# 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:

26 comments Page 3 of 3.
Sushma said:
5 years ago

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

Saikiran said:
4 years ago

I am not getting this, Please the solution clearly.

(2)

Usman Asif NFC said:
4 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.

(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.

Frequency = fr-1200

Fr= 1/2*π√LC.

fr= 4331.

Frequency = f = 4331-1200 = 3131 Hz.

Z= √R^2 + ( Xl-Xc)^2.

Z= 1616.

(11)

Nosheen Memon said:
4 years ago

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

(1)

Shaktimaan said:
11 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.

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)

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