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Agarwal - Foundations of Analog and Digital Electronic Circuits Elsevier pg 96

The problem is solved in book by using basic methods and v1 = -9 V, v2 = 2 V, v3 = -11 V, v0 = 2 V.

However, when I applied mesh current method
loop1 : current in loop1 say i1 such that direction in 2 ohm resistor is same in both loops
I get equations 2 + 2(i1 + i2) = 0 --(1)
and 2(i1 + i2) +3i2 + v3= 0 --(2) where i2 is current in loop2 and v3 is voltage across current source (not to be confused with i1 i2 i3 in given circuit).

Solving these I get v1 = 9 V , v2 = -2 V , v3 = -7 V , v0 = 2 V

What is it that is wrong with these equations?

Transistor
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vector
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  • I think you mixed up the + and - directions. The voltage 2 and 2(i1+i2) are opposing each other so subtract one of them. v1 is also opposing the two other voltages in that loop. – user253751 May 17 '23 at 19:03
  • As to me the equation should be: 2(I1-I2) = 2v --(1) and 2(I2-I1)+V3+3I2 = 0 ---(2) Note that I1 and I2 are the two mesh currents. – Ady May 17 '23 at 19:03

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You write that "current in loop1 say i1 such that direction in 2 ohm resistor is same in both loops". I don't like using number subscripts for the current loops because you already have existing currents with number subscripts. So to avoid any of that kind of confusion I've used \$x\$ and \$y\$ as subscripts for the loop currents:

enter image description here

Just to put it on top of the table and in plain view: It does not matter which directions are picked for the mesh current loop directions. One must simply be consistent in their application, once chosen.

I like to follow the direction indicated by the arrow of each loop, because I'm a cooperative type and don't like to be seen as always going against the grain. And I'll start at the grounded node in each case because I'm ground-biased. What can I say? (These choices are also arbitrary.)

$$\begin{align*} 0\:\text{V}+2\:\text{V}-2\:\Omega\cdot\left(i_x+i_y\right)&=0\:\text{V} \\\\ 0\:\text{V}-v_{_3}-3\:\Omega\cdot i_y-2\:\Omega\cdot\left(i_x+i_y\right)&=0 \:\text{V} \\\\ i_y&=3\:\text{A} \end{align*}$$

The last equation isn't a mesh loop equation but a statement that says \$i_y\$ is in the same direction as the current source and that it must also take on the exact value of the current source. This sets up three equations in three unknowns. (Well, technically, one of them is already known: see last equation.)

The value could replace the use of \$i_y\$ in the loop equations, instead:

$$\begin{align*} 0\:\text{V}+2\:\text{V}-2\:\Omega\cdot\left(i_x+3\:\text{A}\right)&=0\:\text{V} \\\\ 0\:\text{V}-v_{_3}-3\:\Omega\cdot 3\:\text{A}-2\:\Omega\cdot\left(i_x+3\:\text{A}\right)&=0 \:\text{V} \end{align*}$$

That sets up two equations in two unknowns.

Either way, solvable. Solved, the values for all the other unspecified values can be achieved. (Don't forget that from direct inspection \$i_{_2}=i_x+i_y\$, for example.)

Another way to look at it is that you already know what voltage difference is across the \$2\:\Omega\$ resistor. So you know what the net current in it must be. The difference between this and the current source must be what is happening with the current through the voltage source. It all has to balance out.

periblepsis
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  • I can't quite understand that why is the sign of voltage across 3Ω resistor in second equation same as other if v1 is opposing the other two voltages? – vector May 18 '23 at 13:26
  • @vector The more positive side of the resistor is the one which the arrow points into. A teacher can write the signs so that they are opposite this fact. But those are just marks made by someone who wants to confuse you, I suppose. It's still the fact that for the purposes of following the rules for mesh equations, the more positive end of the resistor is where the current enters, not leaves. The diagram shows v1 polarized incorrectly. But that doesn't matter. It only means that v1 will be negative which says the signs were placed in the wrong polarity. – periblepsis May 18 '23 at 14:52