# ENERGY CONSIDERATIONS IN A CIRCUIT(help needed)

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#### steeve_wai

Joined Sep 13, 2007
47
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:

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)

#### beenthere

Joined Apr 20, 2004
15,819
The chemical actions in a battery concentrate positive chage at the anode and negative at the cathode. The amount of charge determines the potential between the terminals.

In a wire, the charge carriers are electrons. Current flows from the cathode to the anode in the external circuit.

Assuming no resistance in the wire, there is no potential difference in the wire (no voltage drop). But, unless we are dealing with a superconductor, that is never the case in the real world. Condider your statement about current in a short circuit - the conducting path opens because of melting. It heats because of internal resistance. That resistance increases because of the heating, so more power is dissipated in it until it melts and opens up.

#### beenthere

Joined Apr 20, 2004
15,819
I am going to lock this thread, as it essentially parallels the one above.

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