Consider a simple circuit with a cell and a bulb.Electrons constantly keep flowing from negative terminal of battery to positive,passing through the wires and bulb.i know that electrons move in the wires,and if it is moving it must have some potential energy.BUT electric potential difference is defined at work done in moving per unit charge.why is it so?if work is done then do we consider it as the product force and distance travelled here.AND IS REALLY ANY WORK DONE HERE TO MOVE ELECTRONS?ELECTRONS IN A CONDUCTOR KEEP FLOWING THEMSELVES.
Voltage is potential difference but I don't thnk that should be defined as work done. Work done would be watts, voltage multiplied by current. It takes energy to create a potential difference, like lifting a weight up. Once it's there it's ready to do work which uses up that energy.
Work done is always equal to energy transferred.work done is not measured in watts,but power is measured in watt AND power is defined as rate of doing work or rate at which energy is transferred by a device .
There is power lost in the transmistion through the wires (except in superconducting wires). You realize that individual electrons move very little in a conductor right?
Work=Force*Distance If you lift a 1 lb weight 6 feet (away from the center of the earth) you will have done 6 ft-lb of work. Of course, work can't be accomplished in zero time, it must have taken time to accomplish the work. Power=Work/Time A watt is a unit of measure of power, like the unit of measure HP is a unit of measure of power. Energy=Power*Time A watt-hour is a unit of measure of energy. Your electric bill states how many watt-hours (or kilowatt-hours) of energy you used. So basically, a light bulb that is operated at 120 volts and draws .5 amps is rated at 60 watts of power (120*.5=60). If you operate that bulb for 1 hour you consume 60 watt-hours of energy.
Vpp: Work is done in an electric circuit one way or another. If it's high-efficiency (ingeniously designed, for example) then most of the work is done on the load (lamp, motor, computer, whatever it is). Otherwise, it's done on the load plus the wasted amounts given up to losses. A common loss is the power lost in wires themselves - as Origin noted - which requires superconductivity in order to eliminate it, or nearly so. You are correct in stating that Work = Force x distance. However, that is only one of several definitions, the one that applies to mechanical work. Just as energy can exist in many forms (electrical, chemical, mechanical, thermal, pneumatic and so on) then work can be done in all of those forms as well. You are correct that we measure power in Watts. For mechanical power, 1 W = 1 kg m s[sup]-2[/sup] x 1 m = 1 kgm[sup]2[/sup]s[sup]-2[/sup]. However, for electrical power 1 W = 1 V x 1 A. This simply means that 1 Watt of power is required to move 1 amp across a gradient of 1 volt. 1 amp = 1 Coulomb per second: 1 A = 1 Cs[sup]-1[/sup], or 6.24E18 electrons/sec (e[sup]-[/sup]s[sup]-1[/sup]). So 1 watt of power is used to move 6.24E18 electrons per second across a gradient of 1 volt. Conversely, if takes (6.24E18)[sup]-1[/sup] watts of power to move 1 electron across a gradient of one volt. That's 1.6E-19 watts to move 1 electron across 1 volt. And yes, electrons are flowing in your simple circuit that connects a battery to a lamp. The amount of current that flows is determined by Ohm's Law, amps = volts รท resistance. That's how many Coulombs per second of current, or how many electrons per second that flow through the loop, and that's how many electrons are moved across the potential difference across the battery terminals. Therefore work is done in an amount equal to the battery voltage times the current, or voltage squared divided by resistance of the lamp. Note, you could get to this also by looking at energy. Each second that the circuit burns 1 W of power it uses 1 Joule of energy: 1 J = 1 W x 1 s = 1 V x 1 A x 1 s = 1 V x 1 C s[sup]-1[/sup] x 1 s = 1 V x 1 C. This just means that once a Coulomb of electrons have been moved across a potential difference of 1 V, then 1 J of energy has been used. When the battery was charged with 1 J of energy (or when 1 J worth of chemical energy was assembled into it at the factory, such as a dry cell) then that amount of energy is transferred to the lamp, and after passing through the lamp, those electrons are returned to their ground potential, 0V, that is, to the "low side" of the battery, having given up their energy to the lamp.
