strange that you would call a pair of wires carrying DC a "transmission line".
what makes you think identical lengths of identical cable with the same resistance R powering identical loads R_L will dissipate more heat when carrying DC than AC?
the total resistance of the pair of wires R and the load R_L form voltage divider
in the DC case: P_cable=RI^2
in the AC case: P_cable=R*(I_RMS)^2
they should dissipate the same heat, you may want to brush up:
>For alternating electric current, RMS is equal to the value of the direct current that would produce the same average power dissipation in a resistive load.
and
>Because of their usefulness in carrying out power calculations, listed voltages for power outlets (e.g., 120 V in the USA, or 230 V in Europe) are almost always quoted in RMS values, and not peak values.
The only association with higher resistance losses would be when using low voltages but high currents... and even then the resistance losses would be equally high with low voltage high current AC since the fraction of energy dissipated in the cable versus the load is the same in both cases I^2 R / R_L
correct, but I don't know how large these are typically, an anecdote: I visited a friend who had a lamp socket (with shade) above his desk, and he was complaining about the fluorescent bulb he put in, in comparison with the resistive filament bulbs he usually put in: it was ticking even though the switch in the wall was turned off. At first I suspected a bad contact, but that didn't really make sense. Then I realized what was probably happening: capacitive leakage, so I ask if the socket has 2 switches? sure enough. the distance between the switches was long enough so that the parallel wires embedded in the wall effectively formed a long capacitor, so in both of the off states AC current was flowing through this unintentional capacitor. AC will also have inductive losses yes.
what makes you think identical lengths of identical cable with the same resistance R powering identical loads R_L will dissipate more heat when carrying DC than AC?
the total resistance of the pair of wires R and the load R_L form voltage divider
in the DC case: P_cable=RI^2
in the AC case: P_cable=R*(I_RMS)^2
they should dissipate the same heat, you may want to brush up:
https://en.wikipedia.org/wiki/Root_mean_square
>For alternating electric current, RMS is equal to the value of the direct current that would produce the same average power dissipation in a resistive load.
and
>Because of their usefulness in carrying out power calculations, listed voltages for power outlets (e.g., 120 V in the USA, or 230 V in Europe) are almost always quoted in RMS values, and not peak values.
The only association with higher resistance losses would be when using low voltages but high currents... and even then the resistance losses would be equally high with low voltage high current AC since the fraction of energy dissipated in the cable versus the load is the same in both cases I^2 R / R_L