Physics/Role of Electric charge and battery in lighting of bulb
Dear Sir, I am a civil engineer working in a hydropower project and only for my interest in electrical energy, I want to be clear about the question as I put it for your valuable response. This question is related to the flow of electrons and the function of a battery (source of emf). Now I would like to ask you about the role of electric charge on electrons in lighting up an electric bulb (or any resistor in the circuit). I have read that the electric bulb lights up because of the electrical energy, transformed from the produced chemical energy by the source (battery) from chemical reactions inside and the electrons take this energy as electrical energy to the resistor for their respective functioning (in this case lighting up of the bulb) and converted to other forms (in this case of bulb converted to light energy and heat energy) and the electrons return back to the battery and the cycles run this way. From here I understand that the electrons are only the transporting agents. If I am correct (may be not) then what is the role of the electric charge (-ve) on the electrons in the lighting up of the bulb. I also know that the current is the flow of charge (electrons with negative charge) and I=Q/t (where Q = number of electrons * elementary charge on each electrons). How can I relate the lighting of the bulb with the charge on electrons.
Or if I am correct, then this chemical energy produced in the battery due to chemical reaction is only responsible for and consumed in the pushing of the electrons, whereas the electric bulb lights up because of the charge of the electron. If this is the case, do the electric charges dissipate in producing the light and heat energy or they remain constant with their electrons. How can I relate the lighting of the bulb with the charge on electrons?
Which one of my understandings is correct. Please clarify.
I have no problem with the summary you wrote before continuing with your first questions. Now, your first question: what is the role of the electric charge (-ve) on the electrons in the lighting up of the bulb?
The reason the electrons flow through the bulb is that they are pushed from the negative terminal(and pulled toward the positive terminal) due to the charge on the electrons and the emf (which generates an electric field). So the role of an electron's charge is that it aids lighting your bulb by causing the electron to move out of the negative terminal and through the circuit toward the positive terminal of the battery.
Your next question: How can I relate the lighting of the bulb with the charge on electrons?
In your summary you discussed the conversion of energy from chemical energy, to electrical energy, and to heat and light. The resistance of the bulb resists the flow of the electrons. Energy must be spent, or dissipated, to get them to go through that resistance. If the resistance of the bulb is R and the current is I, energy is being dissipated (or spent) at a rate (which we call power) that is given by
P = I^2*R
Power is the rate of dissipating energy, (or generating energy) so during a time t, the energy dissipated is P*t. So, substituting the above formula, the energy dissipated in time t is given by
Energy_dissipated = I^2*R*t
So using the formula you gave relating the number of electrons flowing in time t, we see that energy dissipated in the bulb is related to the number of electrons flowing through the bulb in time t. Work is done (energy is spent) to accomplish the act of pumping the electrons through your bulb. The result is that the bulb gets hot enough to glow, giving off light. This all happens because of the action of the electric field on the charges those electrons carry.
Next question: then this chemical energy produced in the battery due to chemical reaction is only responsible for and consumed in the pushing of the electrons?
Yes. If the battery gets weaker so that it does not produce energy at the same rate as before, the rate of consuming energy by the bulb will also have to decrease.
Then you also said: whereas the electric bulb lights up because of the charge of the electron.
That is only part of the story - the electric bulb lights up because of the energy dissipated in the bulb. That energy becomes heat and light.
Next question: If this is the case, do the electric charges dissipate in producing the light and heat energy or they remain constant with their electrons?
No. The electrons continue to have their charge. Without the charge, the electrons would not be pulled back to the positive terminal so those electrons can be recycled through the circuit - if you leave the circuit running.
Last question: How can I relate the lighting of the bulb with the charge on electrons?
Now I hope you see that the electric field and the charge on the electrons is the reason the electrons flow through the circuit and that the energy dissipated in the bulb in this process is the reason for the light.
I hope this helps,