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.
Aruna said:
5 years ago
Q = XL/R.
Ban said:
7 years ago
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.
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.
Harshal said:
7 years ago
@Swapnil,
Can you explain how wL/R =XL/R?
Can you explain how wL/R =XL/R?
(1)
Pawan siddh said:
8 years ago
Q=XL/R.
4000/50 = 80ans.
4000/50 = 80ans.
Ashutosh singh said:
8 years ago
Her, Q = Xl/R.
Ram charan said:
9 years ago
Q = xL/R
= 4k/50
= 4000/50
= 80 ohms.
= 4k/50
= 4000/50
= 80 ohms.
(1)
Swapnil said:
1 decade ago
Q = wL/R = XL/R.
R = 50 OHM.
XL = XC=4K OHM = 4*10^3 OHM.
Q = 4*10^3/50OHM.
Answer = 80 OHM.
R = 50 OHM.
XL = XC=4K OHM = 4*10^3 OHM.
Q = 4*10^3/50OHM.
Answer = 80 OHM.
Abhijit said:
1 decade ago
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.
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.
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.
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.
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