Solve for ##I_x##

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  • #1
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Homework Statement



An image of the circuit:
http://gyazo.com/32ec8c06fb861d115895bec5a15959dd

Homework Equations



##P = ±IV##

The Attempt at a Solution



So the first thing i want to do is solve for all the powers that are absorbed or supplied.

##P_{10V} = (2A)(10V) = 20W##

##P_1 = ##
##P_2 = -(2A)(15W) = -30W##
##P_3 = (2A)(15V) = 30W##

##P_{10VI_x} = -(I_x)(10V) = -10WI_x##
##P_{25V5A} = -(5A)(25V) = -125W##

So by Tellegen's theorem:

##20W + P_1 - 30W +30W - 10WI_x - 125W = 0##
##P_1 - 105W = 10WI_x##

Solving that will give ##I_x##, but I have a problem. I'm not sure about how to find ##P_1##. I know 1A of current will be flowing through it ( my guess is from positive to negative, though I can't justify this ) so that element 1 will absorb power.

The answer is listed as ##I_x = -2A##.

I'm confused here.
 
Last edited:

Answers and Replies

  • #2
CWatters
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Perhaps I've missed something but can't you just apply KCL?
 
  • #3
STEMucator
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Perhaps I've missed something but can't you just apply KCL?

Hm, KCL has not been introduced in the book yet (KCL and KVL are about 2 reading sections away). I honestly believe that not enough information has been given in this problem. If they had at least given the voltage across element 1, it would be possible to deduce what the power across it will be.

Given the work I did in the first post though, ( the only reason I figure this is because I've already been given the answer of -2A. ) it unfairly suggests that ##P_1 = 85W## ( just by comparing the powers I've already found ), which would also suggest that the voltage across element 1 is 85V.

The 1A of current would enter the +ve and leave through the -ve, so the element would be absorbing power.

Assuming all of this would yield:

##P_1 - 105W = 10WI_x##
##85W - 105W = 10WI_x##
##-20W = 10WI_x##
##-2A = I_x##

This does not seem right to me. The way I've done this is not possible by any means. If I wasn't already given the answer, how would I have found ##P_1## without being given any voltage?
 
  • #4
gneill
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Hm, KCL has not been introduced in the book yet (KCL and KVL are about 2 reading sections away). I honestly believe that not enough information has been given in this problem. If they had at least given the voltage across element 1, it would be possible to deduce what the power across it will be.
Components in parallel share the same potential drop. What's in parallel with component [1]?
 
  • #5
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Components in parallel share the same potential drop. What's in parallel with component [1]?

This seems like an interesting and viable approach.

Element 1 is in parallel with the independent current source directly to the right of it (25V and 5A). By what you've said, that would imply that the voltage across element 1 would be 25V. The problem is only 1A of current is going to pass through it. So this still does not give the answer I'm looking for.

This is leading me to believe the supplied answer is incorrect as you would get ##I_x = -8A## if the voltage across element 1 was assumed to be 25V.
 
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  • #6
gneill
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It works out fine for me. Pay close attention to defined current directions at the sources.
 
  • #7
STEMucator
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It works out fine for me. Pay close attention to defined current directions at the sources.

Okay, so presume element 1 has a potential difference of 25V, the positive and negative ends of the element are defined as such: ##[1]^+_-##

I'm thinking that the powers I've calculated are incorrect in some way.

+2A enters the (+) and leaves the (-) :
##P_3 = (2A)(15V) = 30W##

+2A enters the (+) and leaves the (-) :
##P_{10V} = (2A)(10V) = 20W##

+2A enters the (-) and leaves the (+) :
##P_2 = -(2A)(15V) = -30W##

One of the independent current sources:
##P_{10VI_x} = -(I_x)(10V) = -10WI_x##

I'm almost completely certain about these ones ^.

