Question

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

Answer 1

Answer:

C) $12.3$ volts

Explanation:

f = frequency of oscillator = 1 kHz = 1000 Hz

$V_{rms}$ = 12 Volts

L = Inductance of Inductor = 5 mH = 0.005 H

$X_{L}$ = Inductive reactance

Inductive reactance is given as

$X_{L}$ = $2\pi fL$

$X_{L}$ = $2(3.14) (1000)(0.005)$

$X_{L} = 31.4$

$X_{C}$ = Capacitive reactance

Capacitive reactance is given as

$X_{C}=\frac{1}{2\pi fC}$

$X_{C}=\frac{1}{2(3.14)(1000)(4\times 10^{-6})}$

$X_{C} = 39.8$

Impedance of the circuit is given as

$z = \sqrt{R^{2}+(X_{L} - X_{C})^{2}}$

$z = \sqrt{102^{2}+(31.4 - 39.8)^{2}}$

$z = 102.35$

Rms Current flowing is given as

$i = \frac{V_{rms}}{z}$

$i = \frac{12}{102.35}$

$i = 0.12$ A

Rms potential difference across the resistor is given as

$V = i R$

$V = (0.12) (102)$

$V = 12.3$ volts