Georg Simon Ohm, German, a saw’s son, began his career as a Math teacher, publishing a Geometry treaty. But, from 1822 on, excited about the new discoveries of that time, he got dedicated on to the study of electricity. Besides good knowledge in Mathematics, he had the ability as an experimenter developed with his father at the metallurgy.

Ohm theoretically established the law with his name in 1827. He paired the electric current to the movement of a liquid in a canal, comparing the potential difference to the liquid level. Working in a time when the electric phenomena were unknown, when he announced his law, he clearly defined the electric resistance of a conductor. He himself demonstrated that the resistance of a conductor is directly proportional to its length and inversely proportional to the area of its transversal section. He also dedicated himself to the optical and acoustic, but at those areas he didn’t develop works of the same importance, as at the electricity.

Ohm’s Law is a Math formula that establishes the relation between the three main classes of electricity: the current, the resistance and the voltage (voltage: also known as potential difference).

The electric class is represented by symbols (letters), as follows.

Class |
Symbol |
Unit (SI) |

Voltage |
U or V |
Volt (V) |

Current |
I |
Ampère (A) |

Resistance |
R |
Ohm (Ω) |

Power |
P |
Watts (W) |

- The difference of potential among the terminals of a circuit is the same as the result of the resistance of this circuit by the electric current intensity that passes by that circuit.

Example: In an electric circuit, a current of 25 A when passes by a resistor of 6 ohm, provokes a difference of electric potential of 150 V at the resistor.

Mathematically we have,

- The intensity of the electric current that travels the circuit is the same as the division of the potential difference among the terminals of this circuit by the ohmic resistance that this circuit presents to the passage of the electric current.

Example: In a circuit, when we apply a voltage of 220 V upon the terminals of a resistor of 11 ohm, there is an electric current of 20 A.

Mathematically we have,

This law describes the values that influence at the electric resistance of a conductor, as described below:

“The resistance ( R ) of a homogeneous conductor of constant transversal section is directly proportional to its length and is inversely proportional to the area of its transversal section.

Mathematically we have,

where:

A

As the electric resistance unit is the ohm (Ω), so the adopted unit by the SI for the resistivity is:

With a cylindrical conductor of length L and transversal section A, we will see that its electric resistance is bigger when the length L is bigger and the section A is smaller, and the electric resistance will be smaller when the length L is smaller and the section A is bigger, and also depending on the material the conductor is constituted through a coefficient nominated resistivity (ρ).

This way we can Express the 2nd Ohm’s Law as follows:

The electric current is a result of free electron movement. When the electric current circulates the particles that are in movement end up hitting the other parts of the conductor that are in rest, causing a kinetic energy variation that will generate a heating effect. This heating effect is named Joule’s effect.

The heating at the wire can be measured by Joule’s law that is mathematically expressed by:

Where:

i = current intensity

R = conductor resistance

t = time that the current travels the conductor

This relation is valid considering that the intensity of the current is constant during the time interval of its circulation.

The electric power dissipated by a conductor is defined as the amount of thermal energy that passes through it during a time interval.

The unit used for energy is watt

When considering that all the energy transformed in a circuit is result of the Joule’s effect, we admit that the energy transformed in heat is the same as the energy given by a charge

As:

So:

Therefore:

But we know that , and consequently,

Exemplifying:

Which is the current that passes in a shower of 5400 W in a place where the voltage at the grid is 220 V?

I = P/U |
I = 5400/220 |
I = 24,54 A |

By the 1st Ohm’s Law we have that , so we can define two ways that relate the electric power with the resistance.

P = R.i² |
e |
P = U²/R |

For the previous example the ohmic resistance of the shower will be:

R = 220²/5400

Resistors Association

Associate resistors in series means link them in a unique route, in other words:

As there is only one way for the passage of the electric current, this is measured through all the circuit extension.

Now the potential difference between each resistor will vary according to its resistance, in order to obey the 1st Ohm’s Law, like this:

This relation also can be obtained by the analysis of the circuit:

Thus, for

Analyzing this expression, that’s the total voltage and the current intensity are kept, it’s possible to conclude that the total resistance is:

In other words, a way to sum up and remember the properties of a circuit in series is:

Voltage (U) | Divided in each resistor |

Current Intensity (i) | Is the same in all the resistors |

Total Resistance (R) | Algebraic addition of the resistances |

Link a resistor in parallel basically means divide the same current source, in a way that the voltage in each point is kept. In other words:

The parallel associations are usually represented by:

According to the picture, the total intensity

According to the 1st Ohm’s Law:

And by this expression, that the current intensity and the voltage are kept, we can conclude that the total resistance in a circuit in parallel is given by:

For

Re = 60/22

A mixed association consists in a combination, in the same circuit, of series and parallel associations, as an example:

In each circuit section, the voltage (U) and the current intensity are calculated based on what is known about series and parallel circuits, and in order to simplify these calculations it’s possible to reduce the circuits, using resultant resistors for each section, in other words:

As:

R1 = 20 Ω |
R2 = 40 Ω |
R3 = 50 Ω |

1/Re1 = 1/R1 + 1/R2 |
1/Re1 = 1/20 + 1/40 |
therefore: Re1 = 13,33 Ω |

Re = Re1 + R3 |
Re = 13,33 + 50 |
therefore: Re = 63,33 Ω |