How to apply Ohm's Law correctly to a circuit?
The 1.875 Ohm 'resistance' you calculated is based on the 800mA and 1.5V, so this wouldn't be a resistor you add, but is the equivalent resistance of whatever load if drawing that 800mA (which may not be a resistor).  As @Tony Stewart mentioned, the load may not be linear, so e.g. it may draw more current at lower voltage, unlike a resistor which will continue following the V=IR formula as the voltage changes.

If you wanted to step down the 5V source to the 1.5V load voltage, you could use a series dropping resistor, calculated as @Andy aka showed.  So you would have this:







simulate this circuit &ndash; Schematic created using CircuitLab

R2 would be your 'load', and R1 is used to create the required voltage drop.

(Please note that this is OK for educational purposes, but in a practical circuit you would not use a resistor to drop the voltage is it would be very inefficient.)

Many people use the water analogy to think of electrical concepts as it tends to be easier to visualize.  In this particular case you would think of a water source at a high pressure (5V), and requiring a specific flow and pressure (800mA at 1.5V).  If you were to connect the high pressure source directly to the load, instead of 800mA you would get 5 V / 1.875 Ohm = 2.67 A.  But if you put a long narrow hose in between the source and output, this would restrict flow and cause a pressure (voltage) drop.

I hope that analogy helps you think about Ohm's law.