Supermesh Analysis

Neither mesh equation can be written alone. Mesh 1's KVL would read \(9 - 1\cdot I_1 - V_{cs} = 0\), introducing the unknown source voltage \(V_{cs}\) — one new unknown for one new equation, which is no progress at all.

Instead write KVL round the outer boundary, which never crosses the source branch:

That is one equation for two unknowns. The source supplies the second:

Substituting:

Now, and only now, recover the source voltage from mesh 1's KVL — legitimate at last, because everything else in it is known:

The branch carries \(I_1 - I_2 = 3\ \text{A}\) downwards, entering the terminal marked \(+\). The passive sign convention is therefore satisfied and the source absorbs:

Power balance confirms it:

Write both mesh equations honestly, keeping the unknown source voltage \(V_{cs}\) as a symbol. Going clockwise, mesh 1 travels down the shared branch and mesh 2 travels up it, so the source appears with opposite signs in the two equations:

Add them. The \(V_{cs}\) terms cancel exactly:

This is precisely the supermesh equation used in Problem 1 — obtained without ever drawing a boundary or invoking a special technique.

The cancellation is guaranteed, not lucky. Any element in a branch shared by two meshes is traversed in opposite directions by the two clockwise loops, so its voltage always enters the two equations with opposite signs and always vanishes on addition.

Counting confirms nothing has been lost. Two meshes originally offered two KVL equations but carried three unknowns — \(I_1\), \(I_2\) and \(V_{cs}\). Adding the equations eliminates \(V_{cs}\), leaving one equation and two unknowns; the source's own constraint restores the balance:

Tidy the supermesh equation first:

And mesh 2:

Use the constraint \(i_1 = i_3 + 7\) to eliminate \(i_1\) from both. In the supermesh equation:

And in the mesh-2 equation:

Hence

The 3 Ω is shared by meshes 2 and 3, so it carries their difference:

Substitution check on both original equations: \(-7 + (9-2.5) + 3(2-2.5) + 2 = -7 + 6.5 - 1.5 + 2 = 0\;\checkmark\) and \((2.5-9) + 5 + 1.5 = 0\;\checkmark\)

Because mesh 2 is joined to mesh 1 by one source and to mesh 3 by another, no boundary can be drawn that separates any pair of them. All three collapse into a single supermesh, contributing one KVL equation:

The two sources supply the two missing equations:

Substituting the first into the supermesh equation:

Hence

Verify: \(I_1 + I_2 = -5\;\checkmark\), \(I_3 - I_2 = 2\;\checkmark\), \(I_2 - I_1 = 1\;\checkmark\)

The count works out as it must: one supermesh KVL plus two constraints gives three equations for three mesh currents.

Substitute the constraint \(I_1 = I_2 - 2\) into the supermesh equation:

Hence

Both resistors lie on the periphery, so each carries its own mesh current directly:

Check the constraint: \(0.5455 - (-1.4545) = 2\ \text{A}\;\checkmark\), and the supermesh: \(4 + 1.4545 - 5.455 = 0\;\checkmark\)

The supermesh path excludes the entire shared branch, not merely the source symbol within it. The 5 Ω is therefore never traversed, and the KVL equation is unchanged:

The constraint is unchanged too, since a resistor in series with a current source cannot alter the current the source enforces:

So the mesh currents are exactly as before:

The 5 Ω nevertheless carries the full branch current and dissipates:

And the source's terminal voltage has changed. Applying KVL down the shared branch from the top node to the bottom:

With 3 A still entering the terminal previously marked \(+\), the source now has \(P = (-11)(3) = -33\ \text{W}\) — it delivers 33 W, having absorbed 12 W before the resistor was added.

Power balance: the 9 V source still supplies 45 W and the current source now supplies 33 W, giving 78 W. The resistors take \(25 + 8 + 45 = 78\ \text{W}\;\checkmark\)

Walk the supermesh boundary clockwise from the bottom-left. The 26 V is a rise, both resistors are drops, and the 6 V is entered at its positive terminal and so is also a drop:

A voltage source on the supermesh path causes no difficulty whatever — it enters the equation exactly as it would in an ordinary mesh. Only current sources force the boundary to detour.

