I WOULD LIKE TO DISCUSS ELECTRICITY IN TERMS OF ENERGY ELECTRIC FIELD AND POTENTIAL.PLEASE CORRECT ME IF IAM WRONG.
http://www.school-for-champions.com/science/images/elect_ohms_law-dc_circuit.gif
-let battery voltage=10v and 0 internal resistance
-let R=5 ohms
KEY
e- negative charge
e+ positive charge
+ve positive
-ve negative
U potential energy
K kinetic energy
E total energy
now we will discuss the circuit in terms of electrons:
_____A BRIEF DIGRESSION HERE____
-there is a "BATTERY FORCE"(due to chemical reactions),F1 (SAY), that moves an e- from +ve terminal to -ve terminal WHEN the battery is connected to a circuit.
-HENCE THE DIRECTION OF F1,WHISCH ACTS ON AN e-,IS DOWNWARDS
-there is an electric field due to oppositely charged plates of the cell.this field causes an e- to experience a force F2 (say).
-F2 ACTS UPWARDS
-IN EQUILIBRIUM,i.e. WHEN the cell is'nt connected to a circuit,F1=F2.no
movement of charge occurs.
-the battery has set up an electric field in the wire.
-an e- NEAR THE + TERMINAL enters the cell/battery.
-it TENDS to reduce the potential of the +terminal.hence F2 decreases
-now the "battery force" F1 becomes dominant and moves an e- from + terminal to - terminal by doing WORK ON THE e-
-the e- has been moved to a position where its U=10e=U1,LET
-now it is accelerated by the electric field and E=U1+K(x),where k is a function of distance x.
__________CONSIDER THE TWO CASES THAT ARISE_______
---A--- R=0
---B--- R>0
---A--- if R=0,
THEN the e- goes to the + terminal of the cell.its energy profile is as follows:
U=0 (i am a little doubtful about this)
K=k
-the e- will be accelerated indefinitely when R=0
---B--- R>0
-the e- encounters a resistance,its U=U1 will decrease and K will EFFECTIVELY be a constant,say k (remember,that drift velocity is a constant in such cases) .of course the instantaneous values of K may differ from k.
K averages to k because electric field tries to increase K but collisions with the ions in metal decreases K.
-so it is the U1 and the changes in K from k which cause heating of the resistor.
-power(P)=I^2 * R...right.
-hence
P=f(U) for a circuit with a give resistance...
the above equation seems quite reasonable because.U AND R decide the current and the K...
CONCLUSIONS:
1.POTENTIAL DROP OCCURS ONLY WHEN R>0
2.e- ARE ACCELERATED INDEFINITELY IN A SHORT (POWER DISSIPATED WILL BE HIGH BUT NOT INFINITE...THE SHORT SIMPLY MELTS)
NOW...A FEW DOUBTS
IS THE POTENTIAL IN A CURRENT CARRING CONDUCTOR(R=0) CONSTANT AND
NON-ZERO.(IF YES THEN MY ABOVE EXPLANATIONS SEEM TO BE OKAY)
HONESTLY IAM MYSELF NOT CONVINCED ABOUT MY EXPLANATIONS,ESPECIALLY ABOUT POTENTIAL ENERGY...ANYWAY THANKS FOR READING PATIENTLY...
http://www.school-for-champions.com/science/images/elect_ohms_law-dc_circuit.gif
-let battery voltage=10v and 0 internal resistance
-let R=5 ohms
KEY
e- negative charge
e+ positive charge
+ve positive
-ve negative
U potential energy
K kinetic energy
E total energy
now we will discuss the circuit in terms of electrons:
_____A BRIEF DIGRESSION HERE____
-there is a "BATTERY FORCE"(due to chemical reactions),F1 (SAY), that moves an e- from +ve terminal to -ve terminal WHEN the battery is connected to a circuit.
-HENCE THE DIRECTION OF F1,WHISCH ACTS ON AN e-,IS DOWNWARDS
-there is an electric field due to oppositely charged plates of the cell.this field causes an e- to experience a force F2 (say).
-F2 ACTS UPWARDS
-IN EQUILIBRIUM,i.e. WHEN the cell is'nt connected to a circuit,F1=F2.no
movement of charge occurs.
-the battery has set up an electric field in the wire.
-an e- NEAR THE + TERMINAL enters the cell/battery.
-it TENDS to reduce the potential of the +terminal.hence F2 decreases
-now the "battery force" F1 becomes dominant and moves an e- from + terminal to - terminal by doing WORK ON THE e-
-the e- has been moved to a position where its U=10e=U1,LET
-now it is accelerated by the electric field and E=U1+K(x),where k is a function of distance x.
__________CONSIDER THE TWO CASES THAT ARISE_______
---A--- R=0
---B--- R>0
---A--- if R=0,
THEN the e- goes to the + terminal of the cell.its energy profile is as follows:
U=0 (i am a little doubtful about this)
K=k
-the e- will be accelerated indefinitely when R=0
---B--- R>0
-the e- encounters a resistance,its U=U1 will decrease and K will EFFECTIVELY be a constant,say k (remember,that drift velocity is a constant in such cases) .of course the instantaneous values of K may differ from k.
K averages to k because electric field tries to increase K but collisions with the ions in metal decreases K.
-so it is the U1 and the changes in K from k which cause heating of the resistor.
-power(P)=I^2 * R...right.
-hence
P=f(U) for a circuit with a give resistance...
the above equation seems quite reasonable because.U AND R decide the current and the K...
CONCLUSIONS:
1.POTENTIAL DROP OCCURS ONLY WHEN R>0
2.e- ARE ACCELERATED INDEFINITELY IN A SHORT (POWER DISSIPATED WILL BE HIGH BUT NOT INFINITE...THE SHORT SIMPLY MELTS)
NOW...A FEW DOUBTS
IS THE POTENTIAL IN A CURRENT CARRING CONDUCTOR(R=0) CONSTANT AND
NON-ZERO.(IF YES THEN MY ABOVE EXPLANATIONS SEEM TO BE OKAY)
HONESTLY IAM MYSELF NOT CONVINCED ABOUT MY EXPLANATIONS,ESPECIALLY ABOUT POTENTIAL ENERGY...ANYWAY THANKS FOR READING PATIENTLY...