**11**
Tush Watts
**·**2009-09-11 06:03:11
Let the equivalent capacitance b/w points A and B is given by C,

The equivalent capacitance of the ladder b/w A and B is independent of the no of sections in b/w ,

C = [(4 C) / (4+c)] + 2

Therfore C=4 uF

**Anonymous**
**·**2018-06-13 22:40:21
**assuming that the c value dominates (n-1) combinations of capacitors and taking the equivalent capacitance as itself c then you can get the value of c **

**AS C=(4C/4+C) (as c is in series with 4f) +2f(as it is parallel with **(4C/4+C**)**

**3**
msp
**·**2009-09-11 06:20:04
hey tush can u explain for y u have taken C_{eq}=C.

**11**
Tush Watts
**·**2009-09-11 06:42:04
@msp

Sice it is an INFINITE series, so ,there will be negligible change if the arrangement is done as shown in the figure.

**3**
msp
**·**2009-09-11 06:50:16
ya i know dat tush.

where u have explained dat taking C_{eq}=C means dat that the equivalent capacitence of the ladder betweeen A and B becomes independent of the sections in betweennnnn

**11**
Tush Watts
**·**2009-09-11 06:56:54
Since it is an **INFINITE** series, so the equivalent capacitance b/w A and B is independent of the no of sections.(As I told in the earlier post).

**1**
yes no
**·**2009-09-11 08:44:22
Now if the problem is twisted and at subsequent steps(strating from 2nd 4f capacitor) 4F is replaced by 2F,1F,1/2F,1/4F,1/8F respectively,means the value is halved at each step,then can u solve it.??

**1**
RAY
**·**2009-09-11 08:59:30
A finite ladder is constructed by connectiong ...so guys plz dun take it as a infinite ladder!!!!

**19**
Debotosh..
**·**2009-09-11 09:03:42
THE UNKNOWN CAPACITOR SATISFYING THE GIVEN REQUIREMENTS IS (C)

.....the logic is to connect (C) across the terminal ends so that capacitance across the previous terminal also becomes (C)...this reduces the circuit to

thus , {(C+C_{2})C_{1}}/ {C_{1}+C+C_{2}} = C

...THIS GIVES A QUADRATIC IN (C)...which gives (C) as one of its roots....

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