Does voltage and current have an inverse relationship in DC power transmission?
YES, but your assumption is false.
as long as the resistance is constant
When you double the voltage , the resistance must increase by \$R=\dfrac{V^2}{P}\$ for the same power rating or increase x $/\sqrt{2}\$
When you change the voltage for some VA or Watt rating at rated Power, you automatically must choose a different resistance so as not to exceed the Power rating. This would apply to any variable voltage power supply. They may have a V range with I max and Pmax and NONE OF THESE specs may be exceeded by choosing an appropriate DC load R or DCR as it is called on Motors and Inductors . Often this DCR for motors is 8 to 12% of the computed V/I so surge currents with full voltage will be 12 to 8% times the rated current of the motor, then students wonder why their 5A SMPS won’t start their 5A motor!!
Your assumption that voltage may drop as current rises is true when the source impedance is a factor, then the voltage will drop as load current increases.
This not what you asked, but let me explain anyways.
This can be measured for AC large transformers or DC power supplies.
AC source impedance
If we know a transformer has a certain VA or kVA or MVA rating with Vout * Iout for a resistive load, R it will also have an effective series resistance to limit short circuit current, yet minimize power loss and the cost of conductors to improve mutual coupling inductance for the design of the unit. This ratio of short circuit current to rated power current is called Zpu =Imax/Isc or per unit impedance, typically 8 to 12%.
Typically for small transformers we see no load voltage drop 10% at rated power out. This means the effective series resistance is also near 10%or Zpu*Imax = voltage drop from no load. So V,I ratings are always done with max load.for transformers
DC out impedance
For DC it is measured differently. Vout is done with no-load or min load if specified. Unless given with a tolerance which is defined as load regulation error. Here due to the negative feedback gain, Zout is reduced because of Vsensemand internal error gain, which is also a function of rise time of the load (and internal GBW limit).
Typically for any supply a good design is <=1% load regulation error which is defined as \$\Delta V/\Delta I =Zout\$ and not Vout/Imax=Rload !
But the percentage load regulation error is defined by Zout/Rload= \$Zout*Imax/Vout=load~ regulation ~error.\$[%]
This is normally <1% at room temp but may be derated over a wide temp range. This does not define impulse load error , because the load Cap ESR and Zout work together to react to large step load currents. Some regulators may be higher if they use high ohmic resistance series-pass transistors and insufficient feedback gain. So input and output Cap ESR is a critical choice for step load and ripple error specs. In some cases ESR cannot be too low , then the ripple noise feedback gain is reduced and may cause unstable step load overshoot or ringing or worse oscillations, so attention to these details is mandatory.
But you can see Ohm’s Law applies in all the above as an incremental drop in voltage for a rise in current.
Polarity is understood.
We normally speak in absolute positive values yet mathematically a Power source is negative and a load is positive so a negative change in V with a rise in I is a negative Resistance.
If this is an active power source and does not behave like a negative resistance semiconductor to avoid confusion (hopefully) An inverting transistor may have inverting Voltage, can be called a negative resistance but generally not. Usually it is a positive transimpedance, or “gm” or a positive RdsOn or Rce and we generally leave out the polarity with this understanding that the polarity is the same for series pass from emitter to collector or source to drain. We take it for granted when specifying RdsOn or Rce is positive relative to Emitter or Source. Yet if we compared Zout to Zin we could use the fact it inverts voltage out for a rise of some input, but generally do not.
