Engineering Analyzing the inductor current in the circuit

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The discussion focuses on analyzing a circuit involving an inductor and a zener diode, particularly when a switch opens and closes. When the switch is closed, the zener diode becomes reverse biased, while opening the switch results in zero inductor and zener current. The absence of a resistor complicates the analysis of voltage drops in the circuit. The inductor current is influenced by its series resistance and the switch's on-time. A diode across the inductor can limit reverse voltage, affecting the inductor's turn-off time, while the zener allows for a quicker drop in inductor current.
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Homework Statement
Inductor based circuit, the problem statement added
Relevant Equations
V = Ldi/dt
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The above is the problem statement
a. I don't understand this question, i am assuming the switch opens and closes and based on that i have to analyze the circuit and find the voltages and currents in the circuit.
b. When the switch is closed then the zener diode is reverse biased
1744364446017.png

When the switch is open, the inductor current and the zener current is 0. But the main concern for me to analyze the circuit is there is no resistor available for voltage drop and i am not able to analyze the circuit.
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The inductor current is limited by inductor series resistance, or by limiting the on-time of the switch.

The zener will only conduct momentarily after the switch turns off, as the inductor voltage is reversed while the current falls to zero.

A diode across the inductor will limit the inductor reverse voltage, so it will take longer to turn off. The zener allows a greater reverse voltage, so the inductor current falls faster.
 
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