# Mesh Analysis with Current Sources

Applying mesh analysis to circuits containing current sources (dependent or independent) may appear complicated. But it is actually much easier than what we encountered in the previous section, because the presence of the current sources reduces the number of equations. Consider the following two possible cases.
Fig.no.1: A circuit with a current source.
CASE 1 When a current source exists only in one mesh: Consider the circuit in Fig.no.1, for example. We set $i_2 = -5 A$ and write a mesh equation for the other mesh in the usual way, that is, $$-10 + 4i_1 + 6(i_1 - i_2) = 0$$ using $i_2=-5A$ in the above equation, $$i_1 = -2 A$$
Fig.no.2: (a) Two meshes having a current source in common, (b) a supermesh, created by excluding the current source.
CASE 2 When a current source exists between two meshes: Consider the circuit in Fig.no.2(a), for example. We create a supermesh by excluding the current source and any elements connected in series with it, as shown in Fig.no.2(b).

#### What is a supermesh?

A supermesh results when two meshes have a (dependent or independent) current source in common.
As shown in Fig.no.2(b), we create a supermesh as the periphery of the two meshes and treat it differently. (If a circuit has two or more supermeshes that intersect, they should be combined to form a larger supermesh.) Why treat the supermesh differently? Because mesh analysis applies KVL—which requires that we know the voltage across each branch—and we do not know the voltage across a current source in advance.
However, a supermesh must satisfy KVL like any other mesh. Therefore, applying KVL to the supermesh in Fig.no.2(b) gives $$-20 + 6i_1 + 10i_2 + 4i_2 = 0$$ or $$6i_1 + 14i_2 = 20 \tag{1}$$ We apply KCL to a node in the branch where the two meshes intersect. Applying KCL to node 0 in Fig.no.2(a) gives $$i_2 = i_1 + 6 \tag{2}$$ Solving Eqs. (1) and (2), we get $$i_1 = -3.2 A \,, i_2 = 2.8 A$$ Note the following properties of a supermesh:
• The current source in the supermesh is not completely ignored; it provides the constraint equation necessary to solve for the mesh currents.
• A supermesh has no current of its own.
• A supermesh requires the application of both KVL and KCL.
Example 1: For the circuit in Fig.no.3, find i1 to i4 using mesh analysis.
Fig.no.3: For Example 1.
Solution:
Note that meshes 1 and 2 form a supermesh since they have an independent current source in common. Also, meshes 2 and 3 form another supermesh because they have a dependent current source in common. The two supermeshes intersect and form a larger supermesh as shown.
Applying KVL to the larger supermesh, $$2i_1 + 4i_3 + 8(i_3 - i_4) + 6i_2 = 0$$ or $$i_1 + 3i_2 + 6i_3 - 4i_4 = 0 \tag{3}$$ For the independent current source, we apply KCL to node P: $$i_2 = i_1 + 5 \tag{4}$$ For the dependent current source, we apply KCL to node Q: $$i_2 = i_3 + 3i_o$$ But $i_o = -i_4$, hence, $$i_2 = i_3 - 3i_4 \tag{5}$$ Applying KVL in mesh 4, $$2i_4 + 8(i_4 - i_3) + 10 = 0$$ or $$5i_4 - 4i_3 = -5 \tag{6}$$ From Eqs. (3) to (5), $i_1 = -7.5 A$, $i_2 = -2.5 A$, $i_3 = 3.93 A$, $i_4 = 2.143 A$