Superposition, circuit analysis.

In summary, the conversation discusses a circuit analysis problem with a DC voltage source and a current source. The participants go through various steps to find the voltage and current across different resistors in the circuit, including using voltage and current dividers. They also discuss the equivalent circuit and the process of determining it through experience and thought.
  • #1
lam58
29
0
Hi, could check if I have the right answer please. The question (Q 4b) is in the attached image below.

My answer is:First for analysis the DC voltage source.

i2 = 0.82A, i1 = 1.97A

therefore V across 10Ω resistor = 11.5V.

For the analysis of the current source I got a bit confused, however, I think I might have got it right. I shorted the voltage source and found the current across the 50Ω resistor by doing:

Rt= (10//2) + 12 = (41/3)

i50Ω = (Rt*is)/(Rt+50Ω) = 0.86A

Then to find the current across the 10Ω resistor:

Rt = 2Ω hence i10Ω = (2*0.86)/(2+10) = 0.14A.

Thus V10Ω = 1.4V.

Then adding the two separate source responses across the 10Ω resistor the total voltage V10Ω= 1.4 + 11.5 = 12.9V.

Is this correct? :confused:
 

Attachments

  • circuit.png
    circuit.png
    13.9 KB · Views: 528
Last edited:
Physics news on Phys.org
  • #2
There is no 50 ohm resistor.

First thing is to label the resistors and sources. Put in numbers only at the very end.
 
  • #3
rude man said:
There is no 50 ohm resistor.

First thing is to label the resistors and sources. Put in numbers only at the very end.

Woops lol.I'll try again.
 
  • #4
First if we remove current source we have a simply voltage divider:
Vx1 = 110V * (10||14)/(5 + 10||14)
Next we connect a current source but we short a 110 voltage source we have a very simple current divider.
http://en.wikipedia.org/wiki/Current_divider
 
  • #5
Jony130 said:
First if we remove current source we have a simply voltage divider:
Vx1 = 110V * (10||14)/(5 + 10||14)
Next we connect a current source but we short a 110 voltage source we have a very simple current divider.
http://en.wikipedia.org/wiki/Current_divider

The circuit divider doesn't seem that simple to me. ?
 
Last edited:
  • #6

Attachments

  • circuit1.png
    circuit1.png
    9.5 KB · Views: 512
  • circuit.png
    circuit.png
    2.9 KB · Views: 477
  • Like
Likes 1 person
  • #7
Jony130 said:
First we have a voltage divider
http://en.wikipedia.org/wiki/Voltage_divider#Resistive_divider

Vout = Vin * R2/(R1 + R2)

So in our case
R2 = 10Ω||14Ω; And R1 = 5Ω; Vin = 110V;

attachment.php?attachmentid=65541&stc=1&d=1389556772.png


Next we have a two current dividers
http://en.wikipedia.org/wiki/Current_divider
attachment.php?attachmentid=65542&stc=1&d=1389556823.png


The current that is flow through 2Ω resistor is equal to:

I = -4A * 12Ω/(12Ω + (2Ω + 5Ω||10Ω) = -2.76923076A

And 10Ω resistor current is equal to ??

Ix = -2.77A * ??/( ?? + ?? ) = ??

Thanks that's excellent, but how do you know to make the equivalent circuit like that? I was trying to draw it with the 4Amp going into the circuit splitting between the 5ohm resistor and the other branch.
 
  • #8
lam58 said:
Thanks that's excellent, but how do you know to make the equivalent circuit like that?
Hey what can I say, I simply look at the circuit and I see equivalent circuit after some thought.
So all you need is some experience, that's all.
 

FAQ: Superposition, circuit analysis.

What is superposition in circuit analysis?

Superposition is a principle in circuit analysis that states that the total response of a linear circuit is the sum of the individual responses caused by each independent source acting alone.

How is superposition used in circuit analysis?

Superposition is used in circuit analysis by breaking down a complex circuit into simpler parts and analyzing the response of each part separately. The total response can then be found by summing the individual responses.

What is the benefit of using superposition in circuit analysis?

Using superposition in circuit analysis allows for a simplified analysis of complex circuits, making it easier to understand and troubleshoot circuit behavior. It also allows for the calculation of individual source contributions and the overall circuit response.

Are there any limitations to using superposition in circuit analysis?

Yes, superposition can only be applied to linear circuits, meaning that the circuit must follow Ohm's law and Kirchhoff's laws. It also cannot be used for circuits with dependent sources.

Can superposition be used for AC circuits?

Yes, superposition can be used for AC circuits as long as the circuit is linear. However, it should be noted that the analysis may become more complex due to the presence of reactive components such as capacitors and inductors.

Similar threads

Replies
7
Views
1K
Replies
7
Views
2K
Replies
4
Views
1K
Replies
1
Views
2K
Replies
9
Views
4K
Replies
5
Views
2K
Replies
15
Views
6K
Replies
7
Views
2K
Replies
4
Views
2K
Back
Top