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Can anybody answer these 2 questions about networks?

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K2DGR8

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I am attaching these 2 problems herewith.
Can anybody help me out to solve them?
 

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BradtheRad

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First Q #2.

It requires that you find the volt level of the unlabelled node at the bottom.

You do this by first calculating the combined resistance of the two parallel (rightmost) branches.

The rest is using proportional calculations.

**broken link removed**

Now as to Q #1.

Something isn't right here. The most power that can be consumed is 4 W. Not 10 W as it states.

Watts = V * V / R.

So (2 * 2) / 2 + ( 2 * 2 ) / 2 = 4 watts.

Oh. Unless 'x' is a voltage supply. The arrow does point up. As though to indicate current flow. But then it should look like the symbol for a supply, shouldn't it? (Don't you love these nondescript icons professors throw at students who are trying to get a handle on a difficult subject?)

And if 2A is the answer, why would the answer be in amps, if 'x' is an unknown V supply?

It may possible to make 'x' a particular voltage so that rogue current does not go into the battery at right.

It will take further brain-wracking to figure out whether schematic #1 can be seen to make sense the way it was drawn.
 
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lomaxe

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Answer.jpg

R*I1^2+R*I3^2=10
R*I3=E1

I1^2+I3^2=10/R=10/2=5
I3=E1/R=2/2=1

I1^2+1^2=5; I1^2=5-1=4; I1=sqrt(4)=2

I1=2A
 

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Can't argue with the math.

As for me, I was approaching it this way:

We see 2V going through a 2 ohm load in the middle branch. Current is 1 A. Power is 2W.

Now the stated total power is 10W. That means 8 W must be going through the other 2 ohm load at top. THis means 2A goes through it.

To push 2A through a 2 ohm load implies 4V across it.

Where is the 4V supply? We do not see a 4 V source in the circuit. We only see a 2V source.

So I conjectured something the problem doesn't allude to: 'x' must be hiding a voltage supply.

I was skeptical. Could a circuit work the way this problem claimed to? Or was there a mistake?

I used a simulator. I did discover that there has to be a certain voltage coming out of 'x' in order to make the problem make sense (as well as the solution).

I also found out this about the 2V source at right. Current is not flowing out at its top but into it. Thus it does not work as a battery which is the normal meaning of the symbol if we were looking at a normal schematic. (I'm going by convention that the larger plate at top is the positive terminal. And the convention that schematics usually orient a battery with its positive terminal upward.)

So as it turns out, the 2V battery is really a contrivance within the problem to provide oppositional EMF to whatever is the larger voltage source hidden at 'x'. My simulation proves the battery at right pulls current rather than to supply it. However since the 2V battery is not a resistance, no power is consumed.

It puts a bit of a trick into the problem, doesn't it?

Is this a fair problem to give students? Maybe we'd have to say the instructor has a bit of a sense of humor. (Was this for extra credit, by any chance?)

On the one hand it shows us how to use straightforward calculations as one approach to a problem.

On the other hand it forces us to think outside the box when we notice an anomaly.
 

lomaxe

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I think, that the main feature in this problem are the initial conditions: total consumed power, voltage supply and resistance. If we look at this circuit as at separate circuit - the initial conditions are contradictory. But if we imagine that the initial conditions are something that has set after turning the circuit on and that this circuit may be as a part of some large electrical circuit - there is no problem. We must get the initial conditions as a real fact and the problem is easily solved.
I hope you understand what I wonted to say you (because my English is not very good).
 
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lomaxe

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About the second problem:
a0486d38ccd1.jpg


I1=E/R=1.3 (A) where E=10V and R=R1+((R2+R3)*(R4+R5)/(R2+R3+R4+R5))=7.684 (Ohm)

U(CD)=I1*(R2+R3)*(R4+R5)/(R2+R3+R4+R5)=4.789 (V)

I2=U(CD)/(R2+R3)=0.342 (A)

I3=U(CD)/(R4+R5)=0.958 (A)

U(R2)=I2*R2=2.052 (V)

U(R4)=I3*R4=1.916 (V)

U(R2)=φ(A)-φ(C)
U(R4)=φ(B)-φ(C)
φ(A)-φ(B)=U(R2)-U(R4)=0.136 (V)

So, my answer is different from the answer in the problem attached. I don't know why. May be there is mistake in my solving, may be there is mistake in the attached problem answer.
 

BradtheRad

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I can confirm your figures and your answer of .136 V.

My simulator did the math. I didn't have to.

:^)

Which goes to show that instructors (and authors) are only human. Despite our natural tendency to believe what they write in black and white. Especially if we're paying them to prepare us for a career in a certain field.

But then, it does prepare us for what's ahead. There's a 'gotcha' hiding in practically all the projects we'll ever undertake.
 

lomaxe

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I agree with you :) Sometimes there's a "gotcha", hiding in the projects, and sometimes there are just mistakes made by instructors. I came across things like that when studied at the institute.
 
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