# Energy Considerations In A Circuit

Discussion in 'General Electronics Chat' started by steeve_wai, Sep 24, 2007.

1. ### steeve_wai Thread Starter Active Member

Sep 13, 2007
47
0
ENERGY TRANSFERS IN A CIRCUIT:

http://www.school-for-champions.com/science/images/elect_ohms_law-dc_circuit.gif

LET VOLTAGE OF CELL IS 10V.
Consider this:

AN ELECTRIC FIELD IS SETUP IN THE WIRE.
THE FORCE DUE TO THIS FIELD ACCELERATES THE CONDUCTION ELECTRONS IN THE WIRE.
AN ELECTRON(S) NEAR THE + TERMINAL ENTERS THE +TERMINAL OF THE CELL.
THIS TENDS TO DECREASE THE POTENTIAL OF THE POTENTIAL OF THE CELL.
HENCE THE CELL DOES WORK TO MOVE AN ELECTRON FROM + TERMINAL TO – TERMINAL , AGAINST THE INTERNAL ELECTRIC FIELD OF THE CELL.

THIS ELECTRON HAS BEEN “ELEVATED” TO A HIGH POTENTIAL ENERGY AT THE – TERMINAL.

LET THE ENERGY OF THIS ELECTRON BE “E”, E=U+K
WHERE U AND K ARE THE POTENTIAL AND KINETIC ENERGIES OF THE ELECTRON RESPECTIVELY.

AT THE –TERMINAL,U=10e,(e is the charge on an electron)
AND K=0.

THIS K WILL INCREASE DUE TO ACCELERATION BY FORCE DUE TO ELECTRIC FIELD IN THE WIRE.BUT IT IS DECREASED DUE TO COLLISIONS WITHIN THE METAL.

ITS AVERAGE VELOCITY IS (½)MV^2 , WHERE V IS THE DRIFT VELOCITY OF THE ELECTRONS,WHICH IS SMALL IN MAGNITUDE.
LET THIS KINETIC ENERGY BE k1.

HENCE AT ANY POINT IN THE WIRE E=U+k1.

NOW, WHEN THE ELECTRON ENCOUNTERS A RESISTANCE,THE GAINS IN KINETIC ENERGY ABOVE k1 WILL CAUSE HEATING OF THE RESISTOR.
BUT WHAT HAPPENS TO THE POTENTIAL ENERGY U=10e.
DOES THIS U GET LOST IN HEATING UP OF THE RESISTOR OR IS IT SOMETHING ELSE THAT DECREASES THIS U TO 0 ONCE THE ELECTRON LEAVES THE RESISTANCE.

DOES POTENTIAL DROP ACROSS ACROSS A RESISTOR MEAN THE DROP IN POTENTIAL ENERGY OF THE ELECTRON ONCE IT LEAVES THE RESISTANCE OR IS IT SOMETHING ELSE.
IF YES,WHAT CAUSES THE DROP IN POTENTIAL ENERGY.

those INTERESTED,PLEASE COMMENT...DONOT GIVE ANY ANALOGIES OR ANY NON TECHNICAL ANSWERS...I AM EXPECTING A MORE "MATURE" ANSWER.THANK YOU.

2. ### cumesoftware Senior Member

Apr 27, 2007
1,330
15
Actually the electron that goes to the positive terminal of the battery is not promoted. It is rather absorved through a redox reaction, where another electron is released to the negative terminal of the battery. The fact that electrons are being absorved by a reduction reaction, promotes a charge diference that motivates electrical current, which by its turn motivates the oxidation reaction and release of electrons on the negative side. The electrons don't circulate inside the battery, only protons (depending on the battery).

3. ### nanovate Distinguished Member

May 7, 2007
665
1
If I Use All Caps Am I Worthy To Answer Your Question?

4. ### thingmaker3 Retired Moderator

May 16, 2005
5,072
8
No, it has not.

In point of fact the electron in question probably never travels into the supply. It simply bumps another electron out of place in the crystal lattice of the conductor.

Current isn't about single electrons on some epic journey around the circuit, it is about vast swarms of electrons which collectively pass the charge around the circuit. Each one travels a tiny distance before bumping another along to continue the process. Only a fraction of the available electrons move at all. (The volume of copper needed for a coulomb of electrons is only 73 cubic micrometers.) Its not a marathon, its a relay race!

