# please help with problem

Thread Starter

#### terrakota

Joined Feb 8, 2005
67
Hi i need to solve this problem from a book i'm studing, i just need to know the right approach to solve it,
thanks in advance

- What value of coupling capacitor is required in the figure so that the signal voltage at the input of amplifier 2 is at least 70.7% of the signal voltage at the output of amplifier 1 when the frequency is 20Hz ?

i know the answer the cap must be: 0.0796 microF

but i cant find a way to achieve that

what I'm doing is to use the voltage divider formula, (r1/r1+Xc)Vs = vs * 70.7%
and solve for Xc i use a sample voltage of 2v source , then i use the xc = 1/(2)(pi)(f)© and solve for c, but my answer is totally wrong.

thanks for your help and please excuse my poor english.

#### hgmjr

Joined Jan 28, 2005
9,027
Originally posted by terrakota@Aug 2 2005, 12:41 PM
Hi i need to solve this problem from a book i'm studing, i just need to know the right approach to solve it,
thanks in advance

- What value of coupling capacitor is required in the figure so that the signal voltage at the input of amplifier 2 is at least 70.7% of the signal voltage at the output of amplifier 1 when the frequency is 20Hz ?

i know the answer the cap must be: 0.0796 microF

but i cant find a way to achieve that

what I'm doing is to use the voltage divider formula, (r1/r1+Xc)Vs = vs * 70.7%
and solve for Xc i use a sample voltage of 2v source , then i use the xc = 1/(2)(pi)(f)© and solve for c, but my answer is totally wrong.

thanks for your help and please excuse my poor english.
[post=9389]Quoted post[/post]​
what I'm doing is to use the voltage divider formula, (r1/r1+Xc)Vs = vs * 70.7%
The expression you have as your starting point looks good.

i know the answer the cap must be: 0.0796 microF
The value you have calculated for the capacitor computes to a capacitive reactance of 100K at 20 Hz. Plugging 100K into your voltage divider formula would give you 100K/(200K) which calulates to 0.5 or 50% attenuation.

I suspect that you just slipped up in the process of rearranging your starting expression to get the expression for capacitance in terms of the frequency and the value of resistance given in your problem description.

I hope this is helpful in getting you back on the right track.

Good Luck,
hgmjr

Thread Starter

#### terrakota

Joined Feb 8, 2005
67
Originally posted by hgmjr@Aug 2 2005, 03:11 PM
The expression you have as your starting point looks good.
The value you have calculated for the capacitor computes to a capacitive reactance of 100K at 20 Hz. Plugging 100K into your voltage divider formula would give you 100K/(200K) which calulates to 0.5 or 50% attenuation.

I suspect that you just slipped up in the process of rearranging your starting expression to get the expression for capacitance in terms of the frequency and the value of resistance given in your problem description.
[post=9392]Quoted post[/post]​
sorry you lost me there :-(

thanks for reply

#### n9xv

Joined Jan 18, 2005
329
The cap should be a .192-uF or about .2-uF instead of .0796-uF.

The % of voltage drop ratios across each component is proportional to the resistive/reactive ratios of each component. Hence, work it like a proportion problem.

The ratio;

70.7% signal Voltage / 100,000 Ohm resistor

is equal to the ratio;

29.3% signal voltage / (41,442.72 Ohms of capacitive reactance of a .192-uF capacitor at a frequency of 20-Hz.)

#### hgmjr

Joined Jan 28, 2005
9,027
Originally posted by terrakota@Aug 2 2005, 12:41 PM
Hi i need to solve this problem from a book i'm studing, i just need to know the right approach to solve it,
thanks in advance

- What value of coupling capacitor is required in the figure so that the signal voltage at the input of amplifier 2 is at least 70.7% of the signal voltage at the output of amplifier 1 when the frequency is 20Hz ?

i know the answer the cap must be: 0.0796 microF

but i cant find a way to achieve that

what I'm doing is to use the voltage divider formula, (r1/r1+Xc)Vs = vs * 70.7%
and solve for Xc i use a sample voltage of 2v source , then i use the xc = 1/(2)(pi)(f)© and solve for c, but my answer is totally wrong.

thanks for your help and please excuse my poor english.
[post=9389]Quoted post[/post]​
Hi terrakota,

Take a look at this tutorial located on this website.

