Education • All: 3.2: Mesh Analysis

MESH ANALYSIS

Can only be applied to planar circuits.

STEPS:

1. Assign mesh currents.

2. Apply KVL to each mesh.

CASE I: Current source in 1 mesh (Ia = i1)
CASE II: Current source in 2 meshes (Called supermesh where Ia = i1 - i2)

3. Solve.

Example 1 (CASE I)

Find Vo

$\\1.\:i_{1}=4mA\\\\2.\:i_{2}=-2mA\\\\3.\:2k\Omega(i_{3}-i_{1})+4k\Omega (i_{3}-i_{2})+6k\Omega (i_{3} )-3V = 0\\=\:2k\Omega(i_{3}-4mA)+4k\Omega (i_{3}+2mA)+6k\Omega (i_{3} )-3V = 0\\=\:2k\Omega (i_{3})+8V+4k\Omega (i_{3})+8V+6k\Omega (i_{3} )-3V = 0\\=\:12k\Omega (i_{3})=3V\\=\:i_{3}=1/4mA \\\\Vo=6k\Omega(i_{3})-3V\\\\=6k\Omega \times \frac{1}{4}mA-3V=-\frac{3}{2}V$

Example 2 (CASE II) -Supermesh
Find Io

$\\1.\:i_{1}=2mA\\\\2.\:i_{2}-i_{3}=4mA\\\\Supermesh:-6V+1k\Omega (i_{3})+2k\Omega(i_{2})+2k\Omega(i_{2}-i_{1})+1k\Omega(i_{3}-i_{1})=0\\ -6V+1k\Omega(i_{3})+2k\Omega(4mA+i_{3})+2k\Omega(4mA+i_{3}-2mA)+1k\Omega(i_{3}-2mA)=0\\-6V+1k\Omega(i_{3})+8V+2k\Omega(i_{3})+8V+2k\Omega(i_{3})-4V+1k\Omega(i_{3})-2V=0\\6k\Omega(i_{3})=-4V\\i_{3}=-2/3mA=Io$

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3.2: Mesh Analysis

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