Series RLC circuits

[Bob Loblaw]

Homework Statement

A series RLC circuit has a 0.19 mF capacitor, a 16 mH inductor, and a 10.0 resistor, and is connected to an ac source with amplitude 9.0 V and frequency 60 Hz.
(a) Calculate the voltage amplitudes VL, VC, VR, and the phase angle phi .
VL = V
VC = V
VR = V
angle phi= °


VL= IrmsXL
VC=IrmsXC
VR=IrmsR
angle phi =

The Attempt at a Solution

I calculated XL=wL=2pi60Hz*16x10^-3 = 6.03
XC = 1/wC = 1/2pi*60Hz*0.19*10^-3 = 13.96

angle phi =tan-1 (XL-XC/R) which is tan-1(6.03-13.96/10 ohms) = -38.41 degrees.

I have been unable to find VL, VC and VR. I tried translating 9V amplitude into rms and got 6.36V. I tried using Vrms=IrmsR to find Irms but have been unsuccessful. Any ideas?

 

[gneill]

A complete solution is offered.

Let:

E = 9.0 V {We assume given magnitudes are RMS unless otherwise indicated}
R = 6.0 Ω
L = 16 mH
C = 0.19 mF

Calculate the reactances:

$XL = 2\pi f L = 6.032~Ω$
$XC = 1/(2 \pi f C) = 13.96~Ω$

Calculate the current in the series circuit:

$I = \frac{E~∠~0°}{\sqrt{R^2 + (X_L - X_C)^2}~∠~atan \left(\frac{X_L - X_C}{R} \right) }$

$~~= \frac{9~V~∠ 0°}{12.76~Ω~∠~-38.41°} = 0.7052~A ~∠~38.41°$

The phase angle $\phi$ is the angle of the current above: $\phi = 38.41°$

To find the individual voltages across the components we multiply the current by the reactances. Both numbers have phase angles associated with them, so:

$V_L = I~X_L = (0.7052~A~ ∠~ 38.41°)(6.032~Ω~∠~90°) = 4.25~V~∠~128.4°$

$V_C = I~X_C = (0.7052~A~∠~ 38.41°)(13.961~Ω~∠~-90°) = 9.84~V~∠~-51.59°$

$V_R = I~R = (0.7052~A~∠~ 38.41°)(10.0~Ω~∠~0°) = 7.05~V~∠~38.41°$