Electronics - RLC Circuits and Resonance - Discussion
Discussion Forum : 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?
Discussion:
16 comments Page 1 of 2.
Ram charan said:
9 years ago
Q = xL/R
= 4k/50
= 4000/50
= 80 ohms.
= 4k/50
= 4000/50
= 80 ohms.
(1)
Harshal said:
7 years ago
@Swapnil,
Can you explain how wL/R =XL/R?
Can you explain how wL/R =XL/R?
(1)
Neha said:
1 decade ago
Please explain.
ANASWARA said:
1 decade ago
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
=1/50*(L/C)^.5
L=REACTANCE/2*Pi*f
C=1/2*pi*f*REACTANCE
Q=80
Deepak said:
1 decade ago
Q=XL/R
Bharath ks said:
1 decade ago
@Deepak you are absolutely right. Thank you.
Sree said:
1 decade ago
Q = (LW)/(R).
LW = 4KHz.
R = 50ohms.
Q = 80.
LW = 4KHz.
R = 50ohms.
Q = 80.
Yog said:
1 decade ago
Q = XL/R.
= (4*10^3)/50.
= 4000/50.
Q = 80.
= (4*10^3)/50.
= 4000/50.
Q = 80.
Raj said:
1 decade ago
Q = Vl/Vr = Vc/Vr.
=> Q = Xl/R = Xc/R (AT series resonance Xl = Xr).
= (4*10^3)/50.
=> Q = 80.
=> Q = Xl/R = Xc/R (AT series resonance Xl = Xr).
= (4*10^3)/50.
=> Q = 80.
Roy said:
1 decade ago
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.
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.
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