Voltage induced accross an inductor, conceptual confusions
When you throw the switch, the inductor circuit is changing from "being a motor" to "being a generator" and it tries to keep the +1 amp flowing by altering the node C voltage (the only node that it can alter) to ensure that +1 amp still circulates at that instant following switch-over. The only viable voltage at node C that ensures this is -1 volt.

This forces 1 amp (at that instant) to flow through both resistor and inductor in the same direction prior to the switch changing position. The voltage clearly has to be -1 volts (node C) across the resistor to satisfy ohms law for the resistor for 1 amp flowing. This is because one side of the resistor has been connected to 0 volts by the switch changing position.

At the instant the switch changes over, you can assume the inductor to be equivalent to a constant current source of 1 amp and that means that whatever load impedance is connected across it (\$Z_{EXT}\$ = 1 ohm in your example), the voltage produced is 1 amp x \$Z_{EXT}\$. But only for that instant.

There is also one more thing that can be said at that instant; because we know that Faraday's equation is true at all times for an inductor (\$V = L\frac{di}{dt}\$) AND, because the inductor voltage has to be -1 volt, the rate of change of current is now -V/L or -1 volt / inductance. So we know the terminal voltage expressed by the inductor, the current and, the rate of change (fall) of current that will happen at that instant.

What happens hereafter is an exponentially decaying current best described by this picture: -



Picture taken from this slide show (Physics 121 - Electricity and Magnetism, Lecture 12 - Inductance, RL circuits)