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

Answer:

4.89 Ω

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

$r = \frac{R}{3}\\r = \frac{44}{3}$

The three pieces are connected in parallel,

$R_{p}$ = Resistance of parallel combination of three pieces

Resistance of parallel combination is given as

$\frac{1}{R_{p}}= \frac{1}{r} + \frac{1}{r} + \frac{1}{r} \\\frac{1}{R_{p}}= \frac{3}{r}\\R_{p}= \frac{r}{3}\\R_{p} = \frac{\frac{44}{3} }{3}\\R_{p} = \frac{44}{9} \\R_{p} = 4.89 ohm$

Answer 2

Answer:

$\frac{R}{1} = \frac{44}{9}\ohm$

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

$\frac{1}{R} = \frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$

$\frac{1}{R} = \frac{3}{44}+\frac{3}{44}+\frac{3}{44}$

$\frac{1}{R} = \frac{9}{44}$

⇒$\frac{R}{1} = \frac{44}{9}\ohm$