@WBahn
.
While you're waiting - you need a hat..

Goin' for 500

 

@JohnlnTX
Looks more like an altitude than an aptitude query.

 

Do not confuse || with absolute value notation wherein the vertical lines bound an argument e.g. |x|

Got it?

|| sign for parallel an what this mean |x| ?
i have studied it is for +ve value.

 

@JohnlnTX
Looks more like an altitude than an aptitude query.

An aptitude for altitude, I'd say.
36 to go.

 

Aptitude for altitude, I'd say.
36 to go.

|x| Represents the absolute value (i.e. the unsigned magnitude) of the argument x...

 

|x| Represents the absolute value (i.e. the unsigned magnitude) of the argument x...

Non-sequiter but thanks for the info.
34 to go

 

|| sign for parallel an what this mean |x| ?
i have studied it is for +ve value.

It really depends on what 'x' is, but generally it means something related to "the size of x". If x is just a number, the |x| is the absolute value of x, or simply the "positive" version of x. If x is a vector, the |x| is the magnitude of the vector. If x is a set, then |x| is the cardinality of the set, which just means the number of items in the set. If x is a string, then |x| means the length of the string. I'm sure there are many other uses, but the common theme is that it is a positive number that captures the notion of the "size" of x.

 

Non-sequiter...

Oh! I don't think so! Please see post #463

 

@RRITESH KAKKAR

By way of dispelling any confusion please consider the fact that |1±i| =√2 ---- Got it?

(Where "i" = the imaginary unit)

 

Hence:
Rp'q' = "1 Ω + (1 Ω || Req) + 1 Ω"
means:
Rp'q' = 1Ω in series with (1Ω in parallel with Req) in series with 1Ω yes , know.

 

Hence:
Rp'q' = "1 Ω + (1 Ω || Req) + 1 Ω"
means:
Rp'q' = 1Ω in series with (1Ω in parallel with Req) in series with 1Ω yes , know.

So... once again I offer this 'sledgehammer' clue:

Merely write an algebraic equation for: 1Ω ||Req=Req-2Ω

 

So... once again I offer this 'sledgehammer' clue:

What is this?

 

What is this?

It means that I'm almost giving you the answer

 

@WBahn
.View attachment 96475
While you're waiting - you need a hat..

Goin' for 500

djsfantasi actually beat you to it! I just changed my avatar to the one he sent me a few days ago.

 

So, Rpq=Rp'q'
Rp'q' = "1 Ω + (1 Ω || Req) + 1 Ω"


Rpq= 1+.75+1=2.75

If Ro is the resistance between P' and Q' (in the right half of the bottom figure), then what is the resistance, Req, between P and Q. This is a problem that you have already solved. Req is the series sum of two resistors of resistance R plus a third resistor, of resistance R, that is in parallel with Ro. Use that to write an equation for Req in terms of R and Ro.

Rpq is not just the resistance of the three resistors that we split off. Rpq IS the same resistance as Rp'q'.

Merely write an algebraic equation for: 1Ω ||Req=Req-2Ω

1Ω x Req / 1+Req = Req-2
Req =( Req - 2)(1+Req)

 

djsfantasi actually beat you to it! I just changed my avatar to the one he sent me a few days ago.

Looks great! Can @spinnaker 's eyepatch be far behind?
24
Sorry, I'm done.

 

Req =( Req - 2)(1+Req)

Good! -- Now solve for Req

 

Merely write an algebraic equation for: 1Ω ||Req=Req-2Ω

1Ω x Req / 1+Req = Req-2
Req =( Req - 2)(1+Req)

So close! (and, if you are willing to abuse units and the rules of mathematics -- which you obviously are -- then you got it).

1Ω x Req / 1Ω+Req = Req-2Ω

This addresses the units, and is even dimensionally sounds, but is not correct because you are forgetting order of operations. Because multiplication/division are done before addition/subtraction, what you have here is really

((1Ω x Req) / 1Ω) + Req = Req-2Ω

That's not what you intended, but it IS what you wrote. One of your biggest problems has been sloppiness with the math. You need to work on being consistently correct in your presentation of math expressions. What you meant to write was

(1Ω x Req) / (1Ω + Req) = Req - 2Ω

The next step, which you carried out successfully (ignoring units) is

(1Ω x Req) = (Req - 2Ω)(1Ω + Req)

Now just multiply everything out and solve for Req.

 

...if you are willing to abuse units and the rules of mathematics -

@RRITESH KAKKAR --- Please recall that we're dealing with units (CIP Ohms) as opposed to abstract values --- That said you are at the veritable 'doorstep' of victory!

 

Hint: As I've stated earlier this is a second degree function...