Electronics - RLC Circuits and Resonance - Discussion
Discussion Forum : RLC Circuits and Resonance - General Questions (Q.No. 2)
2.
What is the bandwidth of the circuit?
Discussion:
34 comments Page 1 of 4.
LALIT GARG said:
1 decade ago
As Q=B.W/fr
so fr=1/(2pie*underroot of LC)
And Q=UNDERROOT OF L/(R*UNDERROOT OF C)
B.W=?
so fr=1/(2pie*underroot of LC)
And Q=UNDERROOT OF L/(R*UNDERROOT OF C)
B.W=?
Vimal said:
1 decade ago
How?
Maddy said:
1 decade ago
@Lalit from ur formulas I got 142Hz ans.
Is that ri8?
Is that ri8?
AYSHA said:
1 decade ago
fr=BW*Q
Hussain said:
1 decade ago
BW = fr/Q; ---- @Aysha: i agree with u
Q ---- quality factor, higher the quality factor lower the Bandwidth.
Q = (1/R)*(sqrt(L/c));
fr = 1/(2*pi*sqrt(LC)); ---- resonant frequency.
so BW = R/(2*pi*L);
so on substituting the values BW = 31.8 Hz
Q ---- quality factor, higher the quality factor lower the Bandwidth.
Q = (1/R)*(sqrt(L/c));
fr = 1/(2*pi*sqrt(LC)); ---- resonant frequency.
so BW = R/(2*pi*L);
so on substituting the values BW = 31.8 Hz
Oneidensis said:
1 decade ago
My answer is 200... iIve been solving this so many times.. here's the formulas I used,
f=1/(square roof of (LC))
then,
Q= (1/fRC) ; Q= (fL/R)
then,
B=f/Q... B=200
f=1/(square roof of (LC))
then,
Q= (1/fRC) ; Q= (fL/R)
then,
B=f/Q... B=200
Sap said:
1 decade ago
Hussain you are right.
(1)
Seenivasagan said:
1 decade ago
Bandwidth(BW)=fr/Q
Q=(2*pi*fr*L)/R
fr=1/(2*pi*root LC)
by applying this ,
we can get fr=71.21
Q=2.236
BW=fr/Q
=71.21/2.236
=31.84
approximately 31.8
Q=(2*pi*fr*L)/R
fr=1/(2*pi*root LC)
by applying this ,
we can get fr=71.21
Q=2.236
BW=fr/Q
=71.21/2.236
=31.84
approximately 31.8
Sandeep said:
1 decade ago
Q=Fr/BW;
BW=Fr/Q;
Fr=1/2*PI*ROOT(L*C);
Q=(1/R)(ROOT(L/C));
BW=Fr/Q;
Fr=1/2*PI*ROOT(L*C);
Q=(1/R)(ROOT(L/C));
M.V.KRISHNA said:
1 decade ago
BW=fr/Q;
fr=1/(2*pi*underroot(L*C));
Q=(1/R)*underroot(L/C);
BW=R/(2*pi*L);
=>1Kohm/(2*pi*5H);
=32.3Hz
fr=1/(2*pi*underroot(L*C));
Q=(1/R)*underroot(L/C);
BW=R/(2*pi*L);
=>1Kohm/(2*pi*5H);
=32.3Hz
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