# Electronics - RLC Circuits and Resonance - Discussion

### Discussion :: RLC Circuits and Resonance - General Questions (Q.No.5)

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

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.