I don't think this circuit is putting any constraints on Vx. So you can let Vx be whatever you want it to be.

The way I analyzed it, I drew four mesh currents, {I1, I2, I3, I4}. It turned out I4 was determined by that 4A current source, leaving {I1, I2, I3} as variables.

I found three equations for these three unknowns, from Kirchoff's voltage rule around those little mesh loops. Then I rewrote that system of equations as:

A*[I1;I2;I3] = [Vx;Vy; R3*R4]

The 3-by-3 matrix A is just filled with constants from the resistor values, and I found inv(A) exists, which means those 3 equations are independent and there is exactly one solution for [I1;I2;I3], namely what I get if I multiply both sides by inv(A)

That link was produced by a button on the simulator labeled "export link". To humans that url just looks like a long string of funny arguments and stuff. I don't know for sure what you'll get in your browser if you click that link. If everything goes right, you'll see a Java applet running a simulation of that circuit.

I set Vx, what I was imagining to be a voltage source, I set that to Vx=0. I think a zero volt voltage source is the same as a short. So I guess a piece of wire would work there too, for the case of Vx=0.

Strangely, the voltage sources in that simulation do not have labels on them, or at least that's the way I'm seeing it in my browser. The resistors have labels. The current source has a label. But not so for the voltage sources. For those you have to click on them, then choose "edit" to see what voltage they're set to.

Yeah. That link isn't working for me either now, but I remember that simulated circuit looked really pretty when it was running. Anyway, I guess anyone reading this who is interested can go play with that simulator, if it's still around at the time this post is being read.

Basically you can cut out R2, R3 and R4 cause they are not affecting anything in this circuit. Current follows the path of lease resistance. It will not flow into R4 from the 10V battery since the combination of R4 and R2 is greater than the combination of R5 and R1. R3 is larger still so it will avoid that as well. So it will flow from the 10V battery, through R5, through the ammeter, through R1 and into Vx. Since the total resistance is 8 Ohms, 8 multiplied by 4 is 32. So Vx is 22V.

Probably way off base here. Haven't done this sort of thing since 1994.

Thanks mpilchfamily for your interests, but I have concluded another solution with 8V as a value of Vx through that low : I = V-Vx/R-r, but I don't know where'a exactly the correct answer .

The V @ 10V and Vx are the power sources. The I is a current meter attached to the circuit measuring a power draw of 4A. So based off the given circuit and that it's drawing 4A you can calculate what the other voltage source is.

Also the current from Vx will not flow through R2 since the path of least resistance is through the 10V battery. The current is leaving the negative side of the battery and trying to get to the nearest positive source.

The VALUE of Vx will change the currents in the rest of the circuit, but its unconstrained BY the circuit, since its a voltage source, and can supply infinite current.

I don't think this circuit is putting any constraints on Vx. So you can let Vx be whatever you want it to be.

The way I analyzed it, I drew four mesh currents, {I1, I2, I3, I4}. It turned out I4 was determined by that 4A current source, leaving {I1, I2, I3} as variables.

I found three equations for these three unknowns, from Kirchoff's voltage rule around those little mesh loops. Then I rewrote that system of equations as:

A*[I1;I2;I3] = [Vx;Vy; R3*R4]

The 3-by-3 matrix A is just filled with constants from the resistor values, and I found inv(A) exists, which means those 3 equations are independent and there is exactly one solution for [I1;I2;I3], namely what I get if I multiply both sides by inv(A)

inv(A)*A*[I1;I2;I3] = [I1;I2;I3] = inv(A)*[Vx;Vy; R3*R4] = inv(A)*[Vx;10;-80]

A picture I drew this circuit and the equations I wrote for {I1, I2, I3} are attached.

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If I read your sketch right I 4 is in the other direction than you drew it.

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Yes. I4, the way I drew it, is going in the opposite direction of that current source.

So I4 = - 4A ( minus four amperes)

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Ok I was having a hard time seeing it and your calculations on my computer.

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Hey, um, if you want to see some voodoo, I just tried building at simulation of this circuit via that page at falstad.com.

http://www.falstad.com/circuit/#%24+1+5.0E-6+10.20...

That link was produced by a button on the simulator labeled "export link". To humans that url just looks like a long string of funny arguments and stuff. I don't know for sure what you'll get in your browser if you click that link. If everything goes right, you'll see a Java applet running a simulation of that circuit.

I set Vx, what I was imagining to be a voltage source, I set that to Vx=0. I think a zero volt voltage source is the same as a short. So I guess a piece of wire would work there too, for the case of Vx=0.

Strangely, the voltage sources in that simulation do not have labels on them, or at least that's the way I'm seeing it in my browser. The resistors have labels. The current source has a label. But not so for the voltage sources. For those you have to click on them, then choose "edit" to see what voltage they're set to.

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Sorry for so long browser don't like.

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Yeah. That link isn't working for me either now, but I remember that simulated circuit looked really pretty when it was running. Anyway, I guess anyone reading this who is interested can go play with that simulator, if it's still around at the time this post is being read.

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I am wondering is this a question on a math test and your teacher wants you to solve it using a quadrilateral equation or something like that.

When I was in college a math teacher really messed the pooch when he wanted us to solve this problem by factoring.

You have two resistors in parallel with a difference of 4 Ω and a total of 77 Ω what is the value of the two resistors.

He wanted 77 Ω = 7 Ω and 11 Ω

Boy did he get it wrong.

7 Ω + 11 Ω = 4.5 Ω in a parallel circuit.

Joe

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Basically you can cut out R2, R3 and R4 cause they are not affecting anything in this circuit. Current follows the path of lease resistance. It will not flow into R4 from the 10V battery since the combination of R4 and R2 is greater than the combination of R5 and R1. R3 is larger still so it will avoid that as well. So it will flow from the 10V battery, through R5, through the ammeter, through R1 and into Vx. Since the total resistance is 8 Ohms, 8 multiplied by 4 is 32. So Vx is 22V.

Probably way off base here. Haven't done this sort of thing since 1994.

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Thanks mpilchfamily for your interests, but I have concluded another solution with 8V as a value of Vx through that low : I = V-Vx/R-r, but I don't know where'a exactly the correct answer .

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I though I 4 A is a source according to what you said to me.

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The V @ 10V and Vx are the power sources. The I is a current meter attached to the circuit measuring a power draw of 4A. So based off the given circuit and that it's drawing 4A you can calculate what the other voltage source is.

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Not according to what he told me.

I asked him" Is Vx, V and A a source or a meter?

It is not clear in the schematic or your description."

and he said "No, it's independent sources."

Joe

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Also the current from Vx will not flow through R2 since the path of least resistance is through the 10V battery. The current is leaving the negative side of the battery and trying to get to the nearest positive source.

Once again i'm probably off here.

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Just a little loading effect.

What gets me is and I asked V, Vx, and A are all sources and the current source is in the opposite direction of the voltage sources current.

So A is subtracted from V current or vice versa depending on the circuit.

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Is Vx, V and A a source or a meter?

It is not clear in the schematic or your description.

Joe

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No, it's independent sources.

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Then Vx is what ever you make it.

Joe

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I think there is an anonymous current flow throw R5 .

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The VALUE of Vx will change the currents in the rest of the circuit, but its unconstrained BY the circuit, since its a voltage source, and can supply infinite current.

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This is a classic question for using Kirchoff's laws.

1.) The algebraic sum of all the currents entering any node is always zero.

2) The algebraic sum of all the voltage drops in a loop is zero.

Personally, I always preferred to use (2)

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