5th Week: "Mesh Analysis - AC"

 "Mesh Analysis"


Kirchhoff’s voltage law (KVL) forms the basis of mesh analysis.

Example:

The circuit shown is operating in the steady state.
Determine the four mesh currents.

First, convert all values to phasor notation.

Next, perform the mesh analysis.

Since dependent sources are present, define the parameters upon which they depend in terms of the mesh currents.

i₅ = I_C - I_D
V_L = j1K(I_A - I_B)

Since current sources are present, express their values in terms of the mesh currents, then form supermeshes.

500m0° = I_C - I_A
500m0° = - I_A + I_C (Equation 1)
12 i₅ = - I_B
12 (I_C - I_D) = - I_B
0 = I_B + 12 I_C - 12 I_D (Equation 2)

Now write KVL around each remaining mesh and supermesh.
Supermesh A/C

6 V_L + j1K(I_A - I_B) - j200 (I_C - I_D) + j5K I_C + 10K I_C + 4K I_A = 0
6 j1K(I_A - I_B) + j1K(I_A - I_B) - j200 (I_C - I_D) + j5K I_C + 10K I_C + 4K I_A = 0
0 = (4K + j7K) I_A - j 7K I_B + (10K + j4800) I_C + j200 I_D (Equation 3)

Mesh D

0 = - j400 (I_D - I_B) + 1K I_D - j200 (I_D - I_C)
0 = j400 I_B + j200 I_C + (1K - j600) I_D (Equation 4)

Inserting Equations 1 through 4 into the calculator and letting it do the difficult part yields

I_A = 429 m169°
I_B = 337 m127°
I_C = 113 m45.8°
I_D = 120 m59.2°

Converting back to the time domain gives

I_A(t) = 429 cos (10000 t + 169°) mA = - 429 cos (10000 t - 11°) mA (Either form acceptable)
I_B(t) = 337 cos (10000 t + 127°) mA = - 337 cos (10000 t - 53°) mA (Either form acceptable)
I_C(t) = 113 cos (10000 t + 45.8°) mA
I_D(t) = 120 cos (10000 t + 59.2°) mA

Reflection:

• I learned that all the concepts and process in solving the unknowns in Ac Analysis are the same with Dc Analysis. It only involves complex numbers.

Videos:

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GOD Bless! :)

By:

AYALA, ARNY  S.   BSECE -3

ECE 321

Professor:

ENGR. JAY S. VILLAN, MEP - EE