Exponential prob: What is the value of capacitance (C) in a series R-L-C circuit, if

Baxterdmutt

New member
Joined
Aug 26, 2016
Messages
12
Its been a long time since I've had to do math and I'm now trying to complete an advanced radio course. I've been doing not too bad until I came to this question that seems to be giving my an exponential problem!
So here goes:
The equation used to solve the question is:
C=1/(2pi *f)E2 * L


Here is some reference:
C=Capacitance
f=frequency in hertz
L= inductance in henry
1 micro Henry = 1*10E6 Henry
1 Pico Farad = 1*10E-12 farad
1 MHz = 1*10E6 hertz


The sample test question is:
--
What is the value of capacitance (C) in a series R-L-C circuit, if the circuit resonant frequency is 14.25 MHz and L is 2.84 microhenrys?
The answer for C should be 44 picofarads
--


So it's my understanding then that my answer should be 4.4 e-11 but I'm getting 4.4 e-14 which is way smaller a number than it should be.
This is my equation as entered in my calculator:
1/((2pi)E2*(14.25*10E6)E2*(2.84*10E-6)) (Edited this line since posting)
This is my answer:
C=4.392302014589E-14
And that would be the same as C=.0044 picofarads instead of C=44.


I hope I explained this clearly enough.
Thanks in advance for your help.
 
Last edited:
Its been a long time since I've had to do math and I'm now trying to complete an advanced radio course. I've been doing not too bad until I came to this question that seems to be giving my an exponential problem!
So here goes:
The equation used to solve the question is:
C=1/(2pi *f)E2 * L


Here is some reference:
C=Capacitance
f=frequency in hertz
L= inductance in henry
1 Pico Farad = 1*10E-12 farad
1 MHz = 1*10E6 hertz


The sample test question is:
--
What is the value of capacitance (C) in a series R-L-C circuit, if the circuit resonant frequency is 14.25 MHz and L is 2.84 microhenrys?
The answer for C should be 44 picofarads
--


So it's my understanding then that my answer should be 4.4 e-11 but I'm getting 4.4 e-14 which is way smaller a number than it should be.
This is my equation as entered in my calculator:
1/((2pi)E2*(14.25*10E6)E2*(2.84*10E-12)) .... that E2 shouldn't be there.... and 10E-12 should be 10E-6 (micro means one millionth)
This is my answer:
C=4.392302014589E-14
And that would be the same as C=.0044 picofarads instead of C=44.


I hope I explained this clearly enough.
Thanks in advance for your help.

That might solve your problem
 
That might solve your problem
I'm sorry. I entered the question wrong.
It should have read that intered
1/((2pi)E2(14.25*10E6)E2*(2.84*10E-6)) .
questioni I'm not sure why I put -12 on the forum. Just a typo.
(2pi)E2 needs to be entered like that or my calculator doesn't square it.
 
That might solve your problem
Sorry but that doesn't help. I made and error in the forum post. I've corrected My question to what I actually entered in the calculator. I accidentally typed E-12 on the forum and it should have read E-6. Please look again.
 
I'm not sure why but I've posted a couple replies and they aren't showing up. I'll try again.
I entered the equation on the forum wrong. I've edited it. I'm sorry to say then that Harry the cat's solution doesn't help.
 
That might solve your problem

Sorry but there was a typo in my post I don't know why I entered E-12 in my equation. On my calculator I actually entered E-6.
Also 2pi has to be squared. If you noticed I wrote it (2pi)E2 * (14.25*10E6)E2 instead of (2pi * 14.25*10E6)E2 because my calculator doesn't square the insides of the brackets right if I don't break up the equation and do the parts separately.

Thanks


Sent from my iPhone using Tapatalk
 
That might solve your problem

I'm having troubles posting. This I my 6th try. If they all show up at once, I'm sorry.
I've edited my question after your answer. The E-12 in my question was a typo. I've squared Pi separately because my calculator didn't like it all together. My problem still exists. My answer is the same after your suggestion ( because the forum had a typo).


Baxterdmutt
 
That might solve your problem
I'm having a problem posting! This is my 8th try. Let's see if it works.

There was an error in my previous posting. Harry_the_cat you are correct about the E-12 but I entered it wrong on the post. It should have read E-6 so the answer I is still what I posted initially. As far as removing the square after 2pi, that won'r help because if you notice I split the equation into it's 2 parts because my calculator does weird thing when I enter (2pi *14.25*10E6)E2. I can do it that way on another calculator and I still get the answer I posted. Changing the square of the pi portion couldn't be the issue anyway because my answer has the write numbers, just the wrong decimal point.

So I'm sorry I wasted your time because I entered it wrong on the forum, but I still need help please!
 
