Main Effects of Electric Current

The temperature of the conductor rises and communicates heat to its surroundings.
The conductor surrounds itself with a magnetic field and exerts forces on other currents or on magnets.
The current, when passing through certain substances, breaks them down chemically (electrolysis).
Ohm's law: The intensity of the current that passes through a conductor is directly proportional to the difference in potential applied to its ends and inversely proportional to the resistance of the conductor.

l = A
V = V
R = W

l = V / R

EXAMPLE OF OHM'S LAW APPLICATION:

Calculate the current intensity of an electrical circuit that has a voltage of 48 V and a resistance of 25 Ohms.

l = V / R = 48 V / 25 Ohms = 1.92 A

Resistivity: The resistance of any material is called by the ratio of the cross-sectional area between the length of the material.

P = (Ohm) m
R = Ohm
A = m²
L = m

P = R (A / L)

Conductor resistance: It is equal to the product of the resistivity constant of the material and the ratio of the length to the surface of the material.
R = P (L / A)

EXAMPLE OF RESISTANCE APPLICATION:

A rectangular iron bar has a cross section of 2 x 2 cm, a length of 40 cm, and a resistivity constant of 1 x 10 Ohm (m). Calculate its resistance.

R = P (L / A) = (1x10^-7 ohm (m)) 0.40 m / (0.02 m) (0.02 m) = 1x10^-4 Ohms

Electric power: It is called the ratio that exists in the electrical circuit between the work performed at the time it is done or the speed with which energy is consumed or produced. Its units are watts (W)
P = ^ W / ^ t = l²R = V²/R=lV

EXAMPLE OF ELECTRIC POWER APPLICATION:

Calculate the power and resistance that an electric current has inside a conductor, where the work is equal to 3.5x10^-4 J in a time of 14 s, if the intensity of the current is 4.2x10^-3 TO.
First we calculate the electrical power.

P = ^ W / ^ t = 3.5x10^-4 J / 14 s = 2.5x10^-5 W

Now we calculate the resistance:

R = P / l²= 2.5x10-5 W / (4.2x10^-3 TO)²= 1.41 Ohms

Joule's Law: The heat that is obtained from a conductor in the unit of time is directly proportional to the square of the intensity of the current that it carries.

Q = cal
t = s
R = Ohm
l = A

Q / t = 0.239 Rl²

JOULE'S LAW APPLICATION EXAMPLE:

Determine the heat produced in a conductor that has a resistance of 38 Ohms and an electrical current intensity of 1.6 A in a time of 25 s.

Q = 0.239 Rl²t = (0.239) (38 ohms) (25 s) = 227.05 cal