# 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

*f*is_{r}Discussion:

28 comments Page 2 of 3.
Teju said:
1 decade ago

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)

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)

Mrakovic said:
1 decade ago

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.

fr = 4331.648.

f = fr-1200Hz=3131.64.

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

Z = 1617 ohm.

Prashant said:
10 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.

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)

(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.

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.

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