The constraint completes the pair:

Adding and subtracting the two equations:

Check: \(26 - 14 - 6 - 6 = 0\;\checkmark\)

Resistors first, using branch currents. Both lie on the periphery, so each carries its own mesh current:

The 26 V source carries \(I_1 = 7\ \text{A}\) out of its positive terminal, so it delivers:

The 6 V source opposes mesh 2's current, so current enters its positive terminal and it absorbs:

Now the current source. Recover its voltage from mesh 1's KVL:

The branch carries \(I_1 - I_2 = 4\ \text{A}\) downwards into that positive terminal, so the source absorbs:

The balance:

The supermesh path is unchanged, because reversing a source does not move it. So the KVL equation is identical:

Only the constraint flips:

Substituting:

Source voltage from mesh 1:

Now 3 A flows upwards, leaving the positive terminal, so the source delivers:

Balance: the 9 V source supplies \(9 \times 1 = 9\ \text{W}\) and the current source 24 W, total 33 W. The resistors absorb \((1)^2(1) + (4)^2(2) = 1 + 32 = 33\ \text{W}\;\checkmark\)

Side by side:

Source A lies on the outer periphery of mesh 1. Only one mesh current flows through it, so it fixes that current directly:

Source B is shared by meshes 2 and 3. Neither KVL equation can be written, so those two meshes merge:

Source C is peripheral to mesh 4, so like A it simply fixes a mesh current:

Tallying the equations:

Four equations for four mesh currents — exactly right, and only one of them requires any real work.

The mesh count comes from the topology alone:

Each supermesh absorbs two meshes and yields one KVL equation instead of two. With two disjoint supermeshes, four of the five meshes are consumed:

Each current source supplies one constraint:

The total:

Five equations for five mesh currents. The supermesh technique never changes the size of the system — it only changes which equations you write.

The general statement, for \(s\) current sources in shared branches forming disjoint supermeshes:

The supermesh KVL is written exactly as for an independent source — the dependence is irrelevant to the path:

The constraint now involves an unknown on both sides. The 6 Ω is peripheral to mesh 2, so \(i_x = I_2\), and:

Substituting into the supermesh equation:

Verify the constraint: the source value is \(i_x = I_2 = 2\ \text{A}\), and indeed \(I_1 - I_2 = 2\ \text{A}\;\checkmark\)

Recover the source voltage from mesh 1:

The branch carries 2 A downwards into that terminal, so the dependent source absorbs:

Balance: \(24 \times 4 = 96\ \text{W}\) supplied against \((4)^2(3) + (2)^2(6) + 24 = 48 + 24 + 24 = 96\ \text{W}\;\checkmark\)

Substituting the numbers, the three equations become

Substitute \(I_3 = I_2 + 4\) into the first two:

Subtracting gives \(8I_2 = 2\), so

The voltage source carries \(I_1\) out of its positive terminal, so it delivers:

For the current source, first find its terminal voltage by walking mesh 2:

Both sources deliver, so the resistors must absorb the total. Checking:

Collect terms in the first equation:

And the second:

The constraint rearranges to

Substituting into the first two equations:

And into the second:

Hence

The five branch currents:

Check the constraint: \(I_2 - I_1 = -6.143\) and \(4.3(I_3 - I_1) = 4.3(-1.429) = -6.144\;\checkmark\)

Substitute \(I_x = -I_3\) throughout to reduce the unknowns to the four mesh currents:

The two constraints:

From (2) and (4), express \(I_3\) in terms of \(I_1\):

From (3), \(I_2 = I_1 - 3\). Substituting both into (1):

Hence

The required quantities:

Check equation (1): \(-6(3.330) - 8(0.330) + 15(1.508) = -19.98 - 2.64 + 22.62 = 0\;\checkmark\)

Because mesh 2 is tied to mesh 1 by one source and to mesh 3 by another, the two supermeshes overlap and all three meshes merge into a single region. Simplifying its KVL:

The two source constraints:

And mesh 4, which is ordinary:

Use (2) and (3) to express \(i_1\) and \(i_2\) in terms of \(i_3\) and \(i_4\):

Substituting into (1):

