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

C)
`12.3` volts

Step-by-step 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)=(1)/(2\pi fC)`

`X_(C)=(1)/(2(3.14)(1000)(4* 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 = (V_(rms))/(z)`

`i = (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