**Find the value of i in the circuit below using mesh analysis.**

The first thing you should notice about this circuit is that there are two different types of sources: a dependent source (the arrow in the diamond), and an independent source (the arrow in the circle). Independent sources are independent of the circuit–so that source will always push 15 A of current into the circuit, regardless of the circuit elements. The dependent source consists of a gain, in this case 2, and a value of some current or voltage in the circuit, in this case the current (i_a). What current it pushes through the circuit changes depending on (i_a). Since it’s a current source dependent on another current, this is a current controlled current source. You can also have voltage controlled current sources, current controlled voltage sources, and voltage controlled voltage sources.

The second thing to look at is how mesh analysis actually works. Mesh analysis is an application of Kirchhoff’s Voltage Law (KVL), which states that that the summation of the voltages in any loop in a circuit should equal 0.

Mesh analysis is basically just solving KVL equations using arbitrarily defined currents (basically you label what i should be).

NOTE: When applying KVL, it’s typical to have the voltage be positive when the current runs from the negative to positive end of the element (like the voltage source) and negative when it runs from the positive to negative end of the element (like the resistors).

Now to the actual problem. The first thing you want to do when applying mesh analysis is to draw in your mesh currents (a mesh is just a fancy word for loop).

I’ve just called them i1, i2, i3. It’s a good idea to have them all go in the same direction. This will make calculations easier. I chose to have them all go to the right since i1) will match the direction of the two current sources.

Next, write down what we know based on the diagram, and label the positive and negative ends of the resistors (so we know what is positive and what is negative in our equations). Whenever you have two mesh currents running parallel to each other through a wire or element, the actual, total current running through that wire or element is the sum or difference of the mesh currents depending on if they are going in the same (sum) or opposite (difference) directions, here is an example using ohm’s law (V=IR)

The direction you choose for the plus and minus don’t matter, just stay consistent when doing the problem. Since they are resistors I just chose to have the mesh currents run from the positive to the negative ends, this will make future steps easier, but it would work either way.

At this point you would normally write down the mesh equations (since there are three meshes you would have three equations), but in this case you would quickly run into an issue. Mesh analysis is an application of KVL, so we need to know all of the voltages–but we don’t actually have an easy way of figuring out voltages across current sources like we do with resistors, as they only set the current through themselves. Don’t worry though, we can bypass this issue by using a supermesh. A supermesh is just a loop that includes the smaller loops within itself while avoiding the current sources. This is where having all of our smaller mesh currents going in the same direction helps, as we will be “tying” them all together in the supermesh.

The dotted line is our supermesh.

Remember, KVL states that the summation of voltages in *any *loop within a circuit will add up to 0. The supermesh is just a bigger loop, so we can now apply KVL to this supermesh/loop. So the set of equations we will be solving is:

Notice how all three parts/voltages in the KVL equation are negative, since the mesh current is running from the positive to negative ends of the resistors.

The problem is all set up, the rest is just math.

We want to solve for i1, since that is equal to -i. So we solve the first equation for i2 and the second equation for i3.

We can plug these into the third equation and just solve for i1.

And, using one of the first relations :

NOTE: Mesh analysis is a powerful tool for solving more complicated circuits. In general, you want to use mesh analysis when, like this problem, you get an easy set of equations (since you only had to do KVL once due to the presence of a supermesh), or when you can easily solve for one of the mesh currents due to there being a current source on the outside of the circuit:

Since the mesh current, i1, crosses a lone current source, the 5A, right off the bat you know that i1 = 5A.

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