Electrical Engineering - Passive Filters - Discussion

Discussion :: Passive Filters - General Questions (Q.No.12)


An RC high-pass filter consists of an 820 resistor. What is the value of C so that Xc is ten times less than R at an input frequency of 12 kHz?

[A]. 81 F
[B]. 161 F
[C]. 0.161 F
[D]. 220 F

Answer: Option C


No answer description available for this question.

Rajendra said: (Oct 19, 2012)  
For cut off freq Fc=1/(2*Pie*R*C),
Fc=12khz, R=820 ohm.
So by calculating we get 0.161 uF value of capacitor.

Santosh Kumar said: (Jun 17, 2015)  
Please show whole calculation.

Maha said: (Aug 19, 2015)  

R = 820 ohm, f = 12 KHz.

Xc = 8200 ohm. (10 times the value of R).

So from the formula Xc = 1/(2 *pie*f*c) we can calculate c.

The value is 0.161 microfarad.

Mak said: (Apr 6, 2016)  
Thanks for your effort @Maha but I disagree with the value of your reactive capacitance you wouldn't get the answer if your Xc is 8200 ohm. Xc must be 82 ohms only remember Xc is 10 times less than are meaning your Xc must be smaller than your resistor in that case you can get the value of 0.161 uF.

Sofi said: (Sep 21, 2016)  
Except with the value of Xc, it was a nice attempt @Maha.

@Srk said: (Oct 10, 2016)  
The value of Xc should be 82 ohms. But the result differs.

Nethaji said: (Mar 18, 2017)  
Xc = 1/(2 * π * f * c).
R = 820 ohm
Xc = 82 ohm(10 time less than the R); F=12kohm.
ANS:0.161 microfarad.

Rajashekar Balya said: (Jan 3, 2018)  

Majid Mahmood said: (Aug 19, 2019)  
Xc=1/2μ fc.
So c=1/2μfxc, f =12khz.
10xc=R so, xc=R/10,820/10=82.
So final xc=82.

C=1/2*3.1416*12*82 = 0.0161uf.

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