# How to solve this circuit using mesh analysis? [closed]

simulate this circuit – Schematic created using CircuitLab How would I solve this circuit using mesh analysis? I can't make a supermesh because I would then have to know the voltage across the independent current source and I can't have 2 meshes because I have a dependent current source.

## closed as off-topic by hkBattousai, placeholder, Daniel Grillo, Chetan Bhargava, user17592 Nov 5 '14 at 14:57

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• By KCL at the top right node $I_1 = 1\text{A} = 3i_x + i_x = 4i_x$ so $i_x = 0.25$A. – Null Oct 30 '14 at 14:51
• @Null Well that works for this circuit, but if this circuit was part of a larger circuit that had several more meshes you wouldn't be able to solve it using KCL directly. You would have to create the loop currents. So is there a way to solve this by setting up equations involving KVL and loop currents? – dfg Oct 30 '14 at 14:54
• Homework questions with no attempt at a solution are closed. – Leon Heller Oct 30 '14 at 15:14
• @LeonHeller What do you mean no attempt at a solution? I did attempt it, and what I came up with is in the post... I tried a supermesh and 2 meshes and I listed the problems with that in the post. – dfg Oct 30 '14 at 15:19
• You can do a supermesh, but then you still need to define that supermesh current in terms of the dependent current source as the supermesh current branches into the dependent source and R2. – EwokNightmares Oct 30 '14 at 16:09

How would I solve this circuit using mesh analysis?

I assume the controlled source is a CCCS.

The left most clockwise mesh current, $i_a$ is by inspection $i_a = 1A$.

Clearly, the right most clockwise mesh current is just $i_b = i_x$. But $i_b$ cannot be found by KVL (as you've pointed out) so we need an auxiliary constraint to replace the KVL equation.

That constraint is given by the controlled source:

$$i_a - i_b = 3i_x = 3i_b \Rightarrow i_b = \frac{i_a}{4} = 0.25A = i_x$$

Remarkably, the solution does not depend on the resistor values.