Unit of power loss is Wattwhat is the unit of "power loss"
can you provide a circuit diagram ?
The internal resistance and other variables are not provided, so we need to assume them as zero.I think you might need to factor in 2 variables. Change of resistance of the copper wire with temperature.
Internal resistance of the battery - which in this circuit will also be a variable.
where/how will you measure "power loss" and will the measurement involve Time "t" ?
Of course that is wrong, because you cannot assume zero internal resistance because a AA battery has relatively high internal resistance. Internal resistance of a AA is many many times higher than that of the wire.The internal resistance and other variables are not provided, so we need to assume them as zero.
I tried solving it by finding the resistance value first [R=(resistivity*L)/A]=0.0015337ohm, then i added the R value into the power equation (P=VI), but i'm getting a huge value as current which means that i'm wrong somewhere, the power I'm getting is 1.47kW which again i think is wrong
I don't this so, even for an ideal battery, the current cannot be 977A, that's what the value of current I'm getting when i'm substituting the value of R in the Ohm's law equation. (where P = 1.47kW)Assuming an ideal battery you have the correct answer.
I don't *think so..I don't this so, even for an ideal battery, the current cannot be 977A, that's what the value of current I'm getting when i'm substituting the value of R in the Ohm's law equation. (where P = 1.47kW)
It is indeed a retarded question, but someone asked me and I'm confused now.
Those are indeed the correct theoretical values. Which proves how nonsensical the question is. Of course, the battery would go flat in next to no time, if the wire didn't vapourise first .the current cannot be 977A, that's what the value of current I'm getting when i'm substituting the value of R in the Ohm's law equation. (where P = 1.47kW)
Since this ideal battery has zero resistance it will not dissipate power in heating the battery.All of the power in the battery is lost due to heating the wire and the battery itself.