have you read http://forum.allaboutcircuits.com/showthread.php?t=3002 yet?

 

have you read http://forum.allaboutcircuits.com/showthread.php?t=3002 yet?

yes, I did.
So I basically have to put details what I am asking and what I know about it when I ask something?

 

It also said for you to show the steps you've taken and show the work you have done ...

So what do you think happens when you change R1?

(1) What type of filter is C1 and R1 with respect to frequency bands?
(2) What is the formula to find the fc?
(3) Assign a value to C1 and R1. What is the fc? ... Show your work
(4) Decrease R1 by half. What is fc? ... Show your work
(5) Double R1. What is fc? ... Show your work.
(6) Review the results in 3, 4, and 5. What do you conclude from those results?

 

It also said for you to show the steps you've taken and show the work you have done ...

So what do you think happens when you change R1?

(1) What type of filter is C1 and R1 with respect to frequency bands?
(2) What is the formula to find the fc?
(3) Assign a value to C1 and R1. What is the fc? ... Show your work
(4) Decrease R1 by half. What is fc? ... Show your work
(5) Double R1. What is fc? ... Show your work.
(6) Review the results in 3, 4, and 5. What do you conclude from those results?

I am not sure but i think the filter gets narrowed
fc=1/2pifC, Q=XL/R
BW=f0/Q ,
Low pass filter: AV=XC√ (R2+XC2) (-tan-1(R/XC)

High pass filter: AV=R/(R-jXC), AV=R/ (R2+XC2)

 

so basically when I asked question about filters for example, I will put what I know about filters and how it relates to the question I asked right?

 

Where did Xl come from? There are no inductors in your circuit.

According to your diagram, you have C, R, C1, R1. Your inquiry was concerning changing the value of R1.

(7) What kind of a filter does your diagram represent?

And just so we are straight, identify the question number you are answering.

 

no..thts just the formula for quality factor...

 

(7) What kind of a filter does your diagram represent?

And just so we are straight, identify the question number you are answering.

Is it high pass filter..
either high or low..cuse thts what we covered on the lecture so far...

 

(8) C1 and R1 form one filter ... C and R form another. Identify them by name.
(9) What is the formual to find fc?

The quality formula has nothing to do with your inquiry.

 

C1 and R1 form one filter ... C and R form another. Identify them by name.

The quality formula has nothing to do with your inquiry.

yeah..I know I just put basic formulas...for resonant circuit...
so there are tow different filters...so if you change R1 it will affect the whole circuit right?

 

the formula to find fc is 1/2piRC

 

fc is the -3dB point where Xc = R in these filters.

Assign them the following values:

C1 - 0.47 uF
R1 - 10 k
C - 560 pF
R - 10 k

(9) Identify the two filters by their function name.
(10) Compute fc for each filter
(11) Identify the whole filter circuit by it's proper name.

 

fc is the -3dB point where Xc = R in these filters.

Assign them the following values:

C1 - 0.47 uF
R1 - 10 k
C - 560 pF
R - 10 k

(9) Identify the two filters by their function name.
(10) Compute fc for each filter
(11) Identify the whole filter circuit by it's proper name.

fc for 1st one is 33.86Hz
for 2nd one is 28.42Hz
so if decrease the capacitor critical frequency decreases.
dnt know 9 and 11

 

as resistance increases, break frequency decreases...

 

Your fc for the second is incorrect.

Read this http://www.allaboutcircuits.com/vol_2/chpt_8/index.html

 

Your fc for the second is incorrect.

Read this http://www.allaboutcircuits.com/vol_2/chpt_8/index.html

the page just shows a diagram of filters

 

Your fc for the second is incorrect.

so what is the correct one...i think i put wrong numbers on the equation
is 1/2pi*R*C?

 

the page just shows a diagram of filters

Read each of the links on the left.

so what is the correct one...i think i put wrong numbers on the equation

you missed it by a factor of 1000. Since you didn't show your work as you are suppose to, I can only assume you didn't enter in the number properly.

 

Read each of the links on the left.



you missed it by a factor of 1000. Since you didn't show your work as you are suppose to, I can only assume you didn't enter in the number properly.

1/(2pi)(10*10^3)(560*10^-9)=28.42

 

The inductor's impedance increases with increasing frequency.
The capacitor's impedance decreases with increasing frequency.
but what happen to filter ...not ev en sure what kind of filter is the circuit I put...
A filter is an AC circuit that separates some frequencies from others within mixed-frequency signals.
So how does it affect when change the resistance value
i think it shud be either low pass or high pass ...or both?