If we keep the voltage in a circuit constant, then H = V 2 R tso if I increase the resistance, the current must decrease and so the heat will be less. Then why can't we use very good conductors, like copper, for heat formation?

Question

If we keep the voltage in a circuit constant, then  H  =            V      2        R    tso if I increase the resistance, the current must decrease and so the heat will be less. Then why can&#039;t we use very good conductors, like copper, for heat formation?

vihralV5x · Accepted Answer

Because the voltage source and the rest of the circuit has some internal resistance. Therefore if you decrease the resistance of your consumer the voltage across your consumer will decrease, so there will be less heat created in your consumer.Let       R    c   be the resistance of your consumer and       R    i   the internal resistance, then we have      U    c    =      R    c    I      U    i    =      R    i    I  U  =  (      R    c    +      R    i    )  Iand therefore      P    c    =      R    c        I    2    =            R      c              (              R        c            +              R        i                    )        2                  U    2  the maximum of this function with respect to       R    c   is      R    c    =      R    i  Therefore you get the maximum heat dissipation in your consumer if your resistance matches the internal resistance of the circuit. Note however that this is inefficient, as the power dissipated in your internal resistance is equal to the power dissipated in your consumer (where you want the heat to be.) The quotient of consumer power to internal power loss is            P      c              P      i        =            R      c              R      i        ,so the higher your consumer resistance is the more efficient it becomes. Therefore in reality you choose the resistance of your consumer much larger than the internal resistance to make the circuit energy efficient. Then you need a power source with high enough voltage so that you can still match the desired heat output.

If we keep the voltage in a circuit constant, then H=V^2/Rt

Answered question

Answer & Explanation

New Questions in Thermal Physics