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I know voltage is electrical potential difference between two points. So, I think must be calculated:

Voltage = Electrical Potential at A - Electrical Potential at B

Can someone explain where does the ohms law relation: V = IR comes from in this? Is it possible to calculate voltage with electrical potentials given?

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Voltage is "potential difference", so, yes, if you know that point A is at +5V and point B is at +20V, then the voltage between them is -15V (the sign just says which point has the higher potential).

If you know the current trough a resistor is 2A and the resistance is 10Ohm then the voltage is 20V. This is the potential difference. If you say that one wire of the resistor is grounded (potential 0V) then the other wire will be at +20V or -20V. Or, you can say that one wire of the resistor is at +100V then the other wire will be at +120V or at 80V, depending on which wire ("upstream" or "downstream" you chose as reference).

Potentials are always referenced to some "zero" point - usually the ground.

Practical example - a PC power supply provides (among others) +5V and +12V. Between those wires is 7V potential difference and it is sometimes used to power fans if they are too loud at 12V but too slow at 5V.

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    \$\begingroup\$ Just to avoid confusion, the Ground potential is arbitrarily fixed, and only sometimes it's really connected to the actual ground. Being a relative value, the potential has to be referred to a point to have sense, and for simplicity in every circuit the designer refers always to a "Ground". \$\endgroup\$ – clabacchio Jan 22 '12 at 11:11
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Voltage

Voltage is always a relative measure, never absolute. That is if you only consider a singular point there is no concept of Voltage.

In electronics, there is almost always an implicit reference which is generally referred to as 'ground' and, as the reference, is arbitrarily chosen to be 0 Volts. When someone talks about a voltage at a single point, it is generally assumed to be referenced to the circuit ground (whatever that may be).

Ground may be a "Real Ground" that is tied to the Earth, hence the name or it may be a "Virtual Ground" meaning that it is only the reference locally to the circuit.

An example would be a battery powered device. Generally the negative terminal of the battery is chosen as ground, 0 volts, and used as the reference for all other voltages in the circuit. However, if you were to measure the voltage between the negative terminal of the battery (a virtual ground) and a true Earth ground, you'll probably find it isn't zero.

Ohm's Law

V = IR is a formulation of Ohm's Law doesn't doesn't derive directly from the concept of voltage although voltage is certainly a part of it, along with the concepts of resistivity and current.

Ohm's law wasn't actually based on mathematical analysis, it was found experimentally (which is generally true of all 'laws' in science). Its actually only applicable in a fairly narrow set of situations. Specifically 'ohmic conductors' sometimes called 'linear devices' in electronics, things like resistors. 'Non-ohmic conductors' sometimes called 'non-linear devices' do not obey Ohm's law, things like transistor and diodes.

Technically Ohm's law never really applies outside idealized components, that is no real component actually obeys Ohm's law exactly, it is however a reasonable estimate in many applications.

If you want a more technical derivation you can get Ohm's law to drop out of the Drude model pretty easily and they made us do this in one of my college classes.

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Electric potential at a point has little meaning as it needs a reference and so is always a potential difference. We sometimes use the term with this understand and sometimes ( esp in Physics ) use the term to represent the scalar potential which defines the field ( ignoring any component that should be represented as a curl, which is always 0 in the steady state ( i think ) )

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This is called Kirchhoff's voltage law;

http://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws#Kirchhoff.27s_voltage_law_.28KVL.29

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    \$\begingroup\$ This is not so helpful, you could try to spend more effort when answering. \$\endgroup\$ – clabacchio Jan 22 '12 at 11:05

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