relation between the resistance and lost Thermal energy

[ahmeeeeeeeeeed]

Hello

 

What is the relation between the resistance and lost Thermal energy in a circuit ?

 

I mean >>

 

If we put a ((big)) resistance .. the Intensity of the current will lessen

 

 

If we are trying to lessen the lost energy (it is lost as heat) >>

 

what should we do ??

 

use a material that has low Resistivity ?

 

or high Resistivity ??

 

the first will make the intensity of the current increase >> and that it self affects the lost energy as heat

 

and

 

related to this

 

in "coldeg's X-ray tube" why do we use copper (attached to the target of the Wolfram) ??

 

Thanks

[DrRocket]

Hello

 

What is the relation between the resistance and lost Thermal energy in a circuit ?

 

I mean >>

 

If we put a ((big)) resistance .. the Intensity of the current will lessen

 

 

If we are trying to lessen the lost energy (it is lost as heat) >>

 

what should we do ??

 

use a material that has low Resistivity ?

 

or high Resistivity ??

 

the first will make the intensity of the current increase >> and that it self affects the lost energy as heat

 

and

 

related to this

 

in "coldeg's X-ray tube" why do we use copper (attached to the target of the Wolfram) ??

 

Thanks

 

Energy dissipated in a resistor is [math]I^2R[/math]

 

The maximum energy transfer to a resistive load by a voltage source occurs when the load resistance is

equal to the internal resistance of the source.

[alpha2cen]

Hello

 

What is the relation between the resistance and lost Thermal energy in a circuit ?

 

I mean >>

 

If we put a ((big)) resistance .. the Intensity of the current will lessen

 

 

If we are trying to lessen the lost energy (it is lost as heat) >>

 

what should we do ??

 

use a material that has low Resistivity ?

 

or high Resistivity ??

 

the first will make the intensity of the current increase >> and that it self affects the lost energy as heat

 

and

 

related to this

 

in "coldeg's X-ray tube" why do we use copper (attached to the target of the Wolfram) ??

 

Thanks

 

Transfer rate of thermal energy by conduction can be expressed as this equation.

(thermal energy transfer rate)=-(thermal conductivity)(temperature difference)/(distance)

And, we can express thermal resistance like this.

( thermal resistance) =(distance )/(thermal conductivity)

Using thermal resistance equation we can rewrite thermal transfer rate equation like this.

(thermal energy transfer rate) = - (temperature difference)/(thermal resistance)

Above equations materials property is related to thermal conductivity.

Copper has high thermal conductivity, and rubber or cloth has small thermal conductivity.