10.3 Mesh Analysis

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10.3 Mesh Analysis

Kirchhoff’s voltage law (KVL) forms the basis of mesh analysis. The validity of KVL for ac circuits was shown in Section 9.6 and is illustrated in the follo wing examples. Keep in mind that the v ery nature of using mesh analysis is that it is to be applied to planar circuits.

Determine current Io in the circuit of Fig. 10.7 using mesh analysis. Example 10.3

Solution:

Applying KVL to mesh 1, we obtain

$$(8+j10-j2)\mathbf{I}_1 - (-j2)\mathbf{I}_2 - j10\mathbf{I}_3 = 0 \tag{10.3.1}$$

Figure 10.7 For Example 10.3.

For mesh 2,

$$(4 - j2 - j2)I2 - (-j2)I1 - (-j2)I3 + 20 / 90o = 0$$

(10.3.2)

For mesh 3, I3 = 5. Substituting this in Eqs. (10.3.1) and (10.3.2), we get

$$(8+j8)\mathbf{I}_1 + j2\mathbf{I}_2 = j50 \tag{10.3.3}$$ $$j2\mathbf{I}_1 + (4 - j4)\mathbf{I}_2 = -j20 - j10$$

(10.3.4)

Equations (10.3.3) and (10.3.4) can be put in matrix form as

[ ⁸ + j⁸ j2 j2 ⁴ − j⁴ ] [ ^I1 I2 ] = [ ^j50 −j30]

from which we obtain the determinants

$$\Delta = \begin{vmatrix} 8 + j8 & j2 \\ j2 & 4 - j4 \end{vmatrix} = 32(1 + j)(1 - j) + 4 = 68$$ $$\Delta_2 = \begin{vmatrix} 8 + j8 & j50 \\ j2 & -j30 \end{vmatrix} = 340 - j240 = 416.17 \underline{/-35.22^{\circ}}$$ $$\mathbf{I}_2 = \frac{\Delta_2}{\Delta} = \frac{416.17 \underline{/-35.22^{\circ}}}{68} = 6.12 \underline{/-35.22^{\circ}} A$$

The desired current is

$$I_o = -I_2 = 6.12 \underline{144.78}^{\circ} A$$

For Practice Prob. 10.3.

Solution:

As shown in Fig. 10.10, meshes 3 and 4 form a supermesh due to the current source between the meshes. For mesh 1, KVL gives

$$-10 + (8 - j2)\mathbf{I}_1 - (-j2)\mathbf{I}_2 - 8\mathbf{I}_3 = 0$$

or

$$(8 - j2)\mathbf{I}_1 + j2\mathbf{I}_2 - 8\mathbf{I}_3 = 10 \tag{10.4.1}$$

For mesh 2,

$$I_2 = -3 \tag{10.4.2}$$

For the supermesh,

$$(8 - j4)\mathbf{I}_3 - 8\mathbf{I}_1 + (6 + j5)\mathbf{I}_4 - j5\mathbf{I}_2 = 0 \tag{10.4.3}$$

Due to the current source between meshes 3 and 4, at node A,

$$\mathbf{I}_4 = \mathbf{I}_3 + 4 \tag{10.4.4}$$

■ METHOD 1 Instead of solving the above four equations, we re ‑ duce them to two by elimination.

Combining Eqs. (10.4.1) and (10.4.2),

$$(8 - j2)\mathbf{I}_1 - 8\mathbf{I}_3 = 10 + j6 \tag{10.4.5}$$

Combining Eqs. (10.4.2) to (10.4.4),

$$-8I1 + (14 + j)I3 = -24 - j35$$

(10.4.6)

Figure 10.10 Analysis of the circuit in Fig. 10.9.

