# Series Resonant Circuit. (Need help double checking my answers)

Discussion in 'Homework Help' started by libnitz, Feb 20, 2016.

1. ### libnitz Thread Starter New Member

Jan 29, 2016
17
0
Hi,

Here is my solution

2. ### WBahn Moderator

Mar 31, 2012
17,446
4,699
What does the L need to be (in order to have that resonant frequency with that capacitor)?

What will the reactance be at the bandwidth edges?

How does that relate to the resistance?

You need to start building up the ability to check your own work -- it's a critical skill for the engineer.

3. ### libnitz Thread Starter New Member

Jan 29, 2016
17
0
I do understand I need to build skills to be an engineer or a doctor or a plumber (I don't think it takes a genius to realize that); I'm just new to this topic and require clarification.

wo = 1/√LC

so L = 1/((wo^2)C

or L = 2.53 X 10^-9

4. ### WBahn Moderator

Mar 31, 2012
17,446
4,699
Which is why I'm not going to just give you a "correct" or "incorrect" response -- I'm going to try to help you build the skills to verify your own results.

First off, 2.53 X 10^-9 is just a number. It is not an inductance. Just like my height is not 72 or 2, it is 72 inches or 2 feet. 72 and 2 are just numbers and neither of them is a length and they certainly aren't equal. But (72 inches) is identical to (2 feet).

So you need to start properly tracking your units. Most mistakes you make (and you, me, and everyone else will make lots of mistakes) will mess up the units and you can catch them if you always track your units properly.

$
\omega_0 \; = \; \frac{1}{\sqrt{LC}}
$

so

$
L \; = \; \frac{1}{\omega_0^2 C} \; = \; \frac{1}{$$2\pi 10^9 \frac{r}{s}$$^2 \cdot 10 \, pF}
$

Q = CV shows that the unit of capacitance has units of 1 F = 1 C/V = 1 A·s/V
V = L di/dt shows that the unit of inductance has units of 1 H = 1 V·s/A

L = 2.53 nH

Until you get used to units shortcuts, work them out explicitly. By "shortcut" I mean exploiting things like RC has units of time.

So know you have the value of L and the value of C.

What is the frequency at the upper and lower cutoff frequencies?

5. ### libnitz Thread Starter New Member

Jan 29, 2016
17
0
Hah! I forgot to put 1 unit and you are making a big deal of it; I'm sorry but that's just hilarious , you don't have to get so grumpy over that; ofc I know that 72 inches is not the same as 72 lol but thanks for the reminder.

Anyways my half power frequencies are
whigh = wo(1/2Q)* sqrt ((1/2Q)^2 + 1)
= 314MHz

and wlow = whigh - BW
= -5.97 X 10 ^9 Hz

6. ### MrAl Well-Known Member

Jun 17, 2014
2,206
433
Hi,

What Wbahn is saying is that you need to check your work, somehow. Either use units or at least check your results in say the transfer function of the circuit in question. If you learn to do this you will get every answer in the future entirely correct ! If you dont, you will get many answers wrong, i assure you.

If you checked your results you would have found that you get two different answers depending on the way you did it. That would bring up some questions that you could try to answer to determine what went wrong.

My question would be, where did you get the formula:
BW=1/(R*C)
?

Does that actually work?
It's up to you to answer this, and that will help you very much in the future, as well as the other suggestions given here in this thread. Follow them and you'll get better and better at these kinds of circuits, and be very happy in the long run

7. ### libnitz Thread Starter New Member

Jan 29, 2016
17
0
I do understand what he is saying; I did double check my work but I wanted a second opinion of someone pointing if he/she notices an error in my calculation. The reply I received, with all due respect to Wbahn, was childish. I was in a rush and forgot to put units to the inductance and that was it; I also forgot to calculate the high and low frequencies which he requested.

8. ### crutschow Expert

Mar 14, 2008
12,537
3,071
The is a childish response in this thread but it's not from WBahn.

9. ### libnitz Thread Starter New Member

Jan 29, 2016
17
0
Heh...

Maybe childish was the wrong word; but nitpicking for a simple (unit) mistake is just superfluous.

10. ### MrAl Well-Known Member

Jun 17, 2014
2,206
433
Hi,

Ok then it is time to put all this down and get back to work

Did you resolve any issues with your original results yet? The results are not correct yet so maybe you want to look over it again.

Also, could you mention where you got that formula i mentioned before? it does not look right so i was wondering where you got that. If you do use a formula that is not right to begin with, you can never get the right answer obviously.

