# 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:
26 comments Page 1 of 3.

Shaktimaan said:   10 months ago
@All.

Here is my explanation.

To calculate the impedance magnitude at 1,200 Hz below the resonance frequency (fr), we need to consider the components in the series circuit.

The impedance of a resistor (R) in an AC circuit is simply its resistance, which is 10 Ω in this case.

The impedance of an inductor (L) in an AC circuit is given by the formula ZL = jωL, where j is the imaginary unit (√(-1)), ω is the angular frequency, and L is the inductance. Since we are looking for the impedance magnitude, we need to calculate |ZL|.

Substituting the values into the formula, we have:
|ZL| = ωL = 2πfL,
= 2π(1,200 Hz)(90 mH).
≈ 678 Ω.

The impedance of a capacitor (C) in an AC circuit is given by the formula ZC = 1/(jωC). Again, we need to calculate |ZC|.

Substituting the values into the formula, we have:
|ZC| = 1/(ωC) = 1/(2πfC),
= 1/(2π(1,200 Hz)(0.015 F)),
≈ 1.11 Ω.

Now, to find the total impedance, we add the impedances of the three components since they are in series:
Z total = R + ZL + ZC.
= 10 Ω + 678 Ω + 1.11 Ω,
≈ 689.11 Ω.

Therefore, the impedance magnitude at 1,200 Hz below fr is approximately 689.11 Ω.
(2)

Nosheen Memon said:   4 years ago
If Xc is greater than Xl what will be the formula for impedance.
(1)

Azeem said:   4 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.
(10)

Usman Asif NFC said:   4 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).
(2)

Saikiran said:   4 years ago
I am not getting this, Please the solution clearly.
(2)

Sushma said:   5 years ago
1/z = √ [(1/R)^2+(1/Xl-1/Xc)^2] for RLC circuit.

Hassan said:   6 years ago
How we can change the units?

Siddu bhandarkavathe said:   7 years ago
It is 1/2 π √ LC.

Puja said:   7 years ago
Yes, I agree @Prashant.

Dinesh said:   7 years ago
Quicker method needed.