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
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 is1,616
161
3,387
1,771
Discussion:
25 comments Page 1 of 3.
Nosheen Memon said:
3 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.
(4)
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)
Saikiran said:
3 years ago
I am not getting this, Please the solution clearly.
(2)
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.
Shailu said:
8 years ago
Is it correct? Above explained examples?
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