# 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:

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.

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.

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)

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.

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.

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.

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)

Kananya said:
1 decade ago

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)

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.

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.

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.

Manindra said:
1 decade ago

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.

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.

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