Now something about this drawing just didn't sit right with me. Why would 1A be entering the first element? Why not 5A? If I use 5A instead I would get:

+5A enters the (+) and leaves the (-) :
##P_1 = (5A)(25V) = 125W##

The other independent current source:
##P_{25V5A} = -(5A)(25V) = -125W##

So I would get :

##30 + 20 - 30 - 10I_x + 125 - 125 = 0##
##20 = 10I_x##
##2A = I_x##

That's pretty darn close, I know I'm missing something now though since the answer is ##-2A##. I believe it has to do with how i calculated this one:

##P_{10VI_x} = -(I_x)(10V) = -10WI_x##
 
Last edited:
  • #8
gneill
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Okay, so presume element 1 has a potential difference of 25V, the positive and negative ends of the element are defined as such: ##[1]^+_-##

I'm thinking that the powers I've calculated are incorrect in some way.

+2A enters the (+) and leaves the (-) :
##P_3 = (2A)(15V) = 30W##

+2A enters the (+) and leaves the (-) :
##P_{10V} = (2A)(10V) = 20W##
Okay so far. Now here's where it gets tricky...

+2A enters the (-) and leaves the (+) :
##P_2 = -(2A)(15V) = -30W##
Supposedly you don't know yet that Ix is 2A. In fact, it will turn out that Ix is really -2A and flowing from top to bottom in the branch. But leave it in symbolic form for now as Ix. Then you'll have: ##P_2 = -(I_x)(15V)##.
One of the independent current sources:
##P_{10VI_x} = -(I_x)(10V) = -10WI_x##
Leave that one as ##-(I_x)(10V) ## also.
I'm almost completely certain about these ones ^.

Now something about this drawing just didn't sit right with me. Why would 1A be entering the first element? Why not 5A? If I use 5A instead I would get:

+5A enters the (+) and leaves the (-) :
##P_1 = (5A)(25V) = 125W##
It's just that they've defined the component to be drawing 1A at 25V. It could be a 25Ω resistor, for example. In that case, with 25V across it, it MUST draw 1A.
The other independent current source:
##P_{25V5A} = -(5A)(25V) = -125W##

So I would get :

##30 + 20 - 30 - 10I_x + 125 - 125 = 0##
##20 = 10I_x##
##2A = I_x##

That's pretty darn close, I know I'm missing something now though since the answer is ##-2A##. I believe it has to do with how i calculated this one:

##P_{10VI_x} = -(I_x)(10V) = -10WI_x##
I think the problem is with the assumption of the current direction for P2. P2 will turn out to be a positive value once Ix is determined to be -2A.

If you sum up the powers using the "corrected" versions you'll be able to solve for Ix, and then for the "missing" powers in the Ix branch. Something like:

##30W + 20W + 25W + (-125W) + (-I_x)(10V) + (-I_x)(15V) = 0##

Solve for Ix.
 
  • #9
STEMucator
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Okay so far. Now here's where it gets tricky...


Supposedly you don't know yet that Ix is 2A. In fact, it will turn out that Ix is really -2A and flowing from top to bottom in the branch. But leave it in symbolic form for now as Ix. Then you'll have: ##P_2 = -(I_x)(15V)##.

Leave that one as ##-(I_x)(10V) ## also.

It's just that they've defined the component to be drawing 1A at 25V. It could be a 25Ω resistor, for example. In that case, with 25V across it, it MUST draw 1A.

I think the problem is with the assumption of the current direction for P2. P2 will turn out to be a positive value once Ix is determined to be -2A.

If you sum up the powers using the "corrected" versions you'll be able to solve for Ix, and then for the "missing" powers in the Ix branch. Something like:

##30W + 20W + 25W + (-125W) + (-I_x)(25V) + (-I_x)(15V) = 0##

Solve for Ix.

Ahh I see. So I can't "assume" that 2A of current is going through that portion of the circuit because I don't actually know. So the power calculations for those two components have to be done afterwards, but they can both use the same current ##I_x##.

##30W + 20W + 25W + (-125W) + (-I_x)(25V) = 0##

Solving yields ##I_x = -2##.

Tellegen's theorem is satisfied.

I thought the current that would flow across that branch would be the same across the entire portion of the circuit.

So really the current is coming from the top to the bottom in that branch. The independent current supplier and ##[2]^+_-## are both absorbing power, not supplying it.
 
  • #10
gneill
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Yup, that's it.

Once you've been introduced to Kirchhoff's laws you'll be able to determine that current by inspection of the other given currents.
 

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