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

In summary, the effective current through the source in the given circuit can be calculated by using the equations Z= √( R^2 + ( XL - Xc)^2) and Vrms = Irms/ Z. The phaser diagram can also be used to determine the relation between the current in the LC branch and the current in the R branch. The formula Inet = √( (I1)^2 + (I2)^2) can be used to calculate the total current in the circuit. This method is correct but there may not be a simpler way to solve this problem.
  • #1
Shivang kohlii
19
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?
 

Attachments

  • 15431240790771815147712.jpg
    15431240790771815147712.jpg
    34.2 KB · Views: 802
  • 15431241659671431788042.jpg
    15431241659671431788042.jpg
    17 KB · Views: 717
Physics news on Phys.org
  • #2
In relation to applied voltage, what is the phase angle of current thru a:
Resistor?
Capacitor?
Inductor?
 
  • #3
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:

1. What is the purpose of an LC circuit in an AC circuit?

An LC circuit, also known as a tank circuit, is used to create resonance in an AC circuit. This allows for a specific frequency of the alternating current to be amplified and transferred to the load. It can also be used to filter out unwanted frequencies in the circuit.

2. How does the current flow through a source in an AC circuit with R & LC in parallel?

In an AC circuit with R & LC in parallel, the current will flow through the source and split into two paths: one through the resistor (R) and the other through the inductor (L) and capacitor (C) in parallel. The amount of current flowing through each path will depend on the impedance of the components.

3. What is the effect of changing the resistance (R) in an AC circuit with R & LC in parallel?

Changing the resistance (R) in an AC circuit with R & LC in parallel will affect the amount of current flowing through the resistor. This will also change the amount of voltage drop across the resistor and may affect the resonant frequency of the circuit.

4. How does the presence of an inductor (L) and capacitor (C) in parallel affect the current in an AC circuit?

The presence of an inductor (L) and capacitor (C) in parallel will affect the current in an AC circuit by creating a resonant frequency. This means that at a specific frequency, the current flow will increase, and at other frequencies, the current flow will decrease due to the impedance of the components.

5. Can you explain the concept of resonance in an AC circuit with R & LC in parallel?

Resonance in an AC circuit with R & LC in parallel is a phenomenon where the inductive and capacitive reactances cancel each other out, resulting in a lower overall impedance in the circuit. This allows for a specific frequency of the AC current to be amplified and transferred to the load, while other frequencies are blocked. This is useful in tuning circuits and creating filters in electronic devices.

Similar threads

  • Introductory Physics Homework Help
2
Replies
42
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
13
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
923
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
14
Views
4K
  • Introductory Physics Homework Help
Replies
6
Views
1K
Back
Top