Let's suppose that we want to transmit a 1kW via DC generator with the below voltage level. From this we can see that the relationship between the DC voltage and current is inversely proportional,

P = I x V

P=1000, V=110, I = ?, then I = 1000/220 = 4.54 amp.
P=1000, V=220, I = ?, then I = 1000/110 = 9.09 amp.

Hence, higher voltage in this case will lead to lower amp.

My issue here, is how can we combine this fact with what we learned in Ohm's law about the current being proportional to the voltage as long as the resistance is constant? As the theories are kind of conflicting in my opinion.


the current being proportional to the voltage as long as the resistance is constant

This is the false premise: If you have a load that consumes constant power irrespective of the voltage, then that load must vary its resistance (higher voltage + higher resistance = same power).

P = I x U


R = U / I

both hold at the same time. So there is no way that you can keep the power and the resistance constant when either the voltage or the current or both change.

I.o.w., given R = U / I we can say

I = U / R

P = I x U = (U / R) * U = U² / R, so the power changes when the voltage changes, unless the resistance changes too.

  • \$\begingroup\$ With constant resistance, current is proportional to voltage. V = IR, remember? \$\endgroup\$ – Hearth Feb 4 at 15:24

Loads can be divided into two main categories:

  • Constant Impedance: where Z = \$\sqrt{R^2+X^2}\$ is kept constant
  • Constant Power: where the consumed power is kept constant (obviously inside certain limits)

What you are dealing with, in your example is a constant power load. In this case, impedance is NOT constant. So yeah, the higher the voltage, the lower the current, since their product must kept constant.

Even in your example, it is clear that load resistance is not constant

\$\frac{1000W}{110V} = 9.1A \Rightarrow R=\frac{110}{9.1} = 12Ω\$

By increasing the voltage

\$\frac{1000W}{220V} = 4.54A \Rightarrow R = \frac{220}{4.54} = 48.46 Ω\$

On the other hand, if your load was constant impedance, e.g. 12 Ω, by doubling the voltage the power would quadruple since:

\$P = \frac{V^2}{R}\$

You can imagine as constant impedance, loads such as the furnace or a heater and constant power, loads such as motors. Also check this out Constant current, constant power and constant impedance loads


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.

  • \$\begingroup\$ "When you double the voltage , the resistance must also double for the same power rating" are you sure you don't mean "When you double the voltage , the resistance must quadruple for the same power rating" since the relationship is P = V^2/R ? \$\endgroup\$ – rr1303 Feb 5 at 0:36
  • \$\begingroup\$ Thanks for the wake up call... maybe too many Bob Marley adult beverages \$\endgroup\$ – Sunnyskyguy EE75 Feb 5 at 1:04

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.