'Theory of relativity' and 'The principle of conservation energy' Is it possible to distroy energy ?

ajanta said:
The + terminal of battery at the lower potential is the terminal of Y super conductor too. But the battery is at higher potential. Will the electrons not gain potential energy when they flow through the Y super conductor to the battery at higher potential ? But why ?
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I do not understand what you are talking about.
 
exchemist said:
I do not understand what you are talking about.
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Electrons coming down from the battery at higher potential to at lower potential. But when they will return to the battery at higher potential from lower potential then will they gain potential energy ?
 
The resistance of the bulb filament is 12ohm. Potential difference of battery is 12 V. So 1A current can flow through the filament at lower potential in a second. But in order to transform 48J energy of at higher potential to heat and light power at lower potential, 2A current will flow through the 12ohm filament in 1second. How can it possible ?
 
ajanta said:
The resistance of the bulb filament is 12ohm. Potential difference of battery is 12 V. So 1A current can flow through the filament at lower potential in a second. But in order to transform 48J energy of at higher potential to heat and light power at lower potential, 2A current will flow through the 12ohm filament in 1second. How can it possible ?
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You are being silly.
 
The definition of current is the amount of charge passing a point in a circuit per unit time. 1 Ampere means 1 Coulomb of charge passing a point per second.

Let's say you hook up a battery to a light bulb with long superconducting wires, so that the battery is in space and the bulb is at a lower gravitational potential (e.g. on the ground, perhaps). Let's say the battery pumps x Coulombs of charge per second (i.e. x Amps) into the wires (and collects x Amps at its other terminal). What current flows through the bulb?

I think it might well be x Amps. Why? On the one hand, clocks run slower at the bulb than at the battery. But on the other hand, the charge density is also higher at the bulb, I think.
 
Its really horrible to me because the electrons are not gaining potential energy. If the nature does this to obey conservation of energy so its really really horrible !
 
What kind of potential energy are you talking about? As danshawen (I think) pointed out above, for every electron in the circuit that flows down from the battery in space to the bulb on the ground there is another electron that flows up from the bulb to the battery.
 
James R said:
The definition of current is the amount of charge passing a point in a circuit per unit time. 1 Ampere means 1 Coulomb of charge passing a point per second.

Let's say you hook up a battery to a light bulb with long superconducting wires, so that the battery is in space and the bulb is at a lower gravitational potential (e.g. on the ground, perhaps). Let's say the battery pumps x Coulombs of charge per second (i.e. x Amps) into the wires (and collects x Amps at its other terminal). What current flows through the bulb?

I think it might well be x Amps. Why? On the one hand, clocks run slower at the bulb than at the battery. But on the other hand, the charge density is also higher at the bulb, I think.
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I think it should be 2A at lower potential. I thought about it after post my previous reply. Because, at the lower potential if we measure the resistance of the bulb it will be 12 ohm and if we measure the potential difference between two terminal at lower potential it should be 24V. Because only 24V can flow 2A current through the filament to conserve power. And extra potential difference is came from the potential energy of electrons at higher potential.
 
James R said:
What kind of potential energy are you talking about? As danshawen (I think) pointed out above, for every electron in the circuit that flows down from the battery in space to the bulb on the ground there is another electron that flows up from the bulb to the battery.
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Yes I'm talking about 'there is another electron that flows up from the bulb to the battery'. Thanks.
 
A lot of skewed reasoning this thread. A conservative and steady-state i.e 'conveyor-line' transfer of energy from one static elevation to another in a static gravitational field will be tallied differently at the two elevations. The lower elevation recording a greater change in energy δW than at the upper elevation (and naturally of opposite sign). Purely owing to the differing gravitational potentials - the δW ratio corresponding to that of the two (wavelength) redshift values. Doesn't matter if the transfer is via mechanical or EM means. The elevation-dependent elapsed time for a given net energy transfer is the inverse of the δW ratio i.e. shorter at the lower elevation. Hence the ratio of power gain/loss goes as the square of δW ratio i.e as the inverse square of the (frequency) redshift values at the two elevations.

Note this is a 'perceived' discrepancy owing to differential in clock rates and energy is not lost or gained on any net basis - just redistributed within the notionally closed system.
 
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