Electronics - RLC Circuits and Resonance - Discussion

5. 

What is the Q (Quality factor) of a series circuit that resonates at 6 kHz, has equal reactance of 4 kilo-ohms each, and a resistor value of 50 ohms?

[A]. 0.001
[B]. 50
[C]. 80
[D]. 4.0

Answer: Option C

Explanation:

No answer description available for this question.

Neha said: (Oct 18, 2010)  
Please explain.

Anaswara said: (Feb 10, 2011)  
Q=1/R*(L/C)^.5 1/2=.5
=1/50*(L/C)^.5
L=REACTANCE/2*Pi*f
C=1/2*pi*f*REACTANCE
Q=80

Deepak said: (Feb 12, 2011)  
Q=XL/R

Bharath Ks said: (Dec 22, 2011)  
@Deepak you are absolutely right. Thank you.

Sree said: (Jan 6, 2013)  
Q = (LW)/(R).
LW = 4KHz.
R = 50ohms.
Q = 80.

Yog said: (Feb 5, 2013)  
Q = XL/R.

= (4*10^3)/50.

= 4000/50.

Q = 80.

Raj said: (Mar 21, 2013)  
Q = Vl/Vr = Vc/Vr.

=> Q = Xl/R = Xc/R (AT series resonance Xl = Xr).

= (4*10^3)/50.

=> Q = 80.

Roy said: (Oct 8, 2013)  
We know, Q(Quality Factor) = ((L)^1/2)/(R*(C)^1/2)) -------- 1

R = 50 Ohms(Given).

Also,

XL = 2*pi*F*L => L=0.1061(Substituting the values F = 6000 Hz, XL = 4000 Ohms).

XC = 1/2*pi*F*C => C = 6.63*10^-9(Substituting the values F = 6000 Hz, XC = 4000 Ohms).

Now Substituting the values of L, C and R.

We get Q = 80.

Abhijit said: (Jan 10, 2014)  
Fr = 6KHZ= 6000HZ.
XL = XC = 4KOhm = 4000 OHM.
R = 50 OHM.

XL = 2*Pi*Fr*L.
L = XL/(2*Pi*Fr) = 1/3*Pi.
XC = 1/(2*Pi*Fr*C).
C = 1/(2*Pi*Fr*XC)= 1/(48*10^6).

QUALITY FACTOR(Q) = (1/R)*SQRT(L/C).
= (1/50)*SQRT((48*Pi*10^6)/3*Pi).
= 4000/50 = 80.

Swapnil said: (Apr 19, 2014)  
Q = wL/R = XL/R.
R = 50 OHM.

XL = XC=4K OHM = 4*10^3 OHM.
Q = 4*10^3/50OHM.

Answer = 80 OHM.

Ram Charan said: (Jan 19, 2017)  
Q = xL/R
= 4k/50
= 4000/50
= 80 ohms.

Ashutosh Singh said: (Mar 5, 2017)  
Her, Q = Xl/R.

Pawan Siddh said: (Aug 23, 2017)  
Q=XL/R.

4000/50 = 80ans.

Harshal said: (Jun 25, 2018)  
@Swapnil,

Can you explain how wL/R =XL/R?

Ban said: (Sep 14, 2018)  
Quality Factor=(0.159)/f*R*C.

XL=XC.
Xc=1/2* π * f * c.
By using this formula we get the value of C=6631.45pF(6631.45x10^-12),
Substitute the value of C, f and R to the formula of quality factor.
We get Q=79.922 approximately 80.

Aruna said: (May 5, 2020)  
Q = XL/R.

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