# Circuit analysis: I just wanna know, how can I get (Vx) in the attached circuit below? Answered

Please pay attention for the current and voltage direction

Tags:

## Comments

The forums are retiring in 2021 and are now closed for new topics and comments.

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.

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.

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.

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

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.