Current through source in AC circuit with R & LC in parallel

[Shivang kohlii]

Homework Statement

In the circuit diagram shown , Xc = 100 ohm , XL = 200 ohm , R = 100 ohm , the effective current through the source is ?

Homework Equations

Z= √( R^2 + ( XL - Xc)^2)
Vrms = Irms/ Z

The Attempt at a Solution

I tried to draw the phaser diagram and calculate the relation between the current in the LC branch and current in R branch( the phaser diagram is given as attachment)
From it I concluded that Inet = √( (I1)^2 + (I2)^2)
Where I1 = current through R branch
I2 = current through LC branch
And I got ans= 2√2 which is correct
1.What I am confused is that was my answer just by chance
2.And is it even possible to draw phaser diagrams when in parallel.. if yes, then is my attempt correct ?
3.If not , how else can we solve it .. or is there any easier method to solve it?

 

[Tom.G]

In relation to applied voltage, what is the phase angle of current thru a:
Resistor?
Capacitor?
Inductor?

 

[Delta2]

I think your answer (and the associated reasoning) is correct but it can't be solved in a simpler way.

We know that when we have elements in parallel the total impedance is computed in a similar way like when we have resistances in parallel.

In this circuit it would be $$\frac{1}{Z}=\frac{1}{Z_R}+\frac{1}{Z_{L+C}} \text{(1)}$$ where $Z$ is the total impedance of the circuit and $Z_{L+C}$ is the total impedance of the branch that contains the L and C. Because L and C are in series it is $Z_{L+C}=Z_L+Z_C=i{(X_L-X_C)}=i100 Ohm$

Also $Z_R=R=100Ohm$

So by (1) you can calculate $Z$, and then calculate $I_{rms}=V_{rms}/|Z|$