my doubt goes like this...charge flows when there is a potential gradient...and an electric field exists where there is a potential gradient...electric filed exists in a current carrying coductor...so there must exist a potential drop/potential gradient in a conductor of zero resistance in a circuit...so the potential difference between any two points of a zero resistance...NOW DONT AVOID MY DOUBT BY SAYING THAT THERE IS NOTHING SUCH AS A PERFECT CONDUCTOR ...
NOW consider a simple circuit as shown below:
A----/----B--^^^^^D----------C
| S(switch) R=5 OHMS |
| |
__|__ B |
_ V=10 VOLTS |
| |
|F-----------------------------E|
i wish to discuss this in terms of energy only HELP ME IF GET STUCK SOMEWHERE:
-ABBREVIATIONS:
* ckt----circuit
*+ve----positive
*-Ve----negative
*U,u----potential energy
*K,k----kinetic energy
*e-----charge on an electron=1.6 *10^-19 coulombs
*E-----total energy of any particle
-current will be discussed in terms of POSITIVE charges
-i KNOW that current is established instantaneously in a circuit
-ok
-An electric field is immedately set up in the circuit when switch S is turned on.
-a +ve CHARGE enters the - terminal which tends to reduce the potential of the battery.
-the battery DOES WORK to maintain its potential difference.hence an +ve charge leaves the + terminal(i can explain this point in a little more detail if you wish)
-NOW WE DISCUSS THE FATE OF THE CHARGE IN TERMS OF ENERGY AS IT MOVES IN THE CIRCUIT. when i say charge, i mean +ve charge:unless stated otherwise.
-the battery raises a charge to a potential of 10v by doing work on it.
-the potential energy of the charge is 10e = E
-this charge leaves the + terminal of the battery B with U=10e
-it is accelerated by the electric field so its K is increasing and U SHOULD decrease because k+u should be a constant for any ISOLATED mechanical system.
-this K decreases only when it encounters a resistance.
-now the energy of the charge should be a function of distance x
-hence E(x)=U(x)+K(x)
-so the potential energy of the charge depends on how far it is from the + terminal.
-HENCE there is a different potential/potential energy at any point in the circuit...WHETHER it is HAS a zero or a non zero resistance.
-THIS means that there is a potential drop/gradient across those wires whose
resistance = 0 i.e the connecting wires
-but that is CONTRARY to what is observed...there is NO potential drop across a wire with zero resistance...
-consider this current occurs only when there is a potential difference across
a conductor.
-add a short across R in the above diagram
-ALL the current should flow through the short
-but there is no potential drop across the short...so how does the charge flow through the short.....
give some links to websites that cover science seriously and dont just give half baked explanations...thanks for reading...
NOW consider a simple circuit as shown below:
A----/----B--^^^^^D----------C
| S(switch) R=5 OHMS |
| |
__|__ B |
_ V=10 VOLTS |
| |
|F-----------------------------E|
i wish to discuss this in terms of energy only HELP ME IF GET STUCK SOMEWHERE:
-ABBREVIATIONS:
* ckt----circuit
*+ve----positive
*-Ve----negative
*U,u----potential energy
*K,k----kinetic energy
*e-----charge on an electron=1.6 *10^-19 coulombs
*E-----total energy of any particle
-current will be discussed in terms of POSITIVE charges
-i KNOW that current is established instantaneously in a circuit
-ok
-An electric field is immedately set up in the circuit when switch S is turned on.
-a +ve CHARGE enters the - terminal which tends to reduce the potential of the battery.
-the battery DOES WORK to maintain its potential difference.hence an +ve charge leaves the + terminal(i can explain this point in a little more detail if you wish)
-NOW WE DISCUSS THE FATE OF THE CHARGE IN TERMS OF ENERGY AS IT MOVES IN THE CIRCUIT. when i say charge, i mean +ve charge:unless stated otherwise.
-the battery raises a charge to a potential of 10v by doing work on it.
-the potential energy of the charge is 10e = E
-this charge leaves the + terminal of the battery B with U=10e
-it is accelerated by the electric field so its K is increasing and U SHOULD decrease because k+u should be a constant for any ISOLATED mechanical system.
-this K decreases only when it encounters a resistance.
-now the energy of the charge should be a function of distance x
-hence E(x)=U(x)+K(x)
-so the potential energy of the charge depends on how far it is from the + terminal.
-HENCE there is a different potential/potential energy at any point in the circuit...WHETHER it is HAS a zero or a non zero resistance.
-THIS means that there is a potential drop/gradient across those wires whose
resistance = 0 i.e the connecting wires
-but that is CONTRARY to what is observed...there is NO potential drop across a wire with zero resistance...
-consider this current occurs only when there is a potential difference across
a conductor.
-add a short across R in the above diagram
-ALL the current should flow through the short
-but there is no potential drop across the short...so how does the charge flow through the short.....
give some links to websites that cover science seriously and dont just give half baked explanations...thanks for reading...