# energy transfers in an electric circuit

by steeve_wai
Tags: circuit, electric, energy, transfers
 P: 5 ENERGY TRANSFERS IN A CIRCUIT: http://www.school-for-champions.com/...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 (˝)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.
P: 457
Think of electrons as a compressible gas or, better yet, as water flowing through open troughs. That way you can see how kinetic energy (speed of the flow) and potential energy of each particle (height of the water) work together.

EMF (voltage) is electron pressure (analogous to the height of the water). The potential energy for a single electron is, as you say, proportional to the EMF. (Please note that the potential energy for a container of gas is proportional to the square of the pressure, but that is because pressure implies more particles have been pushed into the container of the gas or that the momentum of each particle has increased proportional to the velocity, which varies with the square of kinetic energy. So you should not use the square of voltage to calculate the potential energy in a single electron!)

Current is, of course, Coulombs per second, which is analogous to liters per second for the water.

The battery is a pump that raises the level of the water. The wires are open troughs through which the water can flow. The resistor is a waterfall. Most of the potential energy is wasted in the resistor as heat (turbulence in the water).

With this approach i think you can solve the problem.

By the way...

 AT THE –TERMINAL,U=10e,(e is the charge on an electron)
This is incorrect. An electron volt is a unit of energy, and is not to be confused with the fundamental charge. An electron volt is the kinetic energy gained by an electron that "falls" (accelerated by the electric field) through one volt.
 P: 2,237 THIS REMINDS ME OF MY DAYS AS A STUDENT (MID 70s) WHERE THE COMPUTER PRINTOUT AND TELETYPE PRINTOUT WAS ALWAYS CAPS. ALL MY FORTRAN AND BASIC PROGRAMS HAD NOTHING BUT CAPITAL LETTERS (AND DIGITS AND OPERATORS, ETC). EVERYTHING WE TYPED WAS IN CAPITAL LETTERS. SOMEHOW THAT LOOKED NORMAL TO US AT THE TIME. ___________________ R B-J
P: 5

## energy transfers in an electric circuit

 Quote by rbj THIS REMINDS ME OF MY DAYS AS A STUDENT (MID 70s) WHERE THE COMPUTER PRINTOUT AND TELETYPE PRINTOUT WAS ALWAYS CAPS. ALL MY FORTRAN AND BASIC PROGRAMS HAD NOTHING BUT CAPITAL LETTERS (AND DIGITS AND OPERATORS, ETC). EVERYTHING WE TYPED WAS IN CAPITAL LETTERS. SOMEHOW THAT LOOKED NORMAL TO US AT THE TIME. ___________________ R B-J
IS THAT all YOU can DO ?

i need answers...what you said was of no help

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