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schematic

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.

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

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  • \$\begingroup\$ By KCL at the top right node \$I_1 = 1\text{A} = 3i_x + i_x = 4i_x\$ so \$i_x = 0.25\$A. \$\endgroup\$ – Null Oct 30 '14 at 14:51
  • \$\begingroup\$ @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? \$\endgroup\$ – dfg Oct 30 '14 at 14:54
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    \$\begingroup\$ Homework questions with no attempt at a solution are closed. \$\endgroup\$ – Leon Heller Oct 30 '14 at 15:14
  • \$\begingroup\$ @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. \$\endgroup\$ – dfg Oct 30 '14 at 15:19
  • \$\begingroup\$ 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. \$\endgroup\$ – EwokNightmares Oct 30 '14 at 16:09
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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.

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