Engineering Calculating the effective value and frequency of AC Circuit

[diredragon]

Homework Statement

For AC Circuit shown in the picture, $C=60nF$, $L=100uH$, $X_3=-100Ω$, $Z_2=50Ω$ and $Φ_2=\frac {\pi}{6}$. When the switch is closed the ammeter shows a current $I_1=0.6A$ and the voltmeter shows $U_l=60V$. The measuring instruments mentioned are ideal. Calculate:
a) $w$
b) effective value of $E$
c) ammeter result $I'_1$ when the switch is open.
The Circuit:

Homework Equations

3. The Attempt at a Solution [/B]
The first thing i noticed is that the direction of current and voltage isn't given so I need to guess it. I started from the ammeter and guessed that the flow of current is downwards.
From there, $U_1 = I_1*X1$ = -60V## with a + sign at the top node. Once i have this voltage i can find the current $I_2 = \frac {U_1}{Z_2} = -1.2A$. With the two currents meeting at the top node, the current $I_l = I_c$ of the top node equals $I_l=0.6A$.

Since i have $U_l = I_l*wL$, $w = \frac{U_L}{I_lL} = 10^6$ which is not the correct answer.
The answer should be $w = \frac {\sqrt 3}{3}10^6$. I can't go on solving the other parts if the first is not correct. What did i do wrong?

 

[.Scott]

There are no DC voltages or amperage's in this circuit. Everything is AC.
The 60V and 600mA are presumably rms.

Also, I do not know what your "X₃=−100Ω" refers to.
Unless that is the value for X₁ (using a capacitor symbol).

If X₁ is not a cap, then part of the circuit have a DC component to the current, but still not the volt meter.

 

[diredragon]

There are no DC voltages or amperage's in this circuit. Everything is AC.
The 60V and 600mA are presumably rms.

Also, I do not know what your "X₃=−100Ω" refers to.
Unless that is the value for X₁ (using a capacitor symbol).

If X₁ is not a cap, then part of the circuit have a DC component to the current, but still not the volt meter.

The $X_3$ was supposed to be $X_1$. I solved it now, i thought i could work with effective values of currents only but i guess i needed a complex analysis to solve it.