR_z for simplicity) and

R_1 = \frac{R'}{R_z}, \quad R_2 = \frac{R'}{R_y}, \quad R_3 = \frac{R'}{R_x}.

For this problem you should know Y-Δ transform.

6 лайков