Electrical Engineering - RLC Circuits and Resonance - Discussion

Discussion Forum : RLC Circuits and Resonance - General Questions (Q.No. 1)
1.
A 10 resistor, a 90 mH coil, and a 0.015 F capacitor are in series across an ac source. The impedance magnitude at 1,200 Hz below fr is
1,616
161
3,387
1,771
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
26 comments Page 2 of 3.

Prashant said:   9 years ago
I think result is not coming from above all formula I want real answer from you friends please reply.

Babu said:   1 decade ago
We know that,

Resonance frequency:

Fr = 1/2*pi*sqrt(L*C).

Xl = Wl = 2*3.14*f*l.

Xc = 1/wc.

Dinesh said:   1 decade ago
fr = 1/ 2*Pi*sqrt(L*C)
(or)
XL=2*Pi*F*L

Z=Sqrt of(R^2+(XL-XC)^2)

Nosheen Memon said:   4 years ago
If Xc is greater than Xl what will be the formula for impedance.
(1)

Rahul mishra said:   1 decade ago
IN RLC SERIES CIRCUITS WE KNOW:

Z = (jw)^2+(jw)R/L+1/LC.

Sushma said:   5 years ago
1/z = √ [(1/R)^2+(1/Xl-1/Xc)^2] for RLC circuit.

Saikiran said:   4 years ago
I am not getting this, Please the solution clearly.
(2)

Shwetha said:   9 years ago
Its (xl-xc). Because xl is greater than xc.

Shailu said:   9 years ago
Is it correct? Above explained examples?

RAO said:   1 decade ago
XL=2*Pi*F*L

Z=Sqrt of(R^2+(XL-XC)^2)


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