# 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.
Nosheen Memon said:
2 years ago

If Xc is greater than Xl what will be the formula for impedance.

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.

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.

Saikiran said:
3 years ago

I am not getting this, Please the solution clearly.

Sushma said:
4 years ago

1/z = √ [(1/R)^2+(1/Xl-1/Xc)^2] for RLC circuit.

Hassan said:
5 years ago

How we can change the units?

Siddu bhandarkavathe said:
5 years ago

It is 1/2 π √ LC.

Puja said:
6 years ago

Yes, I agree @Prashant.

Dinesh said:
6 years ago

Quicker method needed.

Shailu said:
7 years ago

Is it correct? Above explained examples?

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