Electrical Engineering - Passive Filters - Discussion

Discussion Forum : Passive Filters - General Questions (Q.No. 12)
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?
81 F
161 F
0.161 F
220 F
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
9 comments Page 1 of 1.

Majid Mahmood said:   6 years ago
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.
(1)

Rajashekar balya said:   8 years ago
10Xc=R.
Xc=1/2μfc.

Nethaji said:   8 years ago
Xc = 1/(2 * π * f * c).
R = 820 ohm
Xc = 82 ohm(10 time less than the R); F=12kohm.
ANS:0.161 microfarad.

@srk said:   9 years ago
The value of Xc should be 82 ohms. But the result differs.

Sofi said:   9 years ago
Except with the value of Xc, it was a nice attempt @Maha.

Mak said:   9 years ago
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.

Maha said:   1 decade ago
Given:

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.

Santosh kumar said:   1 decade ago
Please show whole calculation.

Rajendra said:   1 decade ago
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.

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