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.