Ohm's Law

Back in 1827 a great discovery was revealed by Georg Ohm.

Georg Simon Ohm (1787-1854) was a German physicist (electrician) and mathematician. Making experiments and using his
equipment,

he realized that different conductive materials he used, showed him that current gets through them with different difficulty
according to

the potential difference. *(Voltage)*.

In other words, __the current in a conductor is directly proportional to the voltage and inversely proportional to the resistance.__

he well known ** Ohm's Law**. The name was given as a memorial to Georg Ohm discovery.

Below it's Ohm's life timeline. It was taken from thefamouspeople.com

**Georg Simon Ohm (1787-1854)**

Ok... let's go now more deeper to analyze the Ohms Law.

In circuit analysis, three equivalent expressions of Ohm's law are used interchangeably:

**
(1) I = V / R
(2) V = I * R
(3) R = V / I **

The interchangeability of the equation may be represented by a triangle, where

and the

‟

The line that divides the left and right sections indicate multiplication, and the divider between the top and bottom sections indicates division (hence the division bar).

So... if you want to find voltage, then cover

As we said above,

If you want to find current, then cover

As we also said above, since

If you want to find resistance, then cover

Finally, since

We can now make some examples to see how this law works, on a simple circuit i drew below.

Ok lets's say we have a power source of **5V**

and a resistor value of R1** 500Ohms**

We want to calculate how much currnet will pass through the resistor....

Solving for... **( I )** We have **I = V / R = 5 / 500 = 0.01A ** or **10mA**

Now let's try it in a different approach...

suppose we know the power source of **5V** and the current of **10mA**

We want to calculate how much is the resistance of the resistor....

Solving for... **( R )** We have **R = V / I = 5 / 0.01 = 500Ohms**

And finally, this time we are looking for the voltage.

We know the current that is **10mA** and the resistance of R1 **500Ohms**

Solving for... **( V )** We have **V = I x R = 0.01 x 500 = 5V**

Electrical Power, **(P)** is the amount of energy that is absorbed or produced within a circuit.

A source of energy ** (Voltage)** will produce or deliver power while the connected load absorb it.

Light bulbs for example, absorb power and convert it into heat and light.

Power is the

Now again, three equivalent expressions using of Ohm's law are used interchangeably:

(2) I = P / V

(3) V = P / I

The interchangeability of the equation represented on the triangle, can be solved as the above example that was given before.

It is worth mention another two types for finding Power....

Including Ohms law to this equation, we can create two more equations...

-------------

P = I

P =

-------------

P = V

Now suppose we have this circuit, and we know that values are same as above...
Source: **5V / ** Current: **0.01A / ** R1:**500Ohms **

We can simply calculate how much power dissipation will be on **R1** using

**P = I * V = 0.01 * 5 = 0.05Watts**

**P = I ^{2} * R = 0.0001 * 500 = 0.05Watts**

I believe that right now, we can make a brief report to ** Energy. **

As we said above, Electrical Power, ( **P** ) is the amount of **energy** that is absorbed or produced within a circuit.

*( Taken from wikipedia )*

** ‟ In physics, energy is a property of objects which can be transferred to other objects or
converted into different forms, but cannot be created or destroyed.**

It is becoming very clear why resistors and other components heat up.....

A resistor transforms electric energy into heat. So the above circuit transforms

This energy is called

But even when energy transforms, exploiting energy for the main goal of a component, there'll still be losses to other forms of energy.

For example: LEDs, motors, lamps etc, will still produce heat as a byproduct of their other energy transfers.

Remember... for each component there is a maximum power dissipation rating that they can dissipate.

Keeping these values on low operating ratings, you'll avoid the destruction of the component.