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

  • #1

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?
 

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Answers and Replies

  • #2
Tom.G
Science Advisor
3,245
1,994
In relation to applied voltage, what is the phase angle of current thru a:
Resistor?
Capacitor?
Inductor?
 
  • #3
Delta2
Homework Helper
Insights Author
Gold Member
2,975
1,027
I think your answer (and the associated reasoning) is correct but it cant 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:

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