# 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:
28 comments Page 1 of 3.

Shaktimaan said:   1 year 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)

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)

AGGREY KERE said:   1 decade 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.

Joseph James said:   2 months ago
Given:

R = 10 ohms.
L = 90 mH.
C = 0.015 uF.
F = fr - 1200.

fr = 1/[2π√(LC)].
fr = 4331.65 Hz.

F = 4331.65 - 1200,
F = 3131.65 Hz.

Xl = 2πFL = 1771 ohms.
Xc = 1/(2πFC) = 3388 ohms.

Z= √[R²+ (Xl-Xc)²].
Z= 1617 ohms.

Joseph James said:   2 months ago
Given:

R = 10 ohms.
L = 90 mH,
C = 0.015 uF,
F = fr - 1200.

fr= 1/[2π√(LC)],
fr=4331.65 Hz.

F= 4331.65 - 1200,
F= 3131.65 Hz.

Xl = 2πFL = 1771 ohms,
Xc = 1/(2πFC) = 3388 ohms.

Z= √[R²+ (Xl-Xc)²].
Z= 1617 ohms.
(3)

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)

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.
(12)

ROBERT said:   10 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.

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.

Rajat Dey said:   9 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.