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