An electric current made of tungsten with a length of 15 meters and a cross-sectional area of 0.15 millimeters

An electric current made of tungsten with a length of 15 meters and a cross-sectional area of 0.15 millimeters (squared) flows along a spiral of an electric current equal to 1.1A, what power will be released if two spirals are connected in parallel.

Given:

l = 15 meters – the length of the spiral of the hotplate;

s = 0.15 mm2 – spiral cross-sectional area;

r = 0.055 Ohm * mm2 / m is the resistivity of tungsten;

I = 1.1 Ampere – current strength;

n = 2 is the number of spirals connected in parallel.

It is required to determine W (Watt) – electrical power.

Let’s find the resistance of one spiral of an electric stove:

R = r * l / s = 0.055 * 15 / 0.15 = 0.825 / 0.15 = 5.5 Ohm.

Then the total resistance of two spirals connected in parallel will be equal to:

1 / Rtot = 1 / R + 1 / R;

1 / Rtot = 2 / R;

Rtot = R / 2 = 5.5 / 2 = 2.75 Ohm.

The power will be equal to:

W = U * I = I * I * R = I ^ 2 * R = 1.1 ^ 2 * 2.75 = 1.21 * 2.75 = 3.3 Watts (the result is rounded to one decimal place).

Answer: the power will be equal to 3.3 watts.