@RRITESH KAKKAR

For purposes of this exercise, the "ladder's" 'infinite extent' is significant only in that such results in 'R' compounded with additional stages yet equaling 'R'

 

How to solve this?

 

How to solve this?

Write an equation stating the relationship between the 'first stage' and the semi-infinite network --- Then solve the equation!

 

@RRITESH KAKKAR -- I must sign off -- Please take your time, think about the problem -- and I'll chat with you tomorrow - or next week -- whatever works for you

Best regards
HP

 

OK,
Rpq=1+1+1
Rp'q'=1ohm+1//Req + 1ohm
something like this

 

How to solve this?

If this is the approach that you are going to always take, namely give up and ask how to solve it, then stop wasting your time, money, and effort pursuing a technical education. The reason is quite simple -- anyone that receives a technical education is doing so in preparation for a career of solving problems! That becomes the value that you bring to the table and that you expect others to pay you for -- being able to solve their problems for them. You can't succeed at that unless you can actually solve problems without having to always ask someone to show you how to do it!

 

OK,
Rpq=1+1+1
Rp'q'=1ohm+1//Req + 1ohm
something like this

Yes, something like that, except that Rpq is not 1 Ω + 1 Ω +1 Ω (and I am once again getting tired of having to tell you repeatedly to track your units. 1+1+1 is a number, it is not a resistance).

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

So work with

Rp'q' = 1 Ω + (1 Ω || Req) + 1 Ω

Now apply the formula for combining two resistors in parallel.

Then adjust the equation to account for the fact that Rp'q' is, in fact, equal to Req.

 

Ω

how insert this sign?

 

How did you insert it just now

 

i have paste it.....

 

Rp'q' = 1 Ω + (1 Ω || Req) + 1 Ω

Now apply the formula for combining two resistors in parallel.

Rp'q'= 1ohm + 1ohm || (Req) +1ohm
Rp'q'= 1ohm x 1ohm || (Req=R1xR2/R1+R2) +1ohm
Rp'q'= 1ohm x 1ohm || .75ohm +1ohm
Rp'q'= 1ohm x 1ohm || .75ohm +1ohm

R1=3 ohm
R2=1 ohm
Req=R1xR2/R1+R2
Req= 3/4=.75ohm

 

@RRITESH KAKKAR

Note that the symbol || is 'short hand' for "in parallel with"

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

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

Got it?

 

Ω

how insert this sign?

1) Click on the large, bolded "S" on the format bar -- This will display a 'ribbon' of symbols below the text entry box.
2) Click the Ω symbol on said ribbon -- this will insert the symbol at the cursor's position in the text entry box.


Best regards
HP

 

So work with

Rp'q' = 1 Ω + (1 Ω || Req) + 1 Ω

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

@RRITESH KAKKAR

FWIW here's the same advice stated otherwise:

IF the first stage was actually disconnected from the semi-infinite network then Rpq = 3Ω and Rp'q'=the network resistance.

However the first stage is connected to the semi-infinite network so Rpq=Rp'q'= the network resistance.

The key to the solution is vested in the fact that the network resistance is the same despite the presence or absence of the first stage (and, for that matter, any finite number of preceding stages).

So: Let the network resistance ='R'

2Ω in series with (1Ω in parallel with R) = R ------ All you need do is express the forgoing algebraically, solve the resulting second degree equation for 'R', discern the applicable root and you've got it!

BTW -- Kudos on your eschewal of recourse to analysis! -- Your discipline in that regard is truly inspiring!!!

Sincerely
HP

 

Ω

how insert this sign?

If you have a keyboard with a number pad you can hold down the Alt button and type 2, 3, 4 (written as Alt-234). This only works with a numeric keypad (at least that's been my experience).

You can also click the button with the "S" on the far right of the edit window tool bar and select the symbol you want. This will paste it into the top left corner of the edit window (not at the insertion point, which is very annoying) where you can cut and paste it to where ever you want.

 

1) Click on the large, bolded "S" on the format bar -- This will display a 'ribbon' of symbols below the text entry box.
2) Click the Ω symbol on said ribbon -- this will insert the symbol at the cursor's position in the text entry box.


Best regards
HP

Does it insert the symbol at the cursor position for you? It used to (before the switch from vBulletin), but now (for me) it always inserts it as the first character in the text window and I have to move it manually.

 

Rp'q' = 1 Ω + (1 Ω || Req) + 1 Ω

Now apply the formula for combining two resistors in parallel.

Rp'q'= 1ohm + 1ohm || (Req) +1ohm
Rp'q'= 1ohm x 1ohm || (Req=R1xR2/R1+R2) +1ohm
Rp'q'= 1ohm x 1ohm || .75ohm +1ohm
Rp'q'= 1ohm x 1ohm || .75ohm +1ohm

R1=3 ohm
R2=1 ohm
Req=R1xR2/R1+R2
Req= 3/4=.75ohm

You are STILL insisting on using values that you KNOW are WRONG!

Forget this damn 3 Ω. You KNOW that if the total resistance were three ohms, then that would mean that the total resistance is 2.75 Ω and since you know it can't be both 3 Ω and 2.75 Ω then you KNOW that it is NOT equal to 3 Ω. So STOP using 3 Ω !!!!

Req is an unknown! It IS the value that you are trying to solve for!

Rp'q' is also unknown, but you know that it is ALSO equal to the value that you are trying to solve for!

What is (1 Ω) in parallel with (Req)? Remember, Req is an unknown, so it needs to stay as Req in your equation.

 

Does it insert the symbol at the cursor position for you? It used to (before the switch from vBulletin), but now (for me) it always inserts it as the first character in the text window and I have to move it manually.

FWIW -- In my case symbols are placed at the cursor despite user interface style or method of character selection -- perhaps said behaviour varies with local display settings?

Best regards
HP

 

What is (1 Ω) in parallel with (Req)? Remember, Req is an unknown, so it needs to stay as Req in your equation.

@RRITESH KAKKAR

I wish to add my voice to @WBahn 's in asserting that this is a nice, unencumbered approach -- So... Merely write an algebraic equation for: 1Ω ||Req=Req-2Ω

 

HP why not allow convergence tests? I say you are being the wet blanket to always insist on math being done the hard way! I know the problem has easy algebra solution but that's not point! I say why walk when can ride!