IIT JEE , AIEEE Forum - Physics , Electricity

adwaIIT

Consider a wire of length "L", cross sectional area "A" and resistivty"

". The wires are arranged so as to form an infinite series of equilateral triangles as shown. Find the equivalent resistance between points "a" and "b". (The area of cross section and resistivity is same for all wires)

refer here

pink_ele

is this infinite or finite series dear ?????????

suresan

pink_ele

here ,

make those ends where curren t simply flows be free den

if pointing end of each circle is free

thus

in infinite sereis

one resistance doesnt effect

therefore

if x is eq resistance

n r is resistace of one side den

in parralel

1/r 1=1/(x)+1/(r)

r1=rx/(r+x)

now in series

r2=rx/(r+x)+r=2rx+r ²/(r+x)

finally taking parralel for final side n thus d eq resistance

1/x=(r+x)/2rx+r ²+1/r=3x+2r/(2rx+r ²)

thus solving

x=r/3

hope its correct

adwaIIT

the answers wrong..................

let me clarify the prob.........its an infinite series of triangles (one inside the other)

Give it another try...................
its easy

adwaIIT

the answer's wrong..................

let me clarify the prob.........its an infinite series of triangles (one inside the other)

Give it another try...................
its easy

konichiwa2x

Is the answer

konichiwa2x

Let

.

This problem can done by using the method of symmetry.

(If A and B are connected to battery and AB or it's perpendicular bisector is a line of symmetry, then all points lying on the perpendiculars drawn to AB are at same potential).

In the question, the perpendicular to AB is a line of symmetry. Hence the current through AF is equal to the current throguh FB which means no corrent flows from F to C or D. This implies that even if we break the connection at F, the currents in the circuit will not be affected. : Note that I have drawn the circuit only for the topmost layer of triangles.

Now calculate the resistance between CD.

Now simplifying the circuit, it consists of AB parallel to another branch (of resistance(

) .

Hence,

Now for the next layer of triangles, repeat the above procedure to see that

Now we see a pattern. For any particular layer is the resistance is

, the next layer the resistance is

.

So in effect the circuit is this:

Hence,

ohms

where

.

konichiwa2x

Of course effective resistance cant be negative. So we reject the negative root.

Hence X =

konichiwa2x

dude, is the answer correct?

adwaIIT

that's correct.........

well done

Ask & Discuss Questions with Community & Experts