Question

How does current flow through a simple circuit with one light bulb? Does the light bulb use up the current when it goes through it or is it the same at the end or...? -_-

Answer 1

Current flows through the circuit and the light bulb acts as a resistor, taking up some of the current and converting it into both light/heat (that's why lightbulbs get hot after a while). So no it's not the same at the other end. Current stays the same but you have a different total voltage.

Answer 2

Current is the same, the resistance of the bulb reduced the total current from what would exist if it was a straight wire.

Answer 3

The energy for the light bulb is actually delivered through the air. The metal wires connecting the bulb to the source and the steady current that flows through the wires only set up the conditions for energy to flow through the air from the source to the bulb. Most of the energy actually travels close to the wire but outside it.

This will come as a shock to most because how a circuit actually works is not taught to most EEs and physics students at the undergraduate level. It is one of many great failures of undergraduate education.

Answer 4

The source voltage would remain essentially the same reduced some by the internal resistance of the source.

Answer 5

Regarding the hypothesis that the current flowing through the bulb is delivering the energy, consider that current represents the average velocity of electrons. They have some average kinetic energy. The current entering the bulb is the same as the current leaving the bulb. This means those electrons did not give up any of their energy while travelling through the bulb filament. So where did the energy come from to heat up the element?

Answer 6

In fact the electrons do give up ie. transfer energy to the light bulb. As is generally an simply understood the battery forces a particularly mobile group of electrons (in conductors) through the circuit by creating a difference of potential (voltage). The battery is the source of energy. The electrons forming the current collide with electrons in the filament giving up their kinetic energy . The current is a " river" of electrons drifting from the negative terminal of the battery to the positive. Current is measured in charge units ie Coulombs per second and without reference to their velocity . Do not be shocked, contrary to some beliefs all the energy used by the circuit comes from the battery. All the energy released to the environmnt comes from the battery. Our educational system is working fine.

Answer 7

Yes I should have been more careful in what I said. The energy *is* delivered by collisions of electrons inside the filament and the energy driving the circuit is supplied by the battery. But *none* of that energy is being delivered through the wire.

The act of heating up the filament through collisions means the electrons give up their kinetic energy. If the kinetic energy of electrons flowing in the wire were solely responsible for heating up the element, their speed would fall as they travelled through the element and reach 0 velocity when they got to the - terminal of the battery (for conservation of energy reasons). This is not the case. The average speed of electrons in this DC circuit is constant throughout the wire and inside the bulb filament. What actually happens is energy is being delivered into the wire and filament through the air to maintain the drift velocity of the electrons. This is not taught at an undergraduate level** and is a serious omission IMO.

Circuit analysis is taught using a lumped parameter model which really amounts to a conservation of energy method. It's the right way to analyze low frequency circuits, no question, but it's akin to solving simple mechanics problems using conservation of energy without even talking about F=ma and kinematics. F=ma tells you what is going on, PE=KE does not although it is a very good way to come up with solutions to many problems.

I will post two drawings shortly that show how energy is transferred to the bulb in a dc battery driven circuit.

Answer 8

The first diagram shows the E field in space of an incomplete circuit. One end of the wire is connected to the + terminal and the other end is left open. Initially when the uncharged wire makes contact with the + terminal, electrons are pulled out of the wire and into the + terminal of the battery until the wire reaches the same potential as the + terminal with respect to the - terminal. After this movement is done, there is an electrostatic situation where the wire is at the same potential as the + terminal of the battery. That net +ve charge on the wire is distributed to the wire surface. The potential difference between the wire and the - terminal of the battery means an E field exists and the E field lines will originate on the + charges of the wire and terminate on the - charges of the - battery terminal.

Here's an approximate diagram:



The important characteristic that I drew badly is that the E field lines are perpendicular to the conducting surfaces, like the wire.

The purpose of this diagram is to make it easier to understand the charge distribution on the wires of a complete circuit, which is next.