Solving Transient Response RL Circuit with Switch

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SUMMARY

This discussion focuses on analyzing a transient response in an RL circuit containing an inductor, two resistors, and a diode, with a switch included in the configuration. The time constant is calculated as L/R = 10/20 = 0.5 ms, leading to a steady-state inductor current of 0.5 A when the switch is closed. The behavior of the circuit is clarified, noting that when the switch opens, the inductor current continues through the diode until it reaches zero. Misinterpretations regarding the diode current and the implications of the switch's position are addressed, emphasizing the need for careful analysis in parts (c) and (f) of the problem.

PREREQUISITES
  • Understanding of RL circuit dynamics
  • Knowledge of time constants in electrical circuits
  • Familiarity with the behavior of diodes in transient conditions
  • Ability to analyze series circuits with switches
NEXT STEPS
  • Study the transient response of RL circuits in detail
  • Learn about the role of diodes in circuit protection and current flow
  • Explore the concept of time constants and their impact on circuit behavior
  • Investigate advanced circuit analysis techniques for complex configurations
USEFUL FOR

Electrical engineering students, circuit designers, and anyone studying transient response in RL circuits will benefit from this discussion.

van123
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Homework Statement


I have a circuit containing an inductor, 2 resistors and diode in series however there is also a switch.




The Attempt at a Solution



I thought that when the switch is closed there would be no current flowing through the right side of the circuit (resistor and diode) and therefore for part a) the time constant is L/R=10/20=0.5ms and steady state inductor current is just V/R=0.5A.
However, this can't be the case in order to do the show that question in part C), so I was wondering if someone could just explain the circuit to me when the switch is open and closed. Thank you!
 

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What, in toto, is the question?
Never mind, I got the picture straightened out. Stand by ...
 
Last edited:
How it works: at t=0 swirch S is closed and the current thru the inductor increases. At t = 0.5ms S opens and the inductor current continues thru the diode until iL = 0 at t = TZ. The current stays at zero until S closes again, at which point the cycle repeats.

How about doing (b)?

Part (c) is misleadingly stated. The current thru the diode is -10 mA only if the switch stays open for a long time.

(d) is an important computation. You will need it for part (e).

Part (f) is more challenging because the current thru the diode does not ever equal zero.
 
Reattempt

Thank you rude man so much for explaining, i have tried to answer the entire question again from what I (hope) i now understand. If you have time could you perhaps see my answers and tell me if I'm still not getting it?
Thanks again!
 
van123 said:
Thank you rude man so much for explaining, i have tried to answer the entire question again from what I (hope) i now understand. If you have time could you perhaps see my answers and tell me if I'm still not getting it?
Thanks again!

Certainly. I await your submission.
 

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