I think it will help you get a better understanding of dealing with this type of problem.

Series RC tutorial

hgmjr

#### n9xv

Joined Jan 18, 2005
329
The tutorial is good and all but just realize that phase angle has nothing to do with the circuit example as a simple filter.

#### hgmjr

Joined Jan 28, 2005
9,027
Originally posted by n9xv@Aug 3 2005, 01:50 AM
The tutorial is good and all but just realize that phase angle has nothing to do with the circuit example as a simple filter.
[post=9402]Quoted post[/post]​
Hi n9xv,

You are correct in your statement regarding phase since the only thing that terrakota is looking for is the ratio of the series RC voltage divider. That means that magnitude is the only thing of interest in his particular problem.

After I read the tutorial (kudos to the author of the tutorial) I was reminded that the resistive component and the capacitive reactance component in the denominator do not add algebraically as I INCORRECTLY indicated in my initial reply to this posting. Instead these two components are combined using vector arithmetic.

I believe the voltage divider formula is more accurately written as

(R1/(R1 - jXc)).

This being true then the magnitude of the denominator is actually written as

SQRT(R1^2 + (-jXc)^2)

Plugging in the values from terrakota's circuit yields

SQRT(100K^2 + 100K^2) = SQRT(2 x 10^10) = 141.4K

That gives the voltage divider ratio of 100K/141.4K which calculates to .707 or expressed in percent the voltage divider ratio is 70.7%.

Bottom line, this would make terrakota's value for C of 0.0796 microfarads correct.

I believe this is correct. Did I miss something?

hgmjr

Thread Starter

#### terrakota

Joined Feb 8, 2005
67
thanks to all for your great help

watching your posts pluss studing and thiking a little bit more i finally found the solution, thanks a lot

here's how i finally developed the solution

first find Ztot

using the voltage divider formula and solving for Rtot = Ztot

Ztot = R/(Vd/Vs)

once found the total impedance i just applied the Z formula for reactance and resistance in series

Z = sqrt((r*r)+(Xc*Xc))
and solving for Xc

Xc = sqrt((z*z)-(r*r))

with Xc found use the reactance formula

Xc = 1/2*pi*F*C

solving for C
C = 1/2*pi*F*Xc

and voila
i dont know if is the best way but at least it works know.

thanks again for your great help

#### hgmjr

Joined Jan 28, 2005
9,027
Originally posted by terrakota@Aug 3 2005, 05:25 PM
thanks to all for your great help

watching your posts pluss studing and thiking a little bit more i finally found the solution, thanks a lot

here's how i finally developed the solution

first find Ztot

using the voltage divider formula and solving for Rtot = Ztot

Ztot = R/(Vd/Vs)

once found the total impedance i just applied the Z formula for reactance and resistance in series

Z = sqrt((r*r)+(Xc*Xc))
and solving for Xc

Xc = sqrt((z*z)-(r*r))

with Xc found use the reactance formula

Xc = 1/2*pi*F*C

solving for C
C = 1/2*pi*F*Xc

and voila
i dont know if is the best way but at least it works know.

thanks again for your great help
[post=9414]Quoted post[/post]​
Excellent job. I like your very reasonable approach.

Please forgive me for the confusion I may have caused you with my first reply. For some reason, my critical brain cells failed to line up properly. Just goes to show that old timers like me need a kick in the butt from time to time to bring us back to reality.

Keep up the good work.

Never stop learning.
hgmjr

Thread Starter

#### terrakota

Joined Feb 8, 2005
67
thanks for your words and without your post i would never solve the problem

thanks again