From Eqs. (10.4.5) and (10.4.6), we obtain the matrix equation

$$\begin{bmatrix} 8-j2 & -8 \ -8 & 14+j \end{bmatrix} \begin{bmatrix} I_1 \ I_3 \end{bmatrix} = \begin{bmatrix} 10+j6 \ -24-j35 \end{bmatrix}$$

We obtain the following determinants

$$\Delta = \begin{vmatrix} 8 - j2 & -8 \\ -8 & 14 + j \end{vmatrix} = 112 + j8 - j28 + 2 - 64 = 50 - j20$$ $$\Delta_1 = \begin{vmatrix} 10 + j6 & -8 \\ -24 - j35 & 14 + j \end{vmatrix} = 140 + j10 + j84 - 6 - 192 - j280$$ $$= -58 - j186$$

Current I1 is obtained as

$$\mathbf{I}_1 = \frac{\Delta_1}{\Delta} = \frac{-58 - j186}{50 - j20} = 3.618 \; \underline{/274.5^\circ} \, \text{A}$$

The required voltage Vo is

$$\mathbf{V}_o = -j2(\mathbf{I}_1 - \mathbf{I}_2) = -j2(3.618 \angle 274.5^\circ + 3)$$

= -7.2134 - j6.568 = 9.756 \angle 222.32^\circ \text{V}

■ METHOD 2 We can use MATLAB to solve Eqs. (10.4.1) to (10.4.4). We first cast the equations as

$$\begin{bmatrix} 8-j2 & j2 & -8 & 0 \ 0 & 1 & 0 & 0 \ -8 & -j5 & 8-j4 & 6+j5 \ 0 & 0 & -1 & 1 \ \end{bmatrix} \begin{bmatrix} I_1 \\ I_2 \\ I_3 \\ I_4 \end{bmatrix} = \begin{bmatrix} 10 \\ -3 \\ 0 \\ 4 \end{bmatrix}$$

(10.4.7a)

or

$$AI = B$$

By inverting A, we can obtain I as

$$\mathbf{I} = \mathbf{A}^{-1} \mathbf{B} \tag{10.4.7b}$$

We now apply MATLAB as follows:

>> A = [(8-j*2) j*2 -8 0;
 0 1 0 0;
 -8 -j*5 (8-j*4) (6+j*5);
 0 0 -1 1];
>> B = [10 -3 0 4]';
>> I = inv(A)*B
I =
 0.2828 - 3.6069i
 -3.0000
 -1.8690 - 4.4276i
 2.1310 - 4.4276i
>> Vo = -2*j*(I(1) - I(2))
Vo =
 -7.2138 - 6.5655i

as obtained previously.

Calculate current Io in the circuit of Fig. 10.11.

Answer: 6.089∕5.94° A.

10.4 Superposition Theorem

Since ac circuits are linear, the superposition theorem applies to ac circuits the same way it applies to dc circuits. The theorem becomes important if the circuit has sources operating at different frequencies. In this case, since the impedances depend on frequency, we must have a different frequency domain circuit for each frequency. The total re ‑ sponse must be obtained by adding the individual responses in the time domain. It is incorrect to try to add the responses in the phasor or fre‑ quency domain. Why? Because the exponential factor e^jω^t is implicit in sinusoidal analysis, and that factor would change for every angular frequency ω. It would therefore not make sense to add responses at different frequencies in the phasor domain. Thus, when a circuit has sources operating at different frequencies, one must add the responses due to the individual frequencies in the time domain.

Use the superposition theorem to find Io in the circuit in Fig. 10.7.

Solution:

Let

$$\mathbf{I}_o = \mathbf{I}_o' + \mathbf{I}_o'' \tag{10.5.1}$$

where I′ o and I″ o are due to the voltage and current sources, respectively. To find I′ o, consider the circuit in Fig. 10.12(a). If we let Z be the parallel combination of −j2 and 8 + j10, then

2 and 8 + j10, then
\n

$$\mathbf{Z} = \frac{-j2(8+j10)}{-2j+8+j10} = 0.25 - j2.25$$

and current I′ o is

$$\mathbf{I}'_o = \frac{j20}{4 - j2 + \mathbf{Z}} = \frac{j20}{4.25 - j4.25}$$

or

$$\mathbf{I}'_o = -2.353 + j2.353\tag{10.5.2}$$

To get I″ o, consider the circuit in Fig. 10.12(b). For mesh 1,

$$(8 + j8)\mathbf{I}_1 - j10\mathbf{I}_3 + j2\mathbf{I}_2 = 0 \tag{10.5.3}$$

For mesh 2,

$$(4 - j4)\mathbf{I}_2 + j2\mathbf{I}_1 + j2\mathbf{I}_3 = 0 \tag{10.5.4}$$

For mesh 3,

I3 = 5 (10.5.5)

Figure 10.12 Solution of Example 10.5.