Electronics - RLC Circuits and Resonance - Discussion
Discussion Forum : RLC Circuits and Resonance - General Questions (Q.No. 6)
6.
What is the range between f1 and f2 of an RLC circuit that resonates at 150 kHz and has a Q of 30?
Discussion:
10 comments Page 1 of 1.
Mzr said:
7 years ago
Correct, thanks @Hussain.
Munna said:
8 years ago
Right @Sachin.
Sachin said:
8 years ago
BW = fr/Q = 150/30 = 5khz.
Then Lower frequency= fr - 1/2 BW= 150 - 1/2 x 5= 146.5 kHz.
Higher frequency = fr + 1/2 BW= 150 + 1/2 x 5= 152.5 khz.
Then Lower frequency= fr - 1/2 BW= 150 - 1/2 x 5= 146.5 kHz.
Higher frequency = fr + 1/2 BW= 150 + 1/2 x 5= 152.5 khz.
(1)
Amit said:
9 years ago
You are right @M.V.Krishna.
Amit Zanwar said:
1 decade ago
It can also be solved by taking the mean of the options and checking it with the resonant freq.
So that's why its option (b).
So that's why its option (b).
Noopur khare said:
1 decade ago
MV Krishna is Right
SANJAY said:
1 decade ago
Q = 30; Fr = 150 KHz.
BW = Fr/Q = 150 K/30 = 5KHz.
BW = Fh - Fl = 5KHz.
Fh - Fl = 5 KHz.------------(1)
Fh + Fl = 300 KHz.----------(2)
Solve both equation
Fh=152.5
Fl=147.5
BW = Fr/Q = 150 K/30 = 5KHz.
BW = Fh - Fl = 5KHz.
Fh - Fl = 5 KHz.------------(1)
Fh + Fl = 300 KHz.----------(2)
Solve both equation
Fh=152.5
Fl=147.5
M.V.KRISHNA/palvoncha said:
1 decade ago
We know,
BW=fr/Q;
BW=(f2-f1);
(f2-f1)= fr/Q;
(f2-f1)= 150k/30;
(f2-f1)= 5k;
now checking with options:
A)155KHz-100kHz=55kHz;
B)152.5kHz-147.5kHz=5kHz;
c)295.5 kHz-4500 kHz= -4204.5
D)150,030 Hz-149,970 Hz=60Hz
So option B is correct.
BW=fr/Q;
BW=(f2-f1);
(f2-f1)= fr/Q;
(f2-f1)= 150k/30;
(f2-f1)= 5k;
now checking with options:
A)155KHz-100kHz=55kHz;
B)152.5kHz-147.5kHz=5kHz;
c)295.5 kHz-4500 kHz= -4204.5
D)150,030 Hz-149,970 Hz=60Hz
So option B is correct.
Hussain said:
1 decade ago
Q = 30; Fr = 150 KHz
BW = Fr/Q = 150 K/30 = 5KHz.
BW = Fh - Fl = 5KHz.
so Fh should be above the Fr and Fl should be less than the Fr.
Fh > 150 KHz and Fl<150 KHz and difference will b 5 KHz.
BW = Fr/Q = 150 K/30 = 5KHz.
BW = Fh - Fl = 5KHz.
so Fh should be above the Fr and Fl should be less than the Fr.
Fh > 150 KHz and Fl<150 KHz and difference will b 5 KHz.
(1)
Kumar_abhinav_anand said:
1 decade ago
f1=fr-r/(4*pi*l)
f2=fr+r/(4*pi*l)
f2=fr+r/(4*pi*l)
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