Its been a long time since I've had to do math and I'm now trying to complete an advanced radio course. I've been doing not too bad until I came to this question that seems to be giving my an exponential problem!
So here goes:
The equation used to solve the question is:
C=1/(2pi *f)E2 * L


Here is some reference:
C=Capacitance
f=frequency in hertz
L= inductance in henry
1 micro Henry = 1*10E6 Henry
1 Pico Farad = 1*10E-12 farad
1 MHz = 1*10E6 hertz


The sample test question is:
--
What is the value of capacitance (C) in a series R-L-C circuit, if the circuit resonant frequency is 14.25 MHz and L is 2.84 microhenrys?
The answer for C should be 44 picofarads
--


So it's my understanding then that my answer should be 4.4 e-11 but I'm getting 4.4 e-14 which is way smaller a number than it should be.
This is my equation as entered in my calculator:
1/((2pi)E2*(14.25*10E6)E2*(2.84*10E-6)) (Edited this line since posting)
This is my answer:
C=4.392302014589E-14 <===== 4.392302014589E-11
And that would be the same as C=.0044 picofarads instead of C=44.


I hope I explained this clearly enough.
Thanks in advance for your help.
I'm sorry but I'm not sure what your equation is. What I am used to seeing is the resonance angular frequency of an RLC circuit is
\(\displaystyle \omega\, = \, \dfrac{1}{\sqrt{L\, C}}\)
where the angular frequency is given by
\(\displaystyle \omega\, = \, 2\, \pi\, f\)
If we turn around the frequency equation for capacitance we get
\(\displaystyle C\, =\, \dfrac{1}{\omega^2\, L}\)

If I interpret your E2 as meaning squared and your expression as having everything in the denominator (other than the 1/) I think we have the same formula. So, let's look at it: One of the things I was taught when I had to do these kind of equations on a slide rule was to make (sometimes very) rough computations to make sure of where the decimal point was. So,
2*pi*f~2*3*14*106 ~ 90 106 [remember, very rough approximation] ~ 9 * 107
Square that and get 8 1015. Now
8 1015 * 2.84 10-6 ~ 22 * 109 ~ 2 * 1010
and
1/(2 * 1010) ~ .5 * 10-10 ~ 50 10-12
So your answer should be on the order of 50 picofarads (as opposed to 5 or 500 picofarads). Looks like you are off in the decimal place and possibly put in millihenries (10-3) instead of microhenries (10-6)
 
I'm sorry but I'm not sure what your equation is. What I am used to seeing is the resonance angular frequency of an RLC circuit is
\(\displaystyle \omega\, = \, \dfrac{1}{\sqrt{L\, C}}\)
where the angular frequency is given by
\(\displaystyle \omega\, = \, 2\, \pi\, f\)
If we turn around the frequency equation for capacitance we get
\(\displaystyle C\, =\, \dfrac{1}{\omega^2\, L}\)

If I interpret your E2 as meaning squared and your expression as having everything in the denominator (other than the 1/) I think we have the same formula. So, let's look at it: One of the things I was taught when I had to do these kind of equations on a slide rule was to make (sometimes very) rough computations to make sure of where the decimal point was. So,
2*pi*f~2*3*14*106 ~ 90 106 [remember, very rough approximation] ~ 9 * 107
Square that and get 8 1015. Now
8 1015 * 2.84 10-6 ~ 22 * 109 ~ 2 * 1010
and
1/(2 * 1010) ~ .5 * 10-10 ~ 50 10-12
So your answer should be on the order of 50 picofarads (as opposed to 5 or 500 picofarads). Looks like you are off in the decimal place and possibly put in millihenries (10-3) instead of microhenries (10-6)

Excellent instruction Ishuda - here is your "GET OUT OF CORNER FREE" ticket.
 
Last edited by a moderator:
I'm sorry but I'm not sure what your equation is. What I am used to seeing is the resonance angular frequency of an RLC circuit is
\(\displaystyle \omega\, = \, \dfrac{1}{\sqrt{L\, C}}\)
where the angular frequency is given by
\(\displaystyle \omega\, = \, 2\, \pi\, f\)
If we turn around the frequency equation for capacitance we get
\(\displaystyle C\, =\, \dfrac{1}{\omega^2\, L}\)

If I interpret your E2 as meaning squared and your expression as having everything in the denominator (other than the 1/) I think we have the same formula. So, let's look at it: One of the things I was taught when I had to do these kind of equations on a slide rule was to make (sometimes very) rough computations to make sure of where the decimal point was. So,
2*pi*f~2*3*14*106 ~ 90 106 [remember, very rough approximation] ~ 9 * 107
Square that and get 8 1015. Now
8 1015 * 2.84 10-6 ~ 22 * 109 ~ 2 * 1010
and
1/(2 * 1010) ~ .5 * 10-10 ~ 50 10-12
So your answer should be on the order of 50 picofarads (as opposed to 5 or 500 picofarads). Looks like you are off in the decimal place and possibly put in millihenries (10-3) instead of microhenries (10-6)

The equation I gave is the one given to me and the one I need to solve. If the answer isn't 44 picofarads then it's wrong. It doesn't help me to throw in another equation since I already mentioned I'm weak in math. The equation I gave is the one I need to solve.
I know I'm off by a few decimal points. That was the question. But why?
I do appreciate your attempt though.


