3 ms·
It's been a long time since I did any physics, but explain to me why you can't run the 0.072A into e.g. a 10kohm load and have 51W of power dissipated in the lo
by aidenn0 3y ago
It's been a long time since I did any physics, but explain to me why you can't run the 0.072A into e.g. a 10kohm load and have 51W of power dissipated in the load? Sure you'd be running at several hundred volts, but high voltage is how you get more power through less copper.
[edit]
Upon further reflection your claim about maximum power transfer being when the load and cable are dissipating the same total power doesn't pass the sniff test, because it would imply that a load connected by the same 33awg cable, but only 1m long could only drive a load of 3.5mW since the load would then only be 678 mohm.
- aidenn0 3y agoFor other's curious: the "load resistance matches source resistance" is true for maximum power transfer of fixed voltage sources. In the case of current capacity of a wire, the voltage is variable, but the maximum current is fixed. In that case we are selecting the input voltage for maximum power so (Rs, I are fixed, Rl can vary, and for a purely resistive load, Vin is a function of Rl): Vin=I(Rs+Rl) Vin = Vs + Vl Vl/Vs=Rl/Rs Pl = (I^2*Rl) Clearly we can always select a Rl (and thus a Vin) that gets the desired Pl at a fixed current. Obviously at some point we are limited by shielding of the wire, and DC/DC conversion at the end-point. We eventually may also be limited by interactions between the medium surrounding the wire and EM fields generated by turning the circuit on and off. But, when you can control the voltage, there's no simple calculation from wire impedance to X maximum watts of load.