Super-mesh Analysis | Network Analysis

Tutorial 4 — Super-mesh Analysis

Worked problems on the supermesh technique: when a current source is shared between two meshes, the two meshes are combined into a single supermesh whose KVL equation is paired with the current-source constraint. Each problem is followed by a complete solution with the final result highlighted.

Network Analysis · Electric Circuit Analysis · 8 solved problems

Demonstrative VideoWalkthrough

Problem 1Supermesh

Find the current flowing through \(R_3\).

Solution

Applying KVL in the supermesh (meshes 1 and 3), KVL in mesh 2, and the current-source constraint between meshes 1 and 3:

\[ \begin{aligned} -7 + (i_1 - i_2) + 3(i_3 - i_2) + i_3 &= 0\\ (i_2 - i_1) + 2 i_2 + 3(i_2 - i_3) &= 0\\ i_1 - i_3 &= 7 \end{aligned} \]

Solving the three equations:

\[ i_1 = 9\ \text{A},\quad i_2 = 2.5\ \text{A},\quad i_3 = 2\ \text{A} \]

Answer\(i_1 = 9\ \text{A},\ i_2 = 2.5\ \text{A},\ i_3 = 2\ \text{A}\)

Problem 2Supermesh

Find the current flowing through \(V_2\).

Solution

KVL in the supermesh, together with the two current-source constraints:

\[ \begin{aligned} -10 - 2 I_1 - 2 I_2 &= 0\\ I_3 - I_2 &= 2\\ I_2 - I_1 &= 1 \end{aligned} \]

Solving:

\[ I_1 = -3\ \text{A},\quad I_2 = -2\ \text{A},\quad I_3 = 0\ \text{A} \]

Answer\(I_1 = -3\ \text{A},\ I_2 = -2\ \text{A},\ I_3 = 0\ \text{A}\)

Problem 3Supermesh — Branch Currents

Determine \(I_a\), \(I_b\), \(I_c\), \(I_d\) and \(I_e\).

Solution

The mesh / supermesh equations are:

\[ \begin{aligned} -10 I_1 - 100 I_2 - 50(I_2 - I_3) - 25(I_1 - I_3) &= 0\\ 200 - 25(I_3 - I_1) - 50(I_3 - I_2) &= 0\\ (I_2 - I_1) &= 4.3\,(I_3 - I_1) \end{aligned} \]

Solving for the mesh currents:

\[ I_1 = 7.403\ \text{A},\quad I_2 = 1.26\ \text{A},\quad I_3 = 5.97\ \text{A} \]

The requested branch currents then follow:

\[ \begin{aligned} I_a = I_1 = 7.40\ \text{A},\quad & I_b = I_3 = 5.97\ \text{A},\quad I_c = I_2 = 1.26\ \text{A}\\ I_d = I_3 - I_1 = -1.43\ \text{A},\quad & I_e = I_3 - I_2 = 4.71\ \text{A} \end{aligned} \]

Answer\(I_a = 7.40,\ I_b = 5.97,\ I_c = 1.26,\ I_d = -1.43,\ I_e = 4.71\ \text{A}\)

Problem 4Supermesh + Dependent Source

Find \(V_x\) and \(I_x\).

Solution

The mesh equations, current-source constraints, and dependent-source relations:

\[ \begin{aligned} -6 I_1 - 8 I_2 - 15 I_x &= 0 & I_1 - I_2 &= 3\\ -32(I_3 - I_4) + 15 I_x - 6 I_3 &= 0 & I_x &= -I_3\\ I_4 &= \frac{V_x}{8} & V_x &= 6 I_1 \end{aligned} \]

Solving the system for the mesh currents:

\[ I_1 = 3.33\ \text{A},\quad I_2 = 0.330\ \text{A},\quad I_3 = 1.508\ \text{A},\quad I_4 = 2.497\ \text{A} \]

Hence the required quantities:

\[ I_x = -I_3 = -1.508\ \text{A},\qquad V_x = 6 I_1 = 19.98\ \text{V} \]

Answer\(I_x = -1.508\ \text{A},\quad V_x = 19.98\ \text{V}\)

Problem 5Supermesh

Find the currents in both resistors.

