Engineering Finding Resistance in an RL Circuit for Time Constant Calculation

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To find the resistance value for calculating the time constant in an RL circuit, the relevant equation is τ = L/R. After the switch is closed for a long time, the inductor acts as a short circuit, allowing current to bypass the 50kΩ resistor. The effective resistance for the time constant calculation should consider the Thévenin equivalent resistance of the circuit to the right of the inductor. It is clarified that during the transient state, the 50kΩ resistor still carries current, so the resistance used in the equation should be the parallel combination of the 75kΩ and 50kΩ resistors. Understanding these dynamics is crucial for accurately determining the time constant in the circuit.
izelkay
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Homework Statement


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Homework Equations


τ = L/R

The Attempt at a Solution


So I already know the initial current and final current (-4mA and -1mA, respectively), my question is how I would find the resistance value to use to solve for the time constant.

After the switch has been closed for a very long time, the 200mH inductor will become a short circuit, and the current from the 75 V source will bypass the 50kΩ resistor (because it will look for the path with least resistance) after traveling through the 75kΩ resistor and go through the short circuit from the inductor, and then the short circuit from the switch being closed. What would then be the resistance for the time constant equation?
 
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After the switch closes, the inductor current time constant involves the Thévenin equivalent resistance of the circuit to the right of the inductor.
 
So even though the current bypasses the 50k Ω resistor, the R I would use would still be 75k // 50k?

Or, would it be better to say that after doing the source transformation, the current bypasses both the 50k AND 75k resistor, and still have 50k // 75k as the R value?
 
The inductor bypasses all of the 50k resistor's current only in the steady state, not while current is varying. So, no, the 50k does not have all its current taken by the inductor during the time it is of interest to us in this exercise.
 
Okay I see, thank you.
 

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