Search Results
| Search type | Search syntax |
|---|---|
| Tags | [tag] |
| Exact | "words here" |
| Author |
user:1234 user:me (yours) |
| Score |
score:3 (3+) score:0 (none) |
| Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
| Views | views:250 |
| Code | code:"if (foo != bar)" |
| Sections |
title:apples body:"apples oranges" |
| URL | url:"*.example.com" |
| Saves | in:saves |
| Status |
closed:yes duplicate:no migrated:no wiki:no |
| Types |
is:question is:answer |
| Exclude |
-[tag] -apples |
| For more details on advanced search visit our help page | |
Results tagged with homework-and-exercises
Search options not deleted
user 14113
Applies to questions of primarily educational value - not only questions that arise from actual homework assignments, but any question where it is preferable to guide the asker to the answer rather than giving it away outright. Please READ THE GUIDANCE IN META before asking homework-like questions.
2
votes
Accepted
Explain why the frequency cutoff for RL circuits is given by $\frac{R}{2\pi L}$
This is where the 0.7 comes from:
Single pole low pass frequency responses are in written in the form:
$$ A_v=\dfrac{1}{\sqrt{1+\left( \dfrac{\omega}{\omega_{co}}\right) ^2}}=\dfrac{1}{\sqrt{1+\left( …
- 70