Question

The resistance of a very fine aluminum wire with a 14 μm × 14 μm square cross section is 800 Ω . A 800 Ω resistor is made by wrapping this wire in a spiral around a 3.7-mm-diameter glass core.

How many turns of wire are needed?

Answer 1

Solution :

First of all we find rhe length of the aluminium wire for this use Ohm's law

R = p l/ A

Here l = length and A cross section area and p is resistivity

P for aluminum is 2.65 x 10 ^-8 ohm metre

RePlacing all the given data In the foRmula

We get = length = RA/p

Hence l = 5.92 metres = 5920 mm

Now number of turns = length of wire/ diameter of rod

Hence number of turns = l /2πr

So number of turns = 5920 *2 / 2x3.14x3.7

Which gives= 254.77*2 turns or in round figure approx 509.24 turns.