Electronic – Inductor Voltage and Current After Switches Open and Close

When the series current is zero, the voltage across L1 must equal V1 (by KVL and Ohm's Law).

But, for an (ideal) inductor, we have:

\$v_L = L \dfrac{di_L}{dt}\$

Thus, by KVL and the definition of an ideal inductor, at the moment SW1 closes, the time rate of change of current is:

\$\dfrac{di_L}{dt} = \dfrac{1V}{1H} = 1 \dfrac{A}{sec} \$

So, the crucial insight here is this: there is no current at the moment the switch closes but, at that very moment, the current begins to change.

How do I calculate the rate of change slightly after the first after-SW1-closed rate if R1 is to be taken into consideration?

By solving the differential equation that describes the circuit. By KVL and Ohm's Law, we have:

\$v_L = L \dfrac{di_L}{dt} = v_1 - i_L R \rightarrow \dfrac{di_L}{dt} + \dfrac{R}{L}i_L = v_1\$

This is an easy 1st order ordinary differential equation for the series current \$i_L\$.

The solution, for zero initial condition, is:

\$i_L(t) = \dfrac{v_1}{R}(1 - e^{-\frac{t}{\tau}}) \$

Where

\$\tau = \dfrac{L}{R} \$

When t is "small enough", i.e., right after the switch closes, we have:

\$i_L(t) \approx \dfrac{v_1}{L}t \$

So, in the early moments, the resistance has negligible effect and the current is approximately a ramp. The current begins to significantly deviate from a ramp only after the current becomes large enough such that the voltage drop across the resistor is significant compared to the voltage source.

I would call it "Alternate Action" if the button returns to the same position each time you remove your finger, but the switch position changes on each press.

With a latching button, I'd expect the button to stay partly depressed with the switch in one position, and fully out with the switch in the other position.

I don't think I've ever seen a switch with the delay function you describe. That sort of thing would normally be done with the aid of a microcontroller or time delay circuit.