A series rlc circuit consists of a 231 ω resistor, a 14. 3 mh inductor, a 9. 71 μf capacitor, and an ac source of amplitude 39 v and variable frequency. The voltage source frequency is set to 6. 1 times the resonance frequency. Find the impedance of this circuit.

The opposition to alternating current performed by the combined effect of resistance and reactance in a circuit is known as impedance. To determine the impedance, we can use this formula:

Z = [tex]\sqrt{R^{2} + (XL -XC)^{2} }[/tex]

From the scenario we know that

ω = 231

L = 14.3 Mh

C = 9.71 μf = 9.71 x 10⁻⁶

V = 39 V

F = 6.1

Firstly we calculate the amount of XL using this formula

XL = 2πfL

XL =2 x 3.14 x 6.1 x 14.3

XL = 0.55 Ω

Then, we also determine the value of Xc

Xc = 1/2πfC

Xc = 1/(2 x 3.14 x 6.1 x 9.71)

Xc = 1/ 371.97 = 0.0026 Ω

After that substitue all the value into the formula to find the impedance

Z = [tex]\sqrt{R^{2} + (XL -XC)^{2} }[/tex]

Z = [tex]\sqrt{231^{2}+(0.55-0.0026)^{2} }[/tex]

Z = [tex]\sqrt{53,361-0.300}[/tex]

Z = [tex]\sqrt{53,360}[/tex]

Z = 230,1

So, the value of impedandance is 230,1 ohm.

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