# 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 1 of 3.
Usman Asif NFC said:
3 years ago

Following Value given:

R= 10 ohm.

L= 90mH.

C= 0.015 micro Farad.

f=1200 Hz.

Impedance Formula: Z = √ ( R^2 + (X.L - X.C)^2).

We have value of "R" Now find X.L & X.C

X.L = wL Where as (w: ω = 2*π*f*L).

X.C = 1/w.C.

frequency = 1/ (2*π*L*C).

Now put all the required values in Z = √( R^2 + (X.L - X.C)^2).

We will get the answer.

R= 10 ohm.

L= 90mH.

C= 0.015 micro Farad.

f=1200 Hz.

Impedance Formula: Z = √ ( R^2 + (X.L - X.C)^2).

We have value of "R" Now find X.L & X.C

X.L = wL Where as (w: ω = 2*π*f*L).

X.C = 1/w.C.

frequency = 1/ (2*π*L*C).

Now put all the required values in Z = √( R^2 + (X.L - X.C)^2).

We will get the answer.

(1)

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.

Kananya said:
1 decade ago

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)

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)

Azeem said:
3 years ago

As the given frequency is 1200 Hz less than fr(resonance).

Frequency = fr-1200

Fr= 1/2*π√LC.

fr= 4331.

Frequency = f = 4331-1200 = 3131 Hz.

Z= √R^2 + ( Xl-Xc)^2.

Z= 1616.

Frequency = fr-1200

Fr= 1/2*π√LC.

fr= 4331.

Frequency = f = 4331-1200 = 3131 Hz.

Z= √R^2 + ( Xl-Xc)^2.

Z= 1616.

(4)

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.

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.

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.

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:
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.

Prashant said:
8 years ago

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

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