Mesh Analysis for ac Circuits

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Kirchhoff's voltage law (KVL) is the foundation of mesh analysis. The validity of KVL for ac circuit may be demonstrated in the Mesh Analysis for dc circuit page.
Keep in mind that mesh analysis is intended to be used on a planar circuit.
Planar circuits are circuits that can be drawn on a plane surface with no wires crossing each other.
Before examining the application of the method to ac networks, the student should first review the appropriate sections on mesh analysis in Chapter 6 since the content of this section will be limited to the general conclusions of Chapter 6.
The general approach to mesh analysis for independent sources includes the same sequence of steps appearing in Chapter 6. In fact, throughout this section the only change from the dc coverage will be to substitute impedance for resistance and admittance for conductance in the general procedure.
  • Assign a distinct current in the clockwise direction to each independent closed loop of the network. It is not absolutely necessary to choose the clockwise direction for each loop current. However, it eliminates the need to have to choose a direction for each application. Any direction can be chosen for each loop current with no loss in accuracy as long as the remaining steps are followed properly.
  • Indicate the polarities within each loop for each impedance as determined by the assumed direction of loop current for that loop.
  • Apply Kirchhoff's voltage law around each closed loop in the clockwise direction. Again, the clockwise direction was chosen to establish uniformity and to prepare us for the format approach to follow.
    • If an impedance has two or more assumed currents through it, the total current through the impedance is the assumed current of the loop in which Kirchhoff's voltage law is being applied, plus the assumed currents of the other loops passing through in the same direction, minus the assumed currents passing through in the opposite direction.
    • The polarity of a voltage source is unaffected by the direction of the assigned loop currents.
  • Solve the resulting simultaneous linear equations for the assumed loop currents.
The technique is applied as above for all networks with independent sources or for networks with dependent sources where the controlling variable is not a part of the network under investigation. If the controlling variable is part of the network being examined, a method to be described shortly must be applied.
Example 1: find the current $I_1$ in [Fig. 1] using mesh analysis.
Fig. 1: Example 1.
Solution: When applying these methods to ac circuits, it is good practice to represent the resistors and reactance (or combinations thereof) by subscripted impedances. When the total solution is found in terms of these subscripted impedances, the numerical values can be substituted to find the unknown quantities.
Fig. 2: Assigning the mesh currents and subscripted impedances for the network of Fig. 1.
The network is redrawn in [Fig. 2] with subscripted impedances:
$$Z_1 = j X_L =j 2 Ω \,\, E_1 = 2 V \angle 0^\circ$$
$$Z_2 = R = 4Ω \,\, E_2 =6 V \angle 0^\circ$$
$$Z_3 = - j X_C = - j 1 Ω $$
step 3:
$$ \begin{split} +E_1 - I_1Z_1 - Z_2(I_1 - I_2) &= 0\\ -Z_2(I_2 - I_1) - I_2 Z_3 - E_2 &=0\\ \text{or}\\ E_1 - I_1Z_1 -I_1Z_2 + I_2Z_2 &=0\\ -I_2Z_2 + I_1Z_2 - I_2Z_3 - E_2 &=0\\ \text{so that}\\ I_1(Z_1 + Z_2) - I_2 Z_2 &=E_1\\ I_2(Z_2 + Z_3) - I_1 Z_2 &= -E_2\\ \text{which are rewritten as}\\ I_1(Z_1 + Z_2) - I_2 Z_2 &=E_1\\ - I_1 Z_2 + I_2(Z_2 + Z_3) &= -E_2\\ \end{split} $$
using determinants, we obtain
$$ \Delta = \begin{vmatrix} Z_1 + Z_2 & -Z_2 \\ -Z_2 & Z_2 + Z_3 \end{vmatrix} $$
$$\begin{gather} I_1 = { \begin{vmatrix} E_1 & -Z_2 \\ -E_2 & Z_2+Z_3 \\ \end{vmatrix} \over \Delta}\\ I_1 = 3.61 A \angle -236.30^\circ\\ \end{gather} $$
Example 2: Write the mesh currents for the network of [Fig. 3] having a dependent voltage source.
Fig. 3: Applying mesh analysis to a network with a voltage-controlled voltage source.
Solution:
$$ \begin{gather} E_1 - I_1 R_1 - R_2 (I_1 - I_2) = 0\\ R_2 (I_2 - I_1) + \mu V_x - I_2R_3 = 0\\ \text{then substitute}\, V_x = (I_1 - I_2)R_2\\ R_2 (I_2 - I_1) + \mu [(I_1 - I_2)R_2] - I_2R_3 = 0\\ \text{The result is two equations and two unknowns.}\\ (R_1+R_2)I_1 - I_2R_2 = - E_1 \tag{1}\\ (\mu R_2 - R_2)I_1 +(R_2 - \mu R_2 - R_3)I_2 = 0 \tag{2}\\ \end{gather} $$

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