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.
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.