Question

9. A 6 ohm resistance wire is doubled on itself. Calculate the new resistance of the wire.​

Answer 1

R1=(rho×l)/A..(i)

initial resistance of the wire=6 ohm

when it is doubled on itself then length is halved and area of cross section is doubled then the new resistance is given by

R=(rho×l/2)2A

R=(rho×l)A×1/4

Rnew=(1/4)R1. (from (i) )

Answer 2

The new resistance of the wire is 1.5 ohm.

Step-by-step explanation:

Given : A 6 ohm resistance wire is doubled on itself.

To find : Calculate the new resistance of the wire. ?

Solution :

The resistance is given by,

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

Here, Resistance=6 ohm

$6=\frac{\rho L}{A}$

In second case,

Resistance is $R'=\frac{\rho L'}{A'}$

As the wire is doubled,

So $L'=\frac{L}{2}$ and $A'=2A$

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

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

$R'=\frac{1}{4}\times 6$

$R'=1.5$

Therefore, the new resistance of the wire is 1.5 ohm.

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