Question

3. Consider a wire of length L and area of cross-section A. We cut up the wire into 4 equal
parts (we mean equal lengths) to get wires W1,W2,W3 and WA. Now, each wire W; (where
i = 1,2,3,4) is stretched to i times its length. Then we put the larger three resistances in
parallel and combine this in series with the smallest resistance. If the effective resistance is
Re = ar/b, where R is the resistance of the original wire and a,b are positive integers with
(a,b) = 1, then a - b is:
WW
M
WW​

Answer 1

Solution :

The wire originally has a length L and area of cross section A .

Resistance of the wire :

> ρ L/A

Now, this wire is cut into 4 equal lengths to form the wires , W1, W2, W3 and W4 .

The resistance of each of these 4 wires will be equal .

As it is cut into 4 parts, the length of each piece of the wire becomes 1/4 th of what it originally was .

Resistance Of Each of the wires :

> ρ L/4A

So, Resistance of Wₙ = ρ L/4A

Now, each wire is stretched to i times the original length where i ∈ [ 1, 4 ]

So, the resistance of the wires become :

W₁ = ρ L/4A

W₂ = ( ρ L/4A ) x 2 = ( 2ρ L/4A )

W₃ =  ( ρ L/4A ) x 3 = ( 3ρ L/4A )

W₄ =  ( ρ L/4A ) x 4 = ( 4ρ L/4A )

The three larger resistances are now put into parallel combination .

Thus , W₂, W₃, and W₄ are placed in parallel and their resultant is placed in series with W₁ .

Let us find the equivalent resistance in the parallel combination :

$\dfrac{1}{R_eq}$  = 1/W₂ + 1/W₃ + 1/W₄

> (4A / 2ρ L ) + (4A / 3ρ L ) + (4A / 4ρ L )

> A/ρL [ 4/2 + 4/3 + 4/4 ]

> A/ρL [ 2 + 4/3 + 1]

> 13A/3ρL

So, R_equivalent becomes :

> 3ρL/13A

This is in series with another resistance,  W₁ = ρL/4A

Effective resistance :

> 3ρL/13A +  ρL/4A

>  25ρL/52A

> [ 25/52 ] ρ L/A

Now, the resistance of the original wire is ρ L/A .

So, effective resistance Re =  [ 25/52 ] R

Thus ,

a = 25

b = 52

( a - b ) = -27 .

This is the required answer.

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