How to use series resistors when connecting sensors to microcontrollers via SPI

The 1A rating on the power supply is simply it's maximum, not the actual amount of current drawn by whatever you hook up to it. That is the amount you must ensure the load stays below.

If the "load" is a short circuit, then your calculation is nearly right. $$ R = {{V}\over{I}} = {{5V}\over{250mA}} = 20\Omega$$ If the load is something else, a resistor of lesser resistance is required.

It's simply telling you that the circuit as drawn will never have the given current running through it.

Consider the following situation: Replace the inductor by a capacitor (capacitor discharge might be more familiar to you), and the current function by

$$ I(t) = t $$

Now calculate voltages - they won't add up. Why? Because the current function makes no sense. The complete discharging process of a capacitor through a resistor is completely defined by the capacity, resistance and the voltage across it at time \$t = 0\$.

Similarly, the 'discharging' of an inductor through a resistor is completely determined by the inductance, resistance and the current through it at time \$t = 0\$.

If you are now given a time-dependent current function, you are overspecifying the system. That function may be right, but (like in your question) it may be wrong for the given circuit, so you arrive at contradictory solutions.

Note that there is a way to keep the question/function/values like they are now and make it consistent again, by adding an ideal current source into the circuit which satisfies the given current function. The strange voltage that you couldn't explain is then simply found across this current source:

^{simulate this circuit – Schematic created using CircuitLab}

Alternatively, simply assume that the current function actually is right at \$t=0\$, that is, \$I_0 = 50\$. The discharge current of an inductor through a resistor is $$I(t) = I_0e^{-\frac{R}{L}t}$$, so the correct current function for the circuit as shown in your question is $$I(t) = 50e^{-20000t}$$. Do your voltage calculations again - they will now work out.