Discussion :: RLC Circuits and Resonance  General Questions (Q.No.2)
2.  What is the bandwidth of the circuit? 

Answer: Option A Explanation: No answer description available for this question.

Lalit Garg said: (Nov 1, 2010)  
As Q=B.W/fr so fr=1/(2pie*underroot of LC) And Q=UNDERROOT OF L/(R*UNDERROOT OF C) B.W=? 
Vimal said: (Apr 4, 2011)  
How? 
Maddy said: (Apr 21, 2011)  
@Lalit from ur formulas I got 142Hz ans. Is that ri8? 
Aysha said: (Apr 27, 2011)  
fr=BW*Q 
Hussain said: (Jun 1, 2011)  
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 
Oneidensis said: (Jun 23, 2011)  
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 
Sap said: (Aug 12, 2011)  
Hussain you are right. 
Seenivasagan said: (Sep 1, 2011)  
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 
Sandeep said: (Sep 3, 2011)  
Q=Fr/BW; BW=Fr/Q; Fr=1/2*PI*ROOT(L*C); Q=(1/R)(ROOT(L/C)); 
M.V.Krishna said: (Sep 6, 2011)  
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 
Amrita said: (Sep 17, 2011)  
BANDWIDTH= Resonant freq/ Quality Factor Resonant Frequency= 1/(2*pi*sqrt of(L*C)) Quality Factor= 1/R*(sqrt(L/C)) Using above fomula we can determine, Bandwidth=31.83 . 
Fadi said: (Mar 27, 2012)  
B.W=FR/Q Q=(1/R)SQRT(L/C) FR=1/2*PI*SQRT(LC) B.W=[1/2*PI*SQRT(LC)]/(1/R)SQRT(L/C) B.W=R*SQRT C / 2*PI*L*SQRT C B.W=R/2*PI*L 
Sreeyush Sudhakaran said: (Aug 19, 2012)  
B.W=fr/Q Q=(2*Pi*fr*L)/R B.W = fr/((2*Pi*fr*L)/R) = R/2*Pi*L = 1000/2*3.14*5 = 31.8Hz 
Alka said: (Jan 15, 2013)  
BW=fr/Q Q=(2*pi*fr*L)/R Bw=R/(2*pi*L) =31.83 
Manjunath said: (May 11, 2013)  
Bandwidth = R/2*pi*L . BW= 1/2*3.14*5 = 31.8 Hz. 
Govind Singh Chouhan said: (Nov 25, 2013)  
Data Given: R = 1k. C = 1uF. L = 5H. So, resonance freq. is = 1/2*pi*sqrt(LC). (fo is a resonance freq.) fo = 1/2*3.14*sqrt(5*1*10^6)). fo = 1/2*3.14*sqrt(0.000005). fo = 71.42Hz. we know about the Quality factor "Q", Q = WoL/R. Q = 2*pi*fo*5/1k. Q = 2.24. The relationship between Q,fo and BW is: Q = fo/BW. So BW = fo/Q. BW = 71.42/2.24. BW = 31.88Hz. 
Shah said: (Apr 5, 2014)  
B' = R/L in radians. B' = 1000/5. B' = 200. B = B'/2*pi. B = 200/2*pi. B = 31.83. 
Daks said: (Oct 16, 2014)  
Easy way. BW = Fr/Q BW = 1/2*pi*L/R BW = R/2*pi*L BW = 31.83 
Ruben said: (Mar 27, 2015)  
So, when you decrease the resistance in the circuit Q will increase and bandwidth and will decrease, but phase angle theta will increase. The circuit efficiency will decrease number. 
Dhana said: (Jun 4, 2015)  
XL=wL. XL=(2*pi*f)(5H). XL=2*22/7*5. XL=31.42 app. 
Krupal Gowda Jedrali said: (Jul 12, 2015)  
= r/2.pai.L formula. = 1000/2*pai*5 = 31.8105 hz. 
Gaurav Kumar Yadav said: (Aug 19, 2015)  
As Bandwidth = Resonant frequency/Quality Factor. And Resonant Frequency = 1/(2*pi*Sqareroot LC). And Quality factor = 1/R*squareroot(L/C). Now BW = 1/ (2*pi*Suareroot LC)/1/R*squareroot (L/C) = R/(2*pi*L). BW = 1000/(2*3.14*5) = 31.8 Hz. 
Laurent P .Eranga said: (Nov 23, 2015)  
First find fr = 1/6.28 times square root of LC. You will got 71.2 hz. Then find xl = 6.28 times fl you will got 2.23 k if xl = xc. Then q = xl/r you will got 2.23 k/1 k = 2.235. If the formula for bw = fr/q. fr = 71.2 and q = 2.235 so bw = fr/q will be equals to 31.8 hz. 
Jatin Choudhary said: (Feb 21, 2016)  
f = R/2*3.14*L. 
Ashish said: (May 31, 2017)  
Here, r/2 * π * L = 31.8. 
Vijay said: (Sep 12, 2017)  
Q=(omega*L)/R. BW=f/Q, => BW=(f*R)/(w*L) => BW=R/(2*pi*L) => Bw=31.8. 
Mody Shaban said: (Dec 30, 2017)  
HOW? 
Ayush Purwar said: (Jan 23, 2018)  
Bandwidth= R/(2*L)...Hz. = 1000/(2*3.14*5). = 31.84 Hz. 
Suma said: (Sep 27, 2018)  
Here, R/(2*π*L) = 31.8. 
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