# 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?

#### Attachments

• 15431240790771815147712.jpg
34.2 KB · Views: 760
• 15431241659671431788042.jpg
17 KB · Views: 681

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

Homework Helper
Gold Member
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|##

Last edited: