Mesh Analysis

Mesh 1 contains the 2 Ω and the shared 4 Ω. The 4 Ω carries \(I_1 - I_2\) in the direction of \(I_1\):

Mesh 2 contains the 6 Ω and the same shared 4 Ω, which now carries \(I_2 - I_1\):

Solving the pair. Multiply the first by 5 and the second by 2, then add:

The shared resistor carries the difference, not either mesh current on its own:

Write both equations with clockwise currents, as always. Mesh 1 is straightforward:

In mesh 2 the 18 V source is a drop when traversed clockwise, so it appears on the right-hand side with a negative sign:

Solving. From the first, \(I_1 = (16 + 4I_2)/6\); substituting:

The shared branch carries

Check mesh 2 numerically: \(-4(2) + 10(-1) = -18\;\checkmark\)

Writing KVL clockwise round each mesh in turn:

Mesh 2 contains no source, so its right-hand side is zero. The two 120 V sources appear with opposite signs because one drives its mesh clockwise and the other anticlockwise.

Dividing through by 10 to simplify:

Add the first and third equations:

Substituting into the middle equation gives \(-i_1 + 10(i_1+i_3) - i_3 = 0\), that is \(9i_1 + 9i_3 = 0\), so \(i_3 = -i_1\) and therefore \(i_2 = 0\). The first equation then gives

Check all three equations: \(4(-3) - 0 = -12\;\checkmark\), \(3 + 0 - 3 = 0\;\checkmark\), \(0 + 12 = 12\;\checkmark\)

With all mesh currents clockwise, the equations can be written straight down without drawing a single KVL loop. The rules are:

Applying them term by term:

Meshes 1 and 3 share nothing, so that entry is zero.

The system in matrix form:

From the third row, \(8I_3 = 4I_2\), so \(I_2 = 2I_3\). Substituting into the second row:

And into the first row:

Verify row 1: \(10(3) - 4(2) + 0 = 22\;\checkmark\); row 2: \(-12 + 24 - 4 = 8\;\checkmark\); row 3: \(-8 + 8 = 0\;\checkmark\)

A current source that belongs to one mesh only fixes that mesh current. No KVL equation is written for mesh 1 at all — and none can be, since the source's voltage is unknown:

Three meshes would normally need three equations; the current source has removed one unknown and one equation together, leaving two.

KVL for mesh 2, with \(i_1\) now a known constant:

KVL for mesh 3:

Substituting \(i_2 = 3i_3 - 5\) into the mesh-2 equation:

The shared 4 Ω carries the difference:

First establish the actual current in each element. Only the shared branch is subtle:

Resistor powers:

Source powers. Each source has current leaving its positive terminal, so both deliver:

The balance:

The two counts are fixed entirely by topology:

a\(n - 1 = 4\) against \(b - n + 1 = 3\). Mesh analysis wins — three equations instead of four.

b\(n - 1 = 3\) against \(b - n + 1 = 3\). A tie on count, so the decision falls to the sources and to what the question asks for.

c\(n - 1 = 4\) against \(b - n + 1 = 5\). On count, nodal wins — and the source mix reinforces it, since four voltage sources are far more convenient for nodal analysis than for mesh.

Three considerations beyond the raw count:

Mesh 1 contains no dependent element, so it is written exactly as before:

For mesh 2, write the dependent source as a symbol first:

Now supply the constraint \(i_x = I_1 - I_2\) and expand:

Substituting into mesh 1:

Hence \(i_x = 3 - 1.5 = 1.5\ \text{A}\) and the dependent source stands at \(2i_x = 3\ \text{V}\). Verify mesh 2:

The source is a rise in the direction of \(I_2\), so it delivers:

Full balance: the 12 V source supplies 36 W, the dependent source 4.5 W, and the resistors absorb \(18 + 9 + 13.5 = 40.5\ \text{W}\;\checkmark\)

A current source in mesh 2's own branch fixes that mesh current — but here the value is itself unknown, so instead of a datum we get a relation:

No KVL equation can be written for mesh 2, because the source's voltage is unknown. Mesh 1 gives the only other equation:

Substituting:

Check the constraint: \(2i_x = 2\ \text{A} = I_2\;\checkmark\)

To find the source's voltage, apply KVL round mesh 2 — now legitimate, because the voltage is the only remaining unknown:

(Mesh 2 contains only the shared 6 Ω and the source, so the source must stand at the 6 Ω's drop.)

Note first that the matrix is not symmetric — \(R_{12} = -55\) against \(R_{21} = -35\), and \(R_{13} = -15\) against \(R_{31} = -35\). That is the signature of the dependent source and is expected here, not an error.

Solving by elimination. From rows 2 and 3, subtract to remove \(I_1\):

Eliminate \(I_1\) between rows 1 and 2 by multiplying row 2 by \(90/35\) and subtracting:

Solving (A) and (B) simultaneously:

Back-substituting into row 3:

The dependent source stands at \(20(I_1 - I_2) = 20(0.448) = 8.96\ \text{V}\), with \(I_1 - I_3 = -0.108\ \text{A}\) entering its positive terminal. Under the passive sign convention:

The negative absorbed power means the source is in fact delivering about 0.968 W to the rest of the circuit.

Expand the determinant of the coefficient matrix along the first row:

Replace the first column by the source vector:

Similarly for the second column:

And the third:

These agree exactly with the elimination result of Problem 4.

