# 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:
25 comments Page 1 of 3.

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.
(4)

Saikiran said:   3 years ago
I am not getting this, Please the solution clearly.
(2)

Usman Asif NFC said:   3 years ago
Following Value given:
R= 10 ohm.
L= 90mH.
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).
(1)

Prashant said:   8 years ago
I think result is not coming from above all formula I want real answer from you friends please reply.

Nosheen Memon said:   3 years ago
If Xc is greater than Xl what will be the formula for impedance.

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:   6 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.