anyway you put Rpq=Req right?

 

Remember this?

For ax^2+bx+c=0

x=(-b±√(b^2-4ac))/2a

 

anyway you put Rpq=Req right?

We're way past that! -- Please concentrate! -- You're almost there!!!

 

(1Ω x Req) = (Req - 2Ω)(1Ω + Req)
(1Ω x Req) = Req+Req2-2Ω-2Req
Req-Req = Req2-2Req -2Ω
Req2-2Req -2Ω=0

 

I must be the only 'normal' person posting here! 16 to go.

 

For ax^2+bx+c=0
x=(-b±√(b^2-4ac))/2a
a=2 b=2 c=-2
x=(-2±√(2^2-42x-2))/2 x 2

 

I must be the only 'normal' person posting here! 16 to go.

For what ?

 

anyway you put Rpq=Req right?
We're way past that! -- Please concentrate! -- You're almost there!!!

but why, i forgot. because it is infinite ?

 

TS beat me to it. 11 to go! Your turn JohnInTX.

 

For ax^2+bx+c=0
x=(-b±√(b^2-4ac))/2a
a=2 b=2 c=-2
x=(-2±√(2^2-4x2x-2))/2 x 2

for +ve
x=2.47213

 

@boatsman -- you're not helping!

 

@RRITESH KAKKAR

Please try re-working the problem -- you're very close! I'll give you further assistance if needed...K?

 

My humble apologies HP.

 

@RRITESH KAKKAR

You know that
Req = (Req - 2Ω)(1Ω + Req)

So substitute 'X' with Req in the quadratic equation (shown in post #482)

 

My humble apologies HP.

Pleased to hear it -- I thought maybe I'd have to send you to the head teacher's office!!!!

 

@RRITESH KAKKAR

Please try re-working the problem -- you're very close! I'll give you further assistance if needed...K?

What is wrong part?

 

What is wrong part?

Please see post #494 -- should things 'go astray' again -- we'll 'dig into it' ...K?

 

@RRITESH KAKKAR

To enhance clarity please refrain from using "x" to indicate multiplication -- where you feel the need to explicitly indicate a product please use '*' --- Thanks!

 

Req = (Req - 2Ω)(1Ω + Re
q)

let Req =x
x=(x-2)(x+1)
x=x2 +x -2x -2
x2+x-x-2x-2=0
x2-2x-2=0

For ax^2+bx+c=0
x=(-b±√(b^2-4ac))/2a

a=1, b=-1,c=-1

 

(1Ω x Req) = (Req - 2Ω)(1Ω + Req)
(1Ω x Req) = Req+Req2-2Ω-2Req
Req-Req = Req2-2Req -2Ω
Req2-2Req -2Ω=0

Don't write Req² as Req2. That appears either like a different quantity -- the resistance eq2 -- or as Req multiplied by two. The normal text expression is to use the carat symbol, '^', for exponentiation.

And, once again, start tracking your units!

(1 Ω)(Req) = (Req)(1 Ω) + (Req)^2 - (2 Ω)(1 Ω) - (2 Ω)(Req)
(Req)(1 Ω) + (Req)^2 - (2 Ω²) - (2 Ω)(Req) - (1 Ω)(Req) = 0
(Req)^2 + (Req)(1 Ω) - (2 Ω)(Req) - (1 Ω)(Req) - (2 Ω²) = 0
(Req)^2 + (Req)[1 Ω - 2 Ω - 1 Ω] - (2 Ω²) = 0
(Req)^2 - (2 Ω)(Req) - (2 Ω²) = 0

If I can be bothered to take the time to do it properly, why can't you?

Now apply the quadratic formula, ask yourself which of the two solutions makes sense, and you are done.