# A 1-mile long wire has a resistance of 44 ohm. What will be its new resistance when it is shortened by cutting it in half (using only 1/2 of the original wire) and what will be its resistance if...

A 1-mile long wire has a resistance of 44 ohm. What will be its new resistance when it is shortened by cutting it in half (using only 1/2 of the original wire) and what will be its resistance if the two halves are combined (twisted together) and used as a single wire?

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The wire with a length of 1 mile has a resistance of 44 ohm. If the wire is cut in half, the resistance of the shortened piece of wire is half the resistance of the original piece of wire. The resistance of a half mile long piece of wire is 22 ohm.

Two half mile pieces of wire created from the mile long piece of wire have a resistance of 22 ohm each. If the two pieces are twisted together to create a single wire this is equivalent to joining the two pieces in parallel. The combined resistance of two wires with resistance R1 and R2 is equal to R where 1/R = 1/R1 + 1/R2. As R1 and R2 here are equal to 22 ohm, the resistance of the wire consisting of the two in parallel is 1/(1/22 + 1/22) = 1/(2/22) = 22/2 = 11 ohm.

The resistance of a half mile long piece of wire is 22 ohm. The resistance of a wire created by twisting together two half mile long pieces of wire is 11 ohm.

`R=rhoL/A`

`=44`

since we are cutting wire ,area of cross esction reamain same .only length will reduced to L/2 ,Let `R_1` be the new resitance

`R_1=rho(L/2)/A`

`R_1=rhoL/(2A)`

`R_1=(1/2)(rhoL/A)`

`R_1=(1/2)*44`

`R_1=22`

when length half resistance will be 22 ohm