Question

A length of copper wire has a resistance 44 Ω. The wire is cut into three pieces of equal length, which are then connected as parallel lengths between points A and B. What resistance will this new "wire" of length L0 3 have between points A and B? Answer in units of Ω.

Answer 1

4.89 Ω

Step-by-step explanation:

we know that resistance is directly proportional to length. hence as the wire is cut in three pieces, the resistance of each piece becomes one-third of the original resistance of the wire.

`R` = Resistance of wire = 44 Ω

`r` = resistance of each piece

Resistance of each piece is given as

`r = (R)/(3)\\r = (44)/(3)`

The three pieces are connected in parallel,

`R_(p)` = Resistance of parallel combination of three pieces

Resistance of parallel combination is given as

`(1)/(R_(p))= (1)/(r) +  (1)/(r) +  (1)/(r) \\(1)/(R_(p))= (3)/(r)\\R_(p)= (r)/(3)\\R_(p) = ((44)/(3) )/(3)\\R_(p) = (44)/(9) \\R_(p) = 4.89 ohm`

Answer 2

`(R)/(1) = (44)/(9)\ohm`

Step-by-step explanation:

Let us imagine that there are three wire of length equal length having equal resistances each of 44/3 Ω

Now connect these wires in parallel to so that their equivalent resistance is R.

then

`(1)/(R) = (1)/(R_1)+(1)/(R_2)+(1)/(R_3)`

`(1)/(R) = (3)/(44)+(3)/(44)+(3)/(44)`

`(1)/(R) = (9)/(44)`

⇒
`(R)/(1) = (44)/(9)\ohm`