5. ### steeve_wai Thread Starter Active Member

Sep 13, 2007
47
0

ENERGY TRANSFERS IN A CIRCUIT:
see the image below

http://www.school-for-champions.com/sci  ircuit.gif

LET VOLTAGE OF CELL IS 10V.
Consider this:

AN ELECTRIC FIELD IS SETUP IN THE WIRE.
THE FORCE DUE TO THIS FIELD ACCELERATES THE CONDUCTION ELECTRONS IN THE WIRE.
AN ELECTRON(S) NEAR THE + TERMINAL ENTERS THE +TERMINAL OF THE CELL.
THIS TENDS TO DECREASE THE POTENTIAL OF THE POTENTIAL OF THE CELL.
HENCE THE CELL DOES WORK TO MOVE AN ELECTRON FROM + TERMINAL TO  TERMINAL , AGAINST THE INTERNAL ELECTRIC FIELD OF THE CELL.

THIS ELECTRON HAS BEEN ELEVATED TO A HIGH POTENTIAL ENERGY AT THE  TERMINAL.

LET THE ENERGY OF THIS ELECTRON BE E, E=U+K
WHERE U AND K ARE THE POTENTIAL AND KINETIC ENERGIES OF THE ELECTRON RESPECTIVELY.

AT THE TERMINAL,U=10e,(e is the charge on an electron)
AND K=0.

THIS K WILL INCREASE DUE TO ACCELERATION BY FORCE DUE TO ELECTRIC FIELD IN THE WIRE.BUT IT IS DECREASED DUE TO COLLISIONS WITHIN THE METAL.

ITS AVERAGE VELOCITY IS (½)MV^2 , WHERE V IS THE DRIFT VELOCITY OF THE ELECTRONS,WHICH IS SMALL IN MAGNITUDE.
LET THIS KINETIC ENERGY BE k1.

HENCE AT ANY POINT IN THE WIRE E=U+k1.

NOW, WHEN THE ELECTRON ENCOUNTERS A RESISTANCE,THE GAINS IN KINETIC ENERGY ABOVE k1 WILL CAUSE HEATING OF THE RESISTOR.
BUT WHAT HAPPENS TO THE POTENTIAL ENERGY U=10e.
DOES THIS U GET LOST IN HEATING UP OF THE RESISTOR OR IS IT SOMETHING ELSE THAT DECREASES THIS U TO 0 ONCE THE ELECTRON LEAVES THE RESISTANCE.

DOES POTENTIAL DROP ACROSS ACROSS A RESISTOR MEAN THE DROP IN POTENTIAL ENERGY OF THE ELECTRON ONCE IT LEAVES THE RESISTANCE OR IS IT SOMETHING ELSE.
IF YES,WHAT CAUSES THE DROP IN POTENTIAL ENERGY

6. ### recca02 Senior Member

Apr 2, 2007
1,211
1
Hi Steeve,
it is rather hard to say if work is done against the electric field in case of a battery. the electrons are generated when ions are formed.
about your question regarding the potential energy
the u in the eqn shud change while k might remain constant.
since potential is dropped across the length of the wire(now do not go thinking about superconductivity since i think the laws of physics will be quite different in that case-i think in that case K will keep on increasing)
it is the energy lost by electron not the electron KE that actually turns up as heat energy or manifests itself in form of work in case of other machines.
and where do u think this energy comes from?-the potential energy.
it is true that potential drop across resistance is drop in potential energy.
as to what causes it is something which can be explained via an analogy.
consider a high pressure and a low pressure point.
fluid will flow from hi to lo and pressure drop will always be there whenever a resistance is encountered more will be the drop for more resistance and remaining will be available as velocity head.

7. ### billbehen Active Member

May 10, 2006
39
1
As the electron progresses around the circuit, the potenetial energy is converted to kinetic energy, which is usually dissipated in the resistive load via physical contact with the molecules in the resistive material.

Potential energy is expended as the electron accelerates (F = ma.) The carrier of energy in atomic interactions is the photon: one is emitted by the electrical source's atoms (the chemical reaction mentioned above) and absorbed by the electron, increasing its energy. As the elctron has no excited state at such low energies (that would require quark interactions) the only possibility is an increase in kinetic energy! The e- speeds up, usually encounters an atom in the resistive mat'l and spits up the photon, increasing the K.E. of the atom/molecule it hits. As this atom is trapped mechanically in some resistive compound, it vibrates more rapidly! I.e. it gets slighly warmer....