# 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
Explanation:
No answer description is available. Let's discuss.
Discussion:
26 comments Page 1 of 3.

fr = 1/ 2*Pi*sqrt(L*C)

XL=2*Pi*F*L

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

fr = 1/ 2*Pi*sqrt(L*C)
(or)
XL=2*Pi*F*L

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

We know that formula for Impedence(Z)
Z=Sqr root of (R^2+(Xl-Xc)^2)
Here R=Resistance
Xl=wl (w=2*3.14*f) and l=inductance of a coil
Xc=1/wc Here c=capacitance of capaciter

Z=Sqr root of (R^2+(Xl-Xc)^2)

XL=WL=(2*PI*f)L
XC=1/WC=1/(2*PI*f)C
where,L=90mh,C=.015microF, f=1200hz and R=10ohm

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

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

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

fr = 1/2*pi*sqrt of LC.
fr = 4331.648.

f = fr-1200Hz=3131.64.
Z = sqrt of(R^2+(XL-XC)^2).

Z = 1617 ohm.

We know that,

Resonance frequency:

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

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

Xc = 1/wc.

fr = 1/2*pi*sqrt (L*c).

Manindra said:   10 years ago
Fr = 1/2*pi*sqrt(L*c)=4331.648.
f = fr-1200Hz = 3131.64.

Xl = Wl = 2*3.14*3131.64*l.

Xc = 1/wc.
Z = sqrt of(R^2+(XL-XC)^2) = 1616 ohm.