Let us find Io in the network in Fig. 3.24a.

Question

Let us find [latex]I_{o}[/latex] in the network in Fig. 3.24a.

Accepted Answer

[ h i n t ]: In this case the 4-mA current source is located on the boundary between two meshes. Thus, we will demonstrate two techniques for dealing with this type of situtation. One is a special loop technique, and the other is known as the supermesh approach.

First, we select two loop currents [latex]I_{1}[/latex] and [latex]I_{2}[/latex] such that [latex]I_{1}[/latex] passes directly through the 2-mA source, and [latex]I_{2}[/latex] passes directly through the 4-mA source, as shown in Fig. 3.24b. Therefore, two of our three linearly independent equations are

[latex]I_{1} = 2 × 10^{-3}[/latex]
[latex]I_{2} = 4 × 10^{-3}[/latex]

The remaining loop current [latex]I_{3}[/latex] must pass through the circuit elements not covered by the two previous equations and cannot, of course, pass through the current sources. The path for this remaining loop current can be obtained by open-circuiting the current sources, as shown in Fig. 3.24c. When all currents are labeled on the original circuit, the KVL equation for this last loop, as shown in Fig. 3.24d, is

-6 + 1k[latex]I_{3} + 2k(I_{2} + I_{3}) + 2k(I_{3} + I_{2} - I_{1}) + 1k(I_{3} - I_{1})[/latex] = 0

Solving the equations yields

[latex]I_{3} =\frac{-2}{3}[/latex] mA

and therefore,

[latex]I_{o} = I_{1} - I_{2} - I_{3} =\frac{-4}{3}[/latex] mA

Next consider the supermesh technique. In this case the three mesh currents are specified as shown in Fig. 3.24e, and since the voltage across the 4-mA current source is unknown, it is assumed to be [latex]V_{x}[/latex]. The mesh currents constrained by the current sources are

[latex]I_{1} = 2 × 10^{-3}[/latex]
[latex]I_{2} - I_{3} = 4 × 10^{-3}[/latex]

The KVL equations for meshes 2 and 3, respectively, are

[latex]2kI_{2} + 2k(I_{2} - I_{1}) - V_{x}[/latex] = 0
-6 + 1k[latex]I_{3} + V_{x} + 1k(I_{3} - I_{1})[/latex] = 0

Adding the last two equations yields

[latex]-6 + 1kI_{3} + 2kI_{2} + 2k(I_{2} - I_{1}) + 1k(I_{3} - I_{1})[/latex] = 0

Note that the unknown voltage [latex]V_{x}[/latex] has been eliminated. The two constraint equations, together with this latter equation, yield the desired result.
The purpose of the supermesh approach is to avoid introducing the unknown voltage [latex]V_{x}[/latex]. The supermesh is created by mentally removing the 4-mA current source, as shown in Fig. 3.24f. Then writing the KVL equation around the dotted path, which defines the supermesh, using the original mesh currents as shown in Fig. 3.24e, yields

[latex]-6 + 1kI_{3} + 2kI_{2} + 2k(I_{2} - I_{1}) + 1k(I_{3} - I_{1})[/latex] = 0

Note that this supermesh equation is the same as that obtained earlier by introducing the voltage [latex]V_{x}[/latex].

Question 3.16: Let us find Io in the network in Fig. 3.24a.

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