Superposition, circuit analysis.

7 replies · 3K views · last reply
lam58
Messages
26
Reaction score
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 schematic with a 110V voltage source, a 4A current source, and resistors 5Ω, 2Ω, 10Ω and 12Ω; voltage v is marked across the 10Ω resistor.
    circuit.png
    13.9 KB · Views: 608
Last edited:
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.
 
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:
Attachments
  • Circuit schematic labelled Figure 4: a 110V voltage source feeds a 5Ω resistor, then a 10Ω resistor with node voltage v across it, in parallel with a branch of 2Ω and 12Ω resistors.
    circuit1.png
    9.5 KB · Views: 574
  • Circuit schematic with a 4A current source at top feeding a node; branches of 2Ω in series with a 5Ω and 10Ω combination, and a 12Ω resistor, with voltage v marked across the 10Ω resistor and polarity signs.
    circuit.png
    2.9 KB · Views: 546
  • Like
Likes   Reactions: 1 person
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;

Circuit schematic with a 110V voltage source on the left, a 5 ohm resistor in series at the top, a 10 ohm resistor on the middle branch with voltage v across it, and 2 ohm and 12 ohm resistors in series on the right branch.


Next we have a two current dividers
http://en.wikipedia.org/wiki/Current_divider
Circuit schematic with a 4A current source at top feeding a node, a 2Ω resistor down to a node, a 5Ω resistor and a 10Ω resistor in parallel with voltage v marked across them, and a 12Ω resistor on the right returning to the bottom wire.


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.
 
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.