# Camara PE Sample Exam #18



## jo9el (Oct 13, 2008)

I tried to solve this problem via Superposition but I get the wrong answer. I only get the right answer when I use Loop Analysis. Why is that?


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## benbo (Oct 13, 2008)

jo9el said:


> I tried to solve this problem via Superposition but I get the wrong answer. I only get the right answer when I use Loop Analysis. Why is that?


Any chance you can post the problem in a pdf? I gave this book away. Not that I'd be able to answer it, but I could take a look.


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## jo9el (Oct 13, 2008)

See attached.


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## k2keylargo (Oct 13, 2008)

Did you short the V source when you did superpositon? You must short V sources for superpositon. I worked it by superposition and got it right.


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## jo9el (Oct 13, 2008)

k2keylargo said:


> Did you short the V source when you did superpositon? You must short V sources for superpositon. I worked it by superposition and got it right.


I shorted each source seperately to get two seperate currents.

I used the voltage divider to get the voltage across the resistor for both instances.

I end up getting I1 = 1 A and I2= .5 A totalling 1.5 A (wrong).

could you post a step by step?


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## benbo (Oct 13, 2008)

jo9el said:


> I shorted each source seperately to get two seperate currents.
> I used the voltage divider to get the voltage across the resistor for both instances.
> 
> I end up getting I1 = 1 A and I2= .5 A totalling 1.5 A (wrong).
> ...


I'll try to get the superposition solution, but why in the world would you do it that way?

IMO, the easiest way is to use nodal analysis to get the voltage at the supernode and then use ohms law to get the current. If you use superposition you are essentially working the same problem twice (actually a harder problem). You're going to have to figure the current through the 18 ohm with the 24v shorted, then the same thing with the 12 V shorted and add them together.

I may have made a mistake in these calcs, but I stick by my statement that superposition is hard.

(V-24)/6+(V-12)/6+V/18 = 0

3(V-24)+3(v-12)+V=0

3V-72+3V-36+V = 0

7V = 108

V = 108/7

I = V/18 = 108/(7*18) = 6/7A


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## benbo (Oct 13, 2008)

jo9el said:


> I shorted each source seperately to get two seperate currents.
> I used the voltage divider to get the voltage across the resistor for both instances.
> 
> I end up getting I1 = 1 A and I2= .5 A totalling 1.5 A (wrong).
> ...


You're right.

This is the easy way to do superpoistion (again, totally too hard as far as I'm concerned.)

Basically, when you short the supplies you get the same resitive circuits - Basicaly a 6 ohm resistor in series with a 6 || 18.

Basically, the total works out to around 10.5 ohms (6+4.5).

So, when you short the 12V you get a voltage at the node of 24*4.5/10.5 = 10.28. So the current is 10.28/18 = .57A

When you short the 24V you get a voltage at the node of 12*4.5/10.5 = 5.14 so I = 5.14/18 = .28A

Add them together = .8557A

I think this is correct, but my math can be wrong again.

6/7 = .857 (closest one)


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