Solution

Supermesh KVL and the current-source constraint:

\[ \begin{aligned} 4 - I_1 - 10 I_2 &= 0\\ I_2 - I_1 &= 2 \end{aligned} \]

Solving:

\[ I_1 = -1.455\ \text{A},\qquad I_2 = 0.5455\ \text{A} \]

Answer\(I_1 = -1.455\ \text{A},\ I_2 = 0.5455\ \text{A}\)

Problem 6Supermesh — Source Power

Find the power of the sources, given \(V_S = 10\,\text{V},\ I_S = 4\,\text{A},\ R_1 = 2\,\Omega,\ R_2 = 6\,\Omega,\ R_3 = 1\,\Omega,\ R_4 = 2\,\Omega\).

Solution

Supermesh KVL, the second mesh KVL, and the current-source constraint:

\[ \begin{aligned} V_s - R_1(I_1 - I_2) - R_3(I_1 - I_3) &= 0\\ -R_1(I_2 - I_1) - I_2 R_2 - I_3 R_4 - (I_3 - I_1) R_3 &= 0\\ I_3 - I_2 &= I_s \end{aligned} \]

Solving:

\[ I_1 = 4.92\ \text{A},\quad I_2 = 0.25\ \text{A},\quad I_3 = 4.25\ \text{A} \]

Power of the voltage source (negative ⇒ it delivers power):

\[ P_{V_S} = -V_S \times I_1 = -10 \times 4.92 = -49.2\ \text{W} \]

Voltage across the current source, then its power:

\[ \begin{aligned} V_{I_S} &= 2(0.25 - 4.92) + 6(0.25) = -7.84\ \text{V}\\ P_{I_S} &= V_{I_S} \times I_S = -7.84 \times 4 = -31.36\ \text{W} \end{aligned} \]

Answer\(P_{V_S} = -49.2\ \text{W},\quad P_{I_S} = -31.36\ \text{W}\) (both delivering)

Problem 7Nested Supermesh + Dependent Source

Find the currents \(i_1\) to \(i_4\) using mesh analysis.

Solution

Meshes 1 and 2 form a supermesh through an independent current source; meshes 2 and 3 form a supermesh through a dependent current source. The two overlap into one larger supermesh. KVL around it:

\[ 2 i_1 + 4 i_3 + 8(i_3 - i_4) + 6 i_2 = 0 \;\Rightarrow\; i_1 + 3 i_2 + 6 i_3 - 4 i_4 = 0 \]

KCL at node P, KCL at node Q (with \(I_0 = -i_4\)), and KVL in mesh 4:

\[ \begin{aligned} i_2 &= i_1 + 5\\ i_2 &= i_3 + 3 I_0 = i_3 - 3 i_4\\ 2 i_4 + 8(i_4 - i_3) + 10 &= 0 \;\Rightarrow\; 5 i_4 - 4 i_3 = -5 \end{aligned} \]

Solving the four equations:

\[ i_1 = -7.5\ \text{A},\quad i_2 = -2.5\ \text{A},\quad i_3 = 3.93\ \text{A},\quad i_4 = 2.143\ \text{A} \]

Answer\(i_1 = -7.5,\ i_2 = -2.5,\ i_3 = 3.93,\ i_4 = 2.143\ \text{A}\)

Additional Practice Problem

A fully worked, self-contained supermesh example that shows the complete method step by step: identify the current source shared between two meshes, replace those meshes with a single supermesh, write one KVL equation around its outer boundary, and add the current-source constraint.

Problem 8Supermesh — Worked Method

A 3 A current source is shared between mesh 1 and mesh 2, so the middle branch cannot take an ordinary KVL equation. Using the supermesh method, find \(I_1\) and \(I_2\), and the power delivered by the current source.

Solution

Since a current source sits between the meshes, combine them into a supermesh and write one KVL equation around the outer boundary (skipping the source branch), with both resistor drops:

\[ 9 - 1\cdot I_1 - 2\cdot I_2 = 0 \;\Rightarrow\; I_1 + 2 I_2 = 9 \]

The current source supplies the constraint equation:

\[ I_1 - I_2 = 3 \;\Rightarrow\; I_1 = I_2 + 3 \]

Substituting:

\[ (I_2 + 3) + 2 I_2 = 9 \;\Rightarrow\; 3 I_2 = 6 \;\Rightarrow\; I_2 = 2\ \text{A},\quad I_1 = 5\ \text{A} \]

The voltage across the source (from mesh-1 KVL) gives its delivered power:

\[ V_{cs} = 9 - 1\cdot I_1 = 4\ \text{V},\qquad P_{cs} = V_{cs} \times 3 = 12\ \text{W} \]

Answer\(I_1 = 5\ \text{A},\ I_2 = 2\ \text{A},\ P_{cs} = 12\ \text{W}\)