Question

A 4 ohm resistance wire is doubled on it. Calculate the new resistance of the wire.

Answer 1

Given:

Resistance of the wire, R= 4 ohm

let the length = L

and area = A

$\rm R = \frac{\rho L}{A}$

$\rm 4 = \frac{\rho L}{A}$ ----> (1)

If the wire is doubled on it, then the new length = L/2

as the length decreases, the area is doubled ie. 2A

let the new Resistance = R'

$\rm R' = \frac{\rho L/2}{2A}$

$\rm R' = \frac{\rho L}{4A}$

$\rm R' = \frac{1}{4} (\frac{\rho L}{A})$

$\rm R' = \frac{1}{4} × 4$ {from, eqn (1)}

$\rm R' = \frac{\rho L}{A}$

$\rm R' = 1 ohm$ (answer)

Answer 2

Given :-

• R = 4 ohms.

To be calculate :-

• New resistance of the wire.

Solution :-

We are given ,

R = 4 ohms .

When a wire is doubled on it , it's length would become half and area of cross would double . That is , a wire of length " l " and area of cross section A change to length l/2 and area of cross section 2A $.$

As you know ,

⇒ R = PL/A 

⇒ R' = P(L/2) / 2A

Where R' is the new resistance .

So,

R'/R = {P(L/2) / 2A} / {PL/A} 

After calculating , we get :-

 R' /R = ¼ 

⇒ R' = R/4

⇒ 4/4

⇒ 1 Ohm.

Hence , the required new resistance of the wire is 1 Ohm.

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