Electronics - RLC Circuits and Resonance - Discussion
Discussion Forum : RLC Circuits and Resonance - General Questions (Q.No. 40)
40.
What is the band pass (F1 – F2) of an RLC filter that resonates at 150 kHz and has a coil Q of 30?
Discussion:
3 comments Page 1 of 1.
KIRAN V said:
5 years ago
Here: band pass (F1 " F2) = is nothing but, Bandwidth (B.W).
Given:
fr = 150 kHz; Q = 30; B.W =?
BANDWIDTH (B.W) = (Resonant freq.)/(Quality Factor) B.W = f_r/Q.
B.W = fr/Q.
(f2 - f1) = (150 * 10^3 )/30.
(f2 - f1) = 5 * 10^3 = 5 kHz.
F1 = (fr- f1) = (150 - 2.5) * 10^3 = 147.5 kHz.
F2 = (fr- f2) = (150+2.5) * 10^3 = 152.5 kHz.
Given:
fr = 150 kHz; Q = 30; B.W =?
BANDWIDTH (B.W) = (Resonant freq.)/(Quality Factor) B.W = f_r/Q.
B.W = fr/Q.
(f2 - f1) = (150 * 10^3 )/30.
(f2 - f1) = 5 * 10^3 = 5 kHz.
F1 = (fr- f1) = (150 - 2.5) * 10^3 = 147.5 kHz.
F2 = (fr- f2) = (150+2.5) * 10^3 = 152.5 kHz.
Ankush said:
1 decade ago
Can anybody explain this?
Sreeyush Sudhakaran said:
1 decade ago
B.w=fr/Q
B.W=150K/30 =5KHz
f1+f2=5KHz if f1=f2 the f1=f2=2.5KHz
F1 = fr-f1=150-2.5KHz=147.5KHz
F2 = fr+f1=150+2.5KHz=152.5KHz
B.W=150K/30 =5KHz
f1+f2=5KHz if f1=f2 the f1=f2=2.5KHz
F1 = fr-f1=150-2.5KHz=147.5KHz
F2 = fr+f1=150+2.5KHz=152.5KHz
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