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3.5 Mesh Analysis with Current Sources

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3.5 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.

■ CASE 1 When a current source exists only in one mesh: Consider the circuit in Fig. 3.22, for example. We set i2 = −5 A and write a mesh equation for the other mesh in the usual way; that is,

−10 + 4i1 + 6(i1 − i2) = 0 ⇒ i1 = −2 A (3.17)

■ CASE 2 When a current source exists between two meshes: Consider the circuit in Fig. 3.23(a), for example. We create a supermesh by excluding the current source and any elements connected in series with it, as shown in Fig. 3.23(b). Thus,

A supermesh results when two meshes have a (dependent or independent) current source in common.

(a) Two meshes having a current source in common, (b) a supermesh, created by excluding the current source.

As shown in Fig. 3.23(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. 3.23(b) gives

−20 + 6i1 + 10i2 + 4i2 = 0

6i1+14i2=20(3.18)6i_1 + 14i_2 = 20 \tag{3.18}

We apply KCL to a node in the branch where the two meshes intersect. Applying KCL to node 0 in Fig. 3.23(a) gives

i2=i1+6(3.19)i_2 = i_1 + 6 \tag{3.19}

Solving Eqs. (3.18) and (3.19), we get

i1=−3.2 A,i2=2.8 Ai_1 = -3.2 \text{ A}, \qquad i_2 = 2.8 \text{ A}

(3.20)

Note the following properties of a supermesh:

    1. The current source in the supermesh pro vides the constraint equation necessary to solve for the mesh currents.
    1. A supermesh has no current of its own.
    1. A supermesh requires the application of both KVL and KCL.

For the circuit in Fig. 3.24, find i1 to i4 using mesh analysis. Example 3.7

Figure 3.24

For Example 3.7.

Solution:

Note that meshes 1 and 2 form a supermesh because 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,

2i1 + 4i3 + 8(i3 − i4) + 6i2 = 0

or

i1+3i2+6i3−4i4=0(3.7.1)i_1 + 3i_2 + 6i_3 - 4i_4 = 0 \tag{3.7.1}

For the independent current source, we apply KCL to node P:

i2=i1+5(3.7.2)i_2 = i_1 + 5 \tag{3.7.2}

For the dependent current source, we apply KCL to node Q:

i2=i3+3Ioi_2 = i_3 + 3I_o

But Io = −i4, hence,

or

i2=i3−3i4(3.7.3)i_2 = i_3 - 3i_4 \tag{3.7.3}

Applying KVL in mesh 4,

2i4+8(i4−i3)+10=02i_4 + 8(i_4 - i_3) + 10 = 0 5i4−4i3=−5(3.7.4)5i_4 - 4i_3 = -5 \tag{3.7.4}

From Eqs. (3.7.1) to (3.7.4),

i1=−7.5 A,i2=−2.5 A,i3=3.93 A,i4=2.143 Ai_1 = -7.5 \text{ A}, \qquad i_2 = -2.5 \text{ A}, \qquad i_3 = 3.93 \text{ A}, \qquad i_4 = 2.143 \text{ A}

Figure 3.26 (a) The circuit in Fig. 3.2, (b) the circuit in Fig. 3.17.

(b)

Use mesh analysis to determine i1, i2, and i3 in Fig. 3.25.

Answer:

i1=12.379 A,i2=378.9 mA,i3=3.284 A.i_1 = 12.379 \text{ A}, i_2 = 378.9 \text{ mA}, i_3 = 3.284 \text{ A}.