Ohm’s Law

1. Analogy with hydraulics

When a consumer is subjected to a voltage difference, an electric current passes through it. The current flowing through the consumer depends on its electrical resistance.

The voltage is expressed in Volt (V) which corresponds to the pressure of a hydraulic circuit.

The intensity is expressed in Ampere (A), it is the flow of “electrons”.

The resistance of an electrical receiver is expressed in Ohms (Ω) , it is the mechanical resistance of a hydraulic component.

To measure the voltage, a voltmeter will be used which will be installed as a bypass (parallel) on the powered circuit.

To measure the intensity, an ammeter will be used which will be installed in series on the powered circuit.

To measure the resistance, we will use an Ohmmeter that we will install in parallel on a component disconnected from the circuit.


2. The formula

It was in 1827 that the German physicist Georg Ohm published his formula.

Ohm's law relates the intensity of the electric current passing through an electric dipole to the voltage across it.

The voltage across a resistor is proportional to the current flowing through it. The value of resistance R is constant and does not vary when the voltage or current is modified.


U = R * I 

U : Voltage in Volt.

R : Résistance in Ω.

I : Intensity in Ampere.

La loi de l ohm

Fig. A, a resistor is subjected to a voltage difference of 12 V. The conventional direction of current is from + to -.

The current through the resistor is:

U = R * I

12 = 120 * I

I = 12/120

I = 0.1 A


3. Useful Formula

P = U * I

P : Power in Watts.

U : Voltage in Volt.

I : Intensity in Ampere.


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The settings

This formula is used to calculate the power of an electrical component. Associated with the formula U = R * I, they make it possible to size an electrical circuit.


4. Application

A customer wants to install a work light on his machine. The power of the headlight is 100 W, the voltage of the battery is 12 V. Calculate the intensity to size the power cable and the fuse.

P = U * I

100 = 12 * I

I = 100/12

I = 8.33 A 

Knowing that the copper cable is sized at 7 A /mm², we will take 1.5mm² wire protected by a 10A fuse.


The customer wants to install a light on his dashboard to indicate that the work light is on. This LED indicator consumes 10 mA at 2 Volts. Calculate the limiting resistor to be installed in series for the LED to work correctly.

Application numerique

The resistance must be subjected to 10 Volt. (12V – 2V = 10V)

10 mA = 0.01 A

U = R * I

10 = R * 0.01

R = 10/0.01

R = 1000 Ω



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