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A series RLC circuit is driven by a 1.0-kHz oscillator. The circuit parameters are Vms = 12 V, L = 5.0 mH, C= 4.0 uF, and R=102. Under steady-state conditions, the rms potential difference across the resistor will be A) 5.13 V B) 9.19 V C) 12.3 V D) 14.7 V E) 18.8 V

Respuesta :

Answer:

C) [tex]12.3 [/tex] volts

Explanation:

f = frequency of oscillator = 1 kHz = 1000 Hz

[tex]V_{rms}[/tex] = 12 Volts

L = Inductance of Inductor = 5 mH = 0.005 H

[tex]X_{L}[/tex] = Inductive reactance

Inductive reactance is given as

[tex]X_{L}[/tex] = [tex]2\pi fL[/tex]

[tex]X_{L}[/tex] = [tex]2(3.14) (1000)(0.005)[/tex]

[tex]X_{L} = 31.4 [/tex]

[tex]X_{C}[/tex] = Capacitive reactance

Capacitive reactance is given as

[tex]X_{C}=\frac{1}{2\pi fC}[/tex]

[tex]X_{C}=\frac{1}{2(3.14)(1000)(4\times 10^{-6})}[/tex]

[tex]X_{C} = 39.8 [/tex]

Impedance of the circuit is given as

[tex]z = \sqrt{R^{2}+(X_{L} - X_{C})^{2}}[/tex]

[tex]z = \sqrt{102^{2}+(31.4 - 39.8)^{2}}[/tex]

[tex]z = 102.35 [/tex]

Rms Current flowing is given as

[tex]i = \frac{V_{rms}}{z}[/tex]

[tex]i = \frac{12}{102.35}[/tex]

[tex]i = 0.12[/tex] A

Rms potential difference across the resistor is given as

[tex]V = i R [/tex]

[tex]V = (0.12) (102) [/tex]

[tex]V = 12.3 [/tex] volts