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 2 of 2.
Yog said:
1 decade ago
Q = XL/R.
= (4*10^3)/50.
= 4000/50.
Q = 80.
= (4*10^3)/50.
= 4000/50.
Q = 80.
Sree said:
1 decade ago
Q = (LW)/(R).
LW = 4KHz.
R = 50ohms.
Q = 80.
LW = 4KHz.
R = 50ohms.
Q = 80.
Bharath ks said:
1 decade ago
@Deepak you are absolutely right. Thank you.
Deepak said:
1 decade ago
Q=XL/R
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
Neha said:
1 decade ago
Please explain.
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