Ohmic loss, ohm's law, resistance etc

[fisico30]

Ohmic loss, ohm's law, resistance etc...

hello, just a simple question about power dissipation in the form of heat.

If a conductor has infinite conductivity (zero resistivity) then the voltage across it is zero, so

by the formula P= I^2 R, it dissipates zero power as heat.
but if I use the formula P=V^2/R, would I get an indeterminate form: 0/0 since V=0 and R=0 ?

In a simple circuit, if the load is meant to generate a lot of heat (for heating or cooking) , do we surely want the wires to be very conducting and dissipate almost no heat (small wire gauge_, but do we want the load resistor to large or small?

IF the voltage is constant, it appears that we would need a small load resistor R_L, according to the first formula so more current goes out ( and current is at the 2nd power)...

If the resistance were too big, little current would go out, and little power dissipated.

It seems that the more the resistance, the less the ohmic loss...
something wrong here..

thanks

 

[fisico30]

maybe i can answer myself:

It depends. If two resistors are in series (same current) then the bigger resistor dissipates more heat. If they are in parallel, the smaller resistor makes more heat.

In the case of a fixed voltage, I guess we want to match the heater resistor to the internal resistor of the voltage source. That gives maximum power transfer and therefore dissipation.

But what if, in some ideal case, I had perfectly conducting wires and a perfect battery ( no internal resistance). Would I choose a large or a small resistance for the heater in order to generate the max heat? The voltage would be all across the resistor, no matter if it is small or large. The smaller the resistor the higher the current, the more power.
But there is a threshold. How small can the resistance be ?

 

[Mapes]

You said it: for best power matching, the source and load resistances should be equal. No real-life power source has zero resistance, so if you happen to be using superconducting wires then the optimum heating element resistance should equal the battery resistance.

In your not-real-life thought experiment, though, the load power increases without bound as the load resistance decreases towards zero. The load power switches to zero when the load resistance becomes zero. If this seems surprising it's probably because the situation is unphysical.

 

[fisico30]

thanks Mapes.
Another bugging situation has to do with Ohms law in point form: J= rho*E.
If some finite current flows in a conductor, then rho and E must be finite. The bigger is rho , the smaller is E.If there is a finite current density, and rho "tends" to infinity, then E tends to zero.

In a electrostatics, E is zero inside a conductor because the conductor is an equipotential volume.

In electrodynamics, when a EM field hits a perfect conductor, we say that the field E is zero inside the conductor, otherwise it would cause an infinite current= infinity *E.
But a current still forms, but only on the surface, where the tangential E is still zero.
This current creates the reflect E wave and cancels the one that would go into the material.

Is that surface current caused by a time-changing B field, causing a voltage, which causes a surface current density? But the word voltage, to, me, implies the existence of an E field, which is supposed to be zero everywhere...

A Dc or AC current generates a B field, but there is always an E field involved...

 

[cabraham]

I believe it is J = sigma*E, or E = rho*J.

 

[fisico30]

you are right cabramham, I meant sigma... sorry