Conditioning. A determinant that is large compared with the individual matrix entries indicates a well-conditioned system, in which small changes in a resistance produce correspondingly small changes in the currents. Here \(\Delta = 672\) against entries of order 10, which is comfortable. A determinant approaching zero would signal a near-singular network — physically, one where two meshes are almost perfectly coupled, and numerically one where rounding errors are amplified.

Take three clockwise mesh currents: \(I_1\) round the upper-left triangle \(a\!-\!c\!-\!d\), \(I_2\) round the upper-right triangle \(c\!-\!b\!-\!d\), and \(I_3\) round the outer loop through the source. Writing the equations by inspection:

That is

Divide the first by 6 and the second by 3:

Adding these two eliminates \(I_1\): \(8I_2 = 4I_3\), so \(I_3 = 2I_2\).

Substituting back into the first: \(6I_1 = 3I_2 + 4I_2 = 7I_2\), so \(I_1 = \tfrac{7}{6}I_2\). The third equation then gives

The source current is the outer mesh current \(I_3\), so

The 18 Ω arm is shared by meshes 1 and 2, so it carries their difference:

aA resistor on the periphery of mesh 1 belongs to no other mesh, so it carries the mesh current itself:

bThe shared resistor carries the difference, in the direction of \(I_1\):

Larger than either mesh current, because \(I_2\) is negative and therefore reinforces \(I_1\) in that branch.

cSimilarly, in the direction of \(I_2\):

That is, 5 A flowing in the direction of \(I_3\).

dRearranging \(l = b - n + 1\) with \(l = 3\) meshes and \(n = 5\) nodes:

Neither mesh equation can be written on its own, because the shared branch contains a current source whose voltage is unknown and cannot be expressed in terms of mesh currents. The remedy is to write KVL round a path that avoids the offending branch — the outer loop, or supermesh:

The current source's voltage never appears, because the path never crosses it.

One equation, two unknowns. The missing relation is supplied by the source itself:

Substituting:

To find the source's voltage, now apply KVL to mesh 1 alone — legitimate at last, since everything else in it is known:

Now determine the sign. The middle branch carries \(I_1 - I_2 = 2\ \text{A}\) downwards, and mesh 1's KVL placed the \(+\) mark at the top of the source. Current therefore enters the positive terminal, so the passive sign convention is satisfied and the source absorbs:

Confirming with the full balance — the 10 V source is the only element delivering:

Mesh 1 is fixed outright by its peripheral current source:

Substituting into the mesh-4 equation:

Expanding the supermesh equation with \(I_1 = 4\):

Now use the current-source constraint \(I_2 = I_3 + 1\):

Solving the pair \(4I_4 - I_3 = 14\) and \(3I_3 - I_4 = 2\). From the second, \(I_4 = 3I_3 - 2\):

Converting to branch currents:

Eliminate \(i_2\) immediately using the constraint. In the supermesh equation:

And in the mesh-3 equation:

From (1), \(i_3 = 2i_1 + 72\). Substituting into (2):

Back-substituting:

The required branch current:

Check both original equations: \(6(-46) + 4(-1) - 5(-20) + 180 = -276 - 4 + 100 + 180 = 0\;\checkmark\) and \(46 + 4 - 140 + 90 = 0\;\checkmark\)

A ladder with three shunt branches has three meshes. Writing the equations by inspection:

That is

The third equation gives \(I_3 = \tfrac{2}{3}I_2\). Substituting into the second:

And into the first:

Converting to branch currents and comparing with Set 2, where the same ladder was reduced rung by rung:

aThe source carries the outer mesh current:

bNow element by element, using the branch currents tabulated in Problem 17:

The two figures agree exactly:

With a single source \(V\) in mesh \(j\), the mesh equations are \(\mathbf{R}\,\mathbf{I} = V\mathbf{e}_j\), where \(\mathbf{e}_j\) is the unit vector. The solution is

Move the source to mesh \(k\) and read the current in mesh \(j\):

For a network of resistors and independent sources \(\mathbf{R}\) is symmetric, and the inverse of a symmetric matrix is symmetric. Therefore

This is the reciprocity theorem, and the proof is two lines once the matrix formulation is in place.

Numerical check. Take the matrix of Problem 4 and place a single 22 V source in mesh 1. Solving \(\mathbf{R}\mathbf{I} = (22,0,0)^{\mathsf T}\) gives

Now move the 22 V source to mesh 3 and read mesh 1. Solving \(\mathbf{R}\mathbf{I} = (0,0,22)^{\mathsf T}\):

The two cofactors are equal because the matrix is symmetric — which is precisely the content of the proof.

aWithout the dependent source, mesh 2's equation would read \(-4I_1 + 10I_2 = 0\). The term \(-2i_x = -2(I_1 - I_2)\) adds \(-2\) to \(R_{21}\) and \(+2\) to \(R_{22}\):

bThree casualties:

A fourth casualty, less often noticed: the matrix need no longer be positive definite, so the network can present a negative input resistance at some port.

cLet the gain be \(k\), so the CCVS is \(k\,i_x\). Mesh 2 becomes \(-(4+k)I_1 + (10+k)I_2 = 0\), and the determinant is

The matrix is singular when

Physical meaning. At \(k = -22\) the equations have no unique solution: the circuit either has no steady state or infinitely many. A dependent source with sufficiently large negative gain is injecting energy in proportion to the current it senses — positive feedback — and the model is telling you that the assumption of a bounded DC solution has broken down. In a real circuit the result is an oscillator, a latch-up, or saturation at a supply rail, none of which a linear DC model can represent.