Superposition, circuit analysis.

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lam58
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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:
 
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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. ?
 
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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.
 
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.