Voltage drop across a single conductor is always I x R where I is the current in the wire and R is its resistance. It doesn't matter what voltage is measured at one end of the wire with respect to ground or other common reference.

"Voltage" is a "potential" relative to some reference point and the reference point must always be stated.

For a similar example, if we ask how high buildings are in Denver, Colorado, and our zero reference is sea level, then every building, even a shack out back, is at least a mile high. For most cases we are interested in height from the base of the building, "ground" level if you like. An aviator, however, usually works with a sea level reference and would typically fly at a mile or more altitude.

A useful circuit technique is to establish a "high" reference and a "low" reference. Usually the "low" is called "ground" and "high" would be "power supply". Now replace each snip of wire with a resistor and current.

The perfect power supply's voltage never sags and it can supply infinite current to maintain this voltage. This voltage is reduced by every snip of wire (I x R) in the external circuit until it reaches the target device. The low side ("ground" if you like) at the target device is sneaky to calculate because its measured voltage (relative to our low reference point) rises due to the loss in each snip of wire along the way and these losses add to the voltage measured at the device low side.

Now, lets look at our hypothetical DC device that has a (+) and (-) terminal. For whatever reason lets assume that the sum of the I x R drops along the positive wire is 0.5V, while the sum of the I x R drops along the negative wire is 0.4V. In this case the voltage measured between the (+) and (-) terminals is 0.9V lower than the supply.

Note that I have deliberately chosen different drops for the (+) and (-). This is very common in real circuits and one must be prepared to recognize and deal with this, but it is a subject for a different day.