Solving (4) and (5). From (4), \(i_3 = (5i_4 + 5)/4\); substituting:

Back-substituting:

Check (1): \(-7.5 + 3(-2.5) + 6(3.929) - 4(2.143) = -7.5 - 7.5 + 23.57 - 8.57 = 0\;\checkmark\)

Mesh 2 is joined to both its neighbours by current sources, so no boundary can separate any pair. All three meshes form one supermesh whose path runs round the outside of the whole circuit, crossing only the peripheral elements:

The two constraints chain the currents together:

Substituting both into the supermesh equation:

Hence

Check the supermesh: \(2(4) + 4(2) + 6(-1) = 8 + 8 - 6 = 10\ \text{V}\;\checkmark\)

The general rule this illustrates: a chain of \(s\) current sources linking \(s+1\) meshes produces

Case 1 — the source is peripheral. If the branch belongs to one mesh only, exactly one mesh current flows through it and the source fixes that current outright:

One unknown and one equation vanish together. This is the situation of Problem 10's sources A and C.

Case 2 — the source has a resistor in parallel with it. Then a source transformation converts the pair into a voltage source in series with the same resistance:

The offending current source is gone and ordinary mesh analysis applies. Note the requirement: a resistor in parallel. A resistor in series with the source, as in Problem 6, cannot be transformed and does not help.

Case 3 — nodal analysis is used instead. Current sources are the natural excitation for nodal analysis, entering the equations directly as known injections. A circuit with several current sources and few voltage sources is usually far quicker to solve nodally, and no supermesh arises at all:

A fourth case is worth mentioning although it is a coincidence rather than a technique: if the circuit can be redrawn so that the source branch becomes peripheral, Case 1 applies. Planar circuits can often be re-laid-out in surprising ways, and it costs nothing to look.

The form \(\mathbf{R}\mathbf{I} = \mathbf{V}\) presumes that every row is a KVL equation in which each coefficient is a resistance and each right-hand side is a sum of source voltages. A supermesh circuit violates this in two ways at once.

First, the constraint row is not a KVL equation. The row expressing \(I_1 - I_2 = I_s\) has coefficients \(+1\) and \(-1\), which are dimensionless, and a right-hand side in amperes rather than volts:

The matrix is no longer dimensionally homogeneous, so calling it a resistance matrix is meaningless and asking whether it is symmetric is a category error.

Second, one mesh has lost its own equation. Two meshes now share a single KVL row, so even the resistive part of the system is not square in the way \(\mathbf{R}\) requires.

What replaces it. The honest description is a general linear system \(\mathbf{A}\mathbf{x} = \mathbf{b}\) in which the rows have mixed physical meaning. It is still square, still solvable, and still \(l \times l\) — but it carries none of the structural guarantees of Set 4: no symmetry, no diagonal dominance, and no automatic proof of a unique solution.

An alternative is to eliminate the current sources first, by source transformation where possible (Problem 18, Case 2). The resulting circuit does have a genuine symmetric \(\mathbf{R}\) — which is a good practical reason to prefer transformation over the supermesh when both are available.

Write the KVL equation for each mesh \(k\) in the chain, keeping every unknown source voltage as a symbol. Let \(\Sigma_k\) denote the sum of all resistive and voltage-source terms round mesh \(k\), and let \(V_j\) be the voltage across the \(j\)th current source:

Each source voltage appears exactly twice — once with each sign — because the two meshes flanking it traverse that branch in opposite directions.

Add all \(s+1\) equations. The source voltages telescope:

Leaving

This sum contains only resistive drops and independent voltage sources — and every element in a branch internal to the chain has also cancelled, for the same reason. What survives is precisely the set of elements on the outer perimeter: the supermesh equation.

Counting. Before: \(s+1\) KVL equations carrying \(s+1\) mesh currents plus \(s\) unknown source voltages, so \(2s+1\) unknowns and only \(s+1\) equations — underdetermined by \(s\). After: one KVL equation, and the \(s\) source currents supply \(s\) constraints:

The result also proves that the choice of supermesh boundary is immaterial: any path that avoids all \(s\) source branches yields the same equation, because the derivation never referred to the path — only to the cancellation.