# How could a bulb light up if aĺl the voltage is lost in the resistor?

I recently came up with this slight confusion about voltage loss in resistors/resistance. So for example My circuit has a 12V battery, a 2 Ohms resistor and a Bulb. If we use the V=I.R formula then a current of 6A(amperes) will flow through the circuit, then i remembered that V=I.R is the voltage loss due to a certain resistance, if we try to insert this formula to the resistor, then 12V will be lost in the resistor(6A×2 ohms), so how would the bulb turn on if all the voltage is wasted in the resistor? *I'm still new to this topic, so please give me a clear simplified explanation, any answers would be appreciated.

• The bulb also has resistance. – HiddenBabel Oct 7 '18 at 4:20

if we try to insert this formula to the resistor, then 12V will be lost in the resistor(6A×2 ohms),

Voltage is electric potential energy per unit charge, measured in joules per coulomb ( = volts). It is often referred to as "electric potential", which then must be distinguished from electric potential energy by noting that the "potential" is a "per-unit-charge" quantity.

So the loss of volts means that energy has been deposited on the resistor.This energy following the quantum mechanical behavior of the specific resistor will become heat or light , if it is an incandescent lamp.

Voltage isn't "lost" in the resistor. Voltage provides the driving force to cause electrons to flow through the circuit. The resistance does what it says: it resists the flow of electrons by using atoms in the resistor to temporarily stop the electrons that are accelerating under the electric field that is in the circuit, and that electric field is measured in voltage drop per meter of length. When these accelerating electrons hit atoms in their path, they impart their kinetic energy onto the atom that stopped them, leading to heat in the resistor. The electron is then accelerated again by the electric field, and it continues until it strikes another atom in its path.

For a light bulb, the heat produces an increase in temperature that is high enough to cause the bulb's filament to glow. The power produced by the bulb is given by the voltage drop across the bulb multiplied by the current flowing through the bulb, or $$P=IV$$.

• So what about power lines? Some people told me that voltage is lost in the powerlines due to the resistance – user208870 Oct 7 '18 at 11:47
• I'm somehow still not getting it. Isn't V=I×R , the voltage lost ? So if you try increasing the resistance more voltage(electrical energy be lost)? {I came up with this question because in one of my textbook, there's a circuit with 2 ohms of resistance, a bulb, and the resistance of the bulb is probably negelected} – user208870 Oct 7 '18 at 12:02
• @KedagiSport, no. $V=IR$ is the voltage drop across a given circuit element. This voltage drop results in power production in that circuit element, in the form of heat. – David White Oct 7 '18 at 17:01
• @KedagiSport, are you reading a book on the subject of voltage and power? Are you trying to learn this stuff on your own, or in a class room? – David White Oct 7 '18 at 17:03
• On my own, so after voltage passes a resistor, a voltage drop happens, right? So how will the bulb light on if all the voltage is 'dropped' across the resistor. (Ex before passing the resistor=12V, after passing the resistor=0V? – user208870 Oct 8 '18 at 3:11