Time constant tao Which resistance do I use?

Click For Summary
SUMMARY

The discussion focuses on determining the Thevenin resistance (R) for calculating the time constant (τ) in electrical circuits. Participants confirm that to find R, one must short the voltage source and calculate the equivalent resistance using parallel resistance formulas. Specifically, the Thevenin resistance for the first circuit is calculated as 6kΩ in parallel with 3kΩ, while the second circuit yields a Thevenin resistance of 0.5kΩ in parallel with (2kΩ + 3kΩ). This method is essential for accurately solving time constant problems in circuit analysis.

PREREQUISITES
  • Understanding of Thevenin's theorem
  • Basic knowledge of parallel resistance calculations
  • Familiarity with time constant (τ) in electrical circuits
  • Ability to manipulate circuit diagrams
NEXT STEPS
  • Study Thevenin's theorem in-depth
  • Practice calculating parallel resistances in various circuit configurations
  • Learn about time constant calculations in RL circuits
  • Explore advanced circuit analysis techniques using simulation tools like LTspice
USEFUL FOR

Electrical engineering students, circuit designers, and anyone involved in analyzing RL circuits and their time constants.

frozenguy
Messages
186
Reaction score
0

Homework Statement


The top circuit is an example my professor did in class, and how he solved Tao. What I'm concerned with is how he is figure R for L/R

The second example is one I found online.

How do I figure the R to use? Is it simply the entire thevinin equivalent or a specific equivalent?

The Attempt at a Solution


TCquest.jpg
 
Physics news on Phys.org
Find Thevenin resistance. That's the value you need to put.

In the second case, short the voltage source.
Thevenin resistance = 0.5k || (2k+3K)

In the first case, short the voltage source.
The vertical 6k becomes ineffective as there is a short across it. The 3k and the horizontal 6k at the top are in parallel with regard to the inductor. Hence thevenin resistance = 6k || 3k.
 
asp_physics said:
Find Thevenin resistance. That's the value you need to put.

In the second case, short the voltage source.
Thevenin resistance = 0.5k || (2k+3K)

In the first case, short the voltage source.
The vertical 6k becomes ineffective as there is a short across it. The 3k and the horizontal 6k at the top are in parallel with regard to the inductor. Hence thevenin resistance = 6k || 3k.

Oh man.. I forgot about shorting them out ahhh.. Thanks I appreciate it!
 

Similar threads

  • · Replies 12 ·
Replies
12
Views
4K
  • · Replies 4 ·
Replies
4
Views
835
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 9 ·
Replies
9
Views
11K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 11 ·
Replies
11
Views
8K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
5
Views
3K
  • · Replies 11 ·
Replies
11
Views
3K