Baxterdmutt
 
The equation I gave is the one given to me and the one I need to solve. If the answer isn't 44 picofarads then it's wrong. It doesn't help me to throw in another equation since I already mentioned I'm weak in math. The equation I gave is the one I need to solve.
I know I'm off by a few decimal points. That was the question. But why?
I do appreciate your attempt though.


Baxterdmutt
Hey, read the whole thing. As I said, I think we both have the same equation and the answer should be about 50 picofarads [were I doing it on a slide rule, I probably would have gotten an answer of 43.9 picofarads and, rounding to two significant digits if asked to do so, answered 44 picofarads].

However you are not using grouping symbols as you should and are using your calculators 'button names' as symbols rather than standard mathematical symbols. If you are going to ask questions in a math help forum, then you should probably try to use math symbols rather than make up your own or use your calculators 'button names' for a symbol. If you are not getting the proper answer then you are not inputting the proper information or possibly in the proper manner.
 
Last edited:
Hey, read the whole thing. As I said, I think we both have the same equation and the answer should be about 50 picofarads [were I doing it on a slide rule, I probably would have gotten an answer of 43.9 picofarads and, rounding to two significant digits if asked to do so, answered 44 picofarads].

However you are not using grouping symbols as you should and are using your calculators 'button names' as symbols rather than standard mathematical symbols. If you are going to ask questions in a math help forum, then you should probably try to use math symbols rather than make up your own or use your calculators 'button names' for a symbol. If you are not getting the proper answer then you are not inputting the proper information or possibly in the proper manner.

OK thats making more sense. I was using the stupid TapaTalk app the the site recommended I use on this forum. Once I go onto a real website I can see the symbols properly. So Yes it looks like we are talking about the same thing so let me try and re-enter the equation my book gave me to work with. They aren't using the angular frequency and just using straight frequency, so I'm not coming at it from quite the same direction as you but we got to the same place.
This is the equation given to me. I've tried to enter it properly this time: C=1÷(2π*ƒ)2*L
If I enter the values given in the question I would enter 1÷((2π * 14.25*106)2 * 2.84 * 10-6)
My answer is then C=4.392302014589-14
I've figured out my mistake as I wrote this out. The course material tells me to do it the way I've written it above, but in fact the extra (*10)s are screwing up the answer. It should be entered 1÷((2π * 14.256)2 * 2.84-6) and getting rid of the 10s they put in their manual fixed it, giving me the correct answerC=4.392302014589-11 or 4.39 (44) picofarads.

Thank you all for talking this through with me it helped me get this figured out. If you hadn't rewritten the equation for me I wouldn't have gotten it. :-D
 
OK thats making more sense. I was using the stupid TapaTalk app the the site recommended I use on this forum. Once I go onto a real website I can see the symbols properly. So Yes it looks like we are talking about the same thing so let me try and re-enter the equation my book gave me to work with. They aren't using the angular frequency and just using straight frequency, so I'm not coming at it from quite the same direction as you but we got to the same place.
This is the equation given to me. I've tried to enter it properly this time: C=1÷(2π*ƒ)2*L
If I enter the values given in the question I would enter 1÷((2π * 14.25*106)2 * 2.84 * 10-6)
My answer is then C=4.392302014589-14
I've figured out my mistake as I wrote this out. The course material tells me to do it the way I've written it above, but in fact the extra (*10)s are screwing up the answer. It should be entered 1÷((2π * 14.256)2 * 2.84-6) and getting rid of the 10s they put in their manual fixed it, giving me the correct answerC=4.392302014589-11 or 4.39 (44) picofarads.

Thank you all for talking this through with me it helped me get this figured out. If you hadn't rewritten the equation for me I wouldn't have gotten it. :-D
Something seems wrong with the TapaTalk app when compared to a 'normal' math calculation. How do you tell the difference between 14.256 = 14.25*14.25*14.25*14.25*14.25*14.25 and 14.25 * 106. Maybe it's in the use of parenthesis.
 
Top