Series LC circuit and voltage magnification?

[HMS-776]

I am trying to figure out what the voltage across each component will be. In the cirucit above resonance occurs between inductor L1 (5 Henries) and Capacitor C1 (10nF) The other inductor (L2) is 10 Henries and the circuit's resistance is 1,000 Ohms.

Here's my attempt at figuring the voltage magnification, total current, and voltage across each component.

Fres=711 Hz
XL1= 22.3 K Ohms
XL2= 44.6 K Ohms
XC1= 22.3 K Ohms

The Q of the circuit is XC or XL divided by R

Q= 22,300 / 1,000 = 23

Voltage magnification is Q times Applied Voltage. The applied voltage is 10V so the voltage magnification is 10 X 23 = 230V.

The circuit's total impedance at resonance is 44.6 K ohms + 1 K Ohm = 45,600 Ohms.
The total current is then V/R=I
I= 230 / 45,600 = .005 amps

So the voltage across each component is I X R = V

VL1 = 111 V
VL2 = ?
VC1 = 111 V

Inductor L2 has the highest reactance, so it should have the highest voltage drop right? Using ohms law the voltage drop across inductor L2 is 223 Volts which is obviously wrong.

What am I doing wrong here, and how do I find the current through the circuit and the voltage across each component? PLEASE HELP!

 

[Phrak]

You've left out resistance in adding up you voltages.
And where is the resistance?
Your voltages should add like vectors.

The square wave drive complicates things above the simplicity of a sine wave drive. You can rough the answer by assuming an equivalent sine wave at 711 Hertz and and 10 VAC RMS.

 

[vk6kro]

You can't get resonance with just two of the components like that.

The capacitor will resonate with the two inductors in series but not at 711 Hz. (more like 411 Hz). The circuit inductance is 15 H.

The whole loop, including the output impedance of the function generator, must be considered.

There will be some spurious effects due to the square wave and its harmonics, but consider it as a sinewave first to get the main effects.