# 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:

25 comments Page 2 of 3.
Rajat Dey said:
8 years ago

I know above formula but this is very long process and heavy time span. So how can do it quickly? Please help me.

Shwetha said:
8 years ago

Its (xl-xc). Because xl is greater than xc.

Prashant said:
8 years ago

I think result is not coming from above all formula I want real answer from you friends please reply.

ROBERT said:
8 years ago

XL=2*3.14*F*L THIS IS THE FORMAL.

I WAS ASK TO MAKE SO I CAN FIND UNKNOWN F AND L.

MOVE THING AROUND I CAME UP WITH SO ARE THESE RIGHT.

F=2*PI*L*XL.

L=2*PI*F*XL.

I WAS ASK TO MAKE SO I CAN FIND UNKNOWN F AND L.

MOVE THING AROUND I CAME UP WITH SO ARE THESE RIGHT.

F=2*PI*L*XL.

L=2*PI*F*XL.

AGGREY KERE said:
9 years ago

From R-L-C circuit above, we know that; Z=root of R^2+(XL-XC)^2.

but XL=2pi*fr*L.

XC=1/2Pi*fr*C.

but XL=2pi*fr*L.

XC=1/2Pi*fr*C.

Manindra said:
9 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.

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.

DURAI said:
9 years ago

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

Babu said:
9 years 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.

Mrakovic said:
9 years 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.

Rahul mishra said:
10 years ago

IN RLC SERIES CIRCUITS WE KNOW:

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

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

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