Electronic – Mesh Analysis with a ideal current source

Recall that, for node voltage analysis, a floating voltage source (a voltage source that does not connect to the GND node) poses a problem since you cannot write an equation relating the current through to the voltage across.

What you must do then is enclose the floating voltage source in a supernode, which reduces the number of KCL equations by one, and add the equation relating the voltage difference between the nodes the voltage source is connected to.

Now, the dual of node voltage analysis is mesh current analysis and here we have the dual problem when we have a current source common to two meshes - we can't write an equation relating the current through to the voltage across a current source.

What must be done then is to form a supermesh which reduces the number of KVL equations by one and add the equation relating the difference of the mesh currents to the common current source.

So, write KVL counter-clockwise around the supermesh consisting of the two voltage sources and the two resistors

$$V_1 = I_aR_1 + V_2 + (I_b - I_c)R_2$$

You have, by inspection (no KVL required for this mesh - this is dual to no KCL required for the node connected to a non-floating voltage source)

$$I_c = -1.25A $$

You need one more equation which is the equation relating to difference of the two mesh currents with the common current source.

$$3A = I_a - I_b $$

Now, you have 3 independent equations and 3 unknowns.

This is how i think i would solve the problem:

I don't see it necessary to add a 1V excitation source since there already a 9v in the circuit. The trick when analyzing circuit with dependent sources is to avoid the Short circuiting you would normally do when having a circuit with independent sources alone:

_Things to note that may help

Focus on node A:

define a current from the 9v source direction to node A and also define a current from 6ohm to A

_not Vx =Va

9 - Vx = I1

Vx/6 = I2

and i reckon Vab = 2Vx..

Hope this helps you!