I wish you the best of luck with it.

11. ### WBahn Moderator

Mar 31, 2012
17,446
4,699
You did not just forget to put 1 unit in one place -- that's part of your flawed approach. You think units are just something to be tacked onto the final answer based on what you HOPE the units SHOULD be. Maybe I'm grumpy about it because I've seen people die because they couldn't be bothered to track their units and because they took EXACTLY that approach -- they did all the work with pure numbers and tacked the units they THOUGHT were correct to the final answer. The part was make according to their results and a week later the person was dead because. It took the accident investigators just a couple hours to determine that, had he tracked his units, he would have realized that his answer was off by a factor of √2.54. Airliners have run out of fuel in midflight because of it. Billion dollar space probes have been slammed into planets because of it. With your attitude, don't be surprised if you end up with a civil case or criminal charges against you because your failure to do something as simple as track your units resulted in serious injury or death to someone else. It DOES happen (albeit not as frequently as it is warranted). When the plaintiff/prosecution shows the jury how the simple act of carrying your units in your work would have caught the mistake, the onus will be on you to convince the jury that it's just hilarious that anyone would even comment on your failure to do so.

And you must think it would be doubly-hilarious for someone to point out that it might be worthwhile asking if your answers make sense. For instance, asking if a low cutoff frequency of -5.97 GHz is at all reasonable.

Since you clearly know what is and what is not reasonable to expect from you in the way of due diligence, and since you clearly are unwilling to listen to suggestions in that regard, there is no point in me trying to help education you -- you clearly think you already know it all.

12. ### libnitz Thread Starter New Member

Jan 29, 2016
17
0
If I thought I knew it all, I would not seek help in the first place. Believe me when I tell you I perfectly understand what you mean by using the appropriate units. Again, here is my solution for the problem where I used the proper units,

I admit that I made a stupid mistake by not putting the right unit (and most likely additional errors) when I replied your first comment. I appreciate you even looking at this thread and taking your time to help but if you feel I was rude or out of place, I apologize. However, I still believe you blew my error out of proportion.

Cheers.

13. ### libnitz Thread Starter New Member

Jan 29, 2016
17
0
Yes! I saw the mistake I made and I have corrected the problem. Thanks so much for your concern

Edit: My phone is dead; I will upload a pic of my solution asap.

14. ### WBahn Moderator

Mar 31, 2012
17,446
4,699

You are NOT using units properly, you are just tacking some units onto the end HOPING that you just HAPPEN to be tacking on the correct units.

If you are tracking them properly, then each and every expression throughout your work would have the proper units and those units would flow from one expression to the next as a consequence of manipulating them and not just because you are tacking them on as an afterthought.

Assuming that

$
BW \; = \; \frac{1}{RC}
$

is the correct formula (and you might ask whether the bandwidth getting larger as R gets smaller (and going to infinity as R goes to zero) makes sense).

The you would have

$
R \; = \; \frac{1}{(BW)(C)}
R \; = \; \frac{1}{$$2 \pi 10^8 \, \frac{rad}{s}$$ $$10 \times 10^{-12} \, F$$}
R \; = \; \frac{1}{$$20 \pi 10^{-4} \, \frac{F}{s}$$} \times \frac{$$1 \, F$$}{$$1 \, \frac{As}{V}$$}
R \; = \; 159.2 \, \frac{V}{A} \times \frac{$$1 \, \Omega$$}{$$1 \, \frac{V}{A}$$}
R \; = \; 159.2 \, \Omega
$

As I said originally, once you gain proficiency you can do shortcuts by doing much of this in your head -- but that still means doing it in your head and not just tacking what you "know" the correct units are to the end.

All I did was point out the shortcoming and show you an example of how it should have been done. How is that blowing it out of proportion? After showing you how to do it correctly I then moved on and asked you a couple of questions intended to move you closer toward being able to check your own work.

15. ### libnitz Thread Starter New Member

Jan 29, 2016
17
0
Now I understand what you mean by using the appropriate units. I originally performed the calculations without the units but I attempted to cancel the calculated units in my head. I have grown accustomed to calculating my values this way and I was pretty sure it would work the same. I'm sorry I missed what you were trying to convey; my mistake

Anyways, the stupid mistake was in deriving my R from Q, wo and C.
$
Q \; = \; \frac{1}{(w)(R)(C)}
rearranging
R \; = \; \frac{1}{(w)(C)(Q)}
$

Thanks WBahn and MrAI
Lesson Learnt!