Determine the value of the voltage source Vx

[Castello]

Homework Statement

Find Vx, as the 5A current source provides 50W to the circuit.

Homework Equations

$P = V.I$
$U = R.I$

The Attempt at a Solution

[/B]
At first, as 5A current source provides 50W, by $p=vi$ there should be 10v through it. Then, i applied KVL and found four equations including Vx:
$2i1-2i2=5$
$-2i1+5i2-2i3-i4=0$
$i3=-5A$
$-i2+3i4+Vx=0$

So, if solving this is the right way, i need one more equation, probably related to the path from Vx to the current source, but I'm not sure how to do this.

I guess it could be: $Vx+2i4+(i1-i4)-i3=0$

Is it right? I mean, probably there's a better way to solve this question, but i couldn't with superposition or source transformation.
P.S. I don't have the answer

 

[tech99]

Is the circuit correct? Is Vx an actual voltmeter, because if so no current will flow through it.

 

[Castello]

Yes, it's correct. The Vx in the circuit is a voltage source.

Is the circuit correct? Is Vx an actual voltmeter, because if so no current will flow through it.

 

[berkeman]

Is Vx an actual voltmeter

That confused me as well. It looks to me that it's a voltage source that you have to adjust to get the 50W number for the current source.

Yes, it's correct. The Vx in the circuit is a voltage source.

The question was whether Vx was a voltmeter, so your reply is backwards. The correct reply would be "No, it is not a Voltmeter with infinite input impedance, it is a Voltage Source".

'll try to look again tomorrow morning.

 

[ehild]

You are right, the voltage across the terminals of the current source is 10 V. What are the potentials at the nodes A and B then? What current does flow between A and B? What are the currents through the "vertical" 2 ohm resistors? What is the current through the voltage source Vx then?

 

[ehild]

The Attempt at a Solution

[/B]
At first, as 5A current source provides 50W, by $p=vi$ there should be 10v through it. Then, i applied KVL and found four equations including Vx:
$2i1-2i2=5$
$-2i1+5i2-2i3-i4=0$
$i3=-5A$
$-i2+3i4+Vx=0$

So, if solving this is the right way, i need one more equation,

It is the KVL equation for the rightmost loop.

 

[Castello]

It is the KVL equation for the rightmost loop.

Really helped... Thank's!

$5+(VB-VA)/1=VA/2+(VA-VB)/2$
$5/2+(VA-VB)/2=VB/2+(VB-VA)/2$

$VA=7,5V$
$VB=20/3V$

By the loops:
$(VA-VB)/2=-i4$
$VB-VA=i2-i4$

and relating Vx with i2 and i4:

$Vx=-2,49V$ and $0,83A$ through it.

So, i4 was on wrong direction, which changed the voltage to negative, i think.

 

[ehild]

$5+(VB-VA)/1=VA/2+(VA-VB)/2$
$5/2+(VA-VB)/2=VB/2+(VB-VA)/2$

No, these equations are not correct. And I meant equation for the voltages in the loop in orange.

 

[Castello]

So... as the 5A current source is in parallel with the 2ohm resistor, now i can see that it could be transformed in 10v voltage source in series with this resistor. I tried again with KVL and found this:

$-5+2(i1-i3)=0$
$2(i3-i4)+2(i3-i1)+i3-i2=0$
$2i2+Vx+i2-i3=0$
$2(i4-i3)+i4=0$
Considering $i3=-5a$ and substituting, i found $Vx=-165v$

Is it right?

 

[ehild]

So... as the 5A current source is in parallel with the 2ohm resistor, now i can see that it could be transformed in 10v voltage source in series with this resistor.

, No, the current source is not parallel with the 2 ohm resistor, but it has to supply 50 W power to the circuit, so it must have 10 V across its terminals.

I tried again with KVL and found this:
View attachment 237559
$-5+2(i1-i3)=0$
$2(i3-i4)+2(i3-i1)+i3-i2=0$
$2i2+Vx+i2-i3=0$
$2(i4-i3)+i4=0$
Considering $i3=-5a$ and substituting, i found $Vx=-165v$.
Is it right?

No, it is wrong. I do not understand your logic when you write up a KVL equations. Choose separate loops which do not contain other loop. See picture. You have 4 loops, I numbered them, Start form the bottom left corner follow the red arrow and write the change of potential across each element.

In the first loop, the potential increases when you go across the 5 V source from - to +, and decreases when you go across the 2 ohm resistor by 2(i1-i2), as i1 flows in the direction of the arrow and i2 flows in the opposite direction, so the net current is i1-i2.. The change of potential around the whole loop is 5-2(i1-i2)=0 Write the change of potential, following the red arrow, in the other loops. In the 4th loop, the potential decreases by 10 V across the current source.

 

[Castello]

Ok, the current source is not in parallel, but in series with the 1ohm resistor, and to check if there is 10v between the current source, it should be on KVL i4 equation. I think i did similar before, as you mentioned for the first loop, but i'll explain how i got there: One equation for each separated loop, four currents. As I'm thinking clockwise, the first loop gets -5V by the minus sign voltage source and +2(i1-i2) for the resistor between meshes 1 and 2 (positive because the current flows from the higher to the lower potential on the resistor, notation that i was following by the Fundamentals of Electric Circuits book). It gets (using your picture for the currents on KVL):

loop 1: $-5+2(i1-i2)=0$
loop 2: $+2(i2-i1)+1(i2-i3)+2(i2-i4)=0$
loop 3: $+1(i3-i2)+2i3+Vx=0$
loop 4: $+2(i4-i2)+2i4+10=0$
by the current source, $i4=-5A$

Sorry, i know it's taking too long

 

[ehild]

I re-labelled the loops, because it was somewhat confusing. Write the equations for the 3rd and 4th loops again, and try to solve the system of equations.

 

[Castello]

I re-labelled the loops, because it was somewhat confusing. Write the equations for the 3rd and 4th loops again, and try to solve the system of equations.
View attachment 237581

It gets:

$2(i1-i2)-5=0$
$2(i2-i3)+2(i2-i1)+i2-i4=0$
$2(i3-i2)+i3+10=0$
$2i4+Vx+i4-i2=0$

$i1=0$
$i2=-2,5$
$i3=-5$
$i4=-2,5$

$Vx=5v$

 

[ehild]

It gets:

$2(i1-i2)-5=0$
$2(i2-i3)+2(i2-i1)+i2-i4=0$
$2(i3-i2)+i3+10=0$
$2i4+Vx+i4-i2=0$
$i1=0$
$i2=-2,5$
$i3=-5$
$i4=-2,5$
$Vx=5v$

Correct !

 

[Castello]

Correct !

Thank's!

 

[ehild]

Thank's!

You are welcome.

There is a much simpler method to find Vx. The positive terminal of the 5 V voltage source is connected to point B, so B is at 5 V. On the right side, the upper terminal of the current source is at 10 V, and 5 A flows out from it through the 1 Ω resistor, causing 5 V potential drop. Therefore, the potential of A is 10-5=5V.
There is no potential difference between A and B, so no current flows through the 1 ohm resistor between them.
5 A flows into node A, and 5/2=2.5 A flows downward on the vertical 2 Ω resistor, so there is 2.5 A flowing through Vx and the series 2 Ω resistor Going from A to B, to potential is 5V+Vx-2Ω*2.5A=5V , so Vx= 5V.