Mesh Analysis | Network Analysis

Tutorial 3 — Mesh Analysis

Worked problems on the mesh-current method: writing KVL equations per mesh, handling current sources with the supermesh technique, and incorporating dependent sources. Each problem is followed by a complete solution with the final result highlighted.

Network Analysis · Electric Circuit Analysis · 10 solved problems

Demonstrative VideoWalkthrough

Problem 1Two-Mesh KVL

Using mesh analysis, find \(V_0\).

Solution

KVL in the first mesh:

\[ \begin{aligned} 12 &= 4(I_1 - I_2) + 8(I_1 - I_2) + 6 I_1\\ 12 &= 18 I_1 - 12 I_2 \;\Rightarrow\; 2 = 3 I_1 - 2 I_2 \quad (1) \end{aligned} \]

KVL in the second mesh:

\[ \begin{aligned} 24 &= 4 I_2 + 8(I_2 - I_1) + 4(I_2 - I_1)\\ 24 &= 16 I_2 - 12 I_1 \;\Rightarrow\; 6 = 4 I_2 - 3 I_1 \quad (2) \end{aligned} \]

Solving (1) and (2):

\[ I_1 = \frac{10}{3},\quad I_2 = 4,\qquad V_0 = 4(I_2 - I_1) = 4\!\left(4 - \frac{10}{3}\right) = \frac{8}{3} = 2.667\ \text{V} \]

Answer\(V_0 = \tfrac{8}{3} \approx 2.667\ \text{V}\)

Problem 2Supermesh

Apply mesh analysis to the circuit to obtain \(I_0\).

Solution

Meshes 1 and 2 form a supermesh. The supermesh KVL is:

\[ 6 i_1 + 4 i_2 - 5 i_3 + 180 = 0 \]

KVL for loop 3, and the current-source constraint between meshes 1 and 2:

\[ -i_1 - 4 i_2 + 7 i_3 + 90 = 0,\qquad i_2 = 45 + i_1 \]

Solving the three relations:

\[ i_1 = -46\ \text{A},\quad i_3 = -20\ \text{A},\qquad I_0 = i_1 - i_3 = -26\ \text{A} \]

Answer\(I_0 = -26\ \text{A}\)

Problem 3Source Power

Find the power delivered by the source in the circuit shown.

Solution

Applying KVL around the outer loop, with the two 2 Ω branches carrying \((3+i)\) and \((2+i)\):

\[ \begin{aligned} (3+i)\,2 + (2+i)\,2 &= 10\\ 6 + 2i + 4 + 2i &= 10 \;\Rightarrow\; 4i = 0 \;\Rightarrow\; i = 0 \end{aligned} \]

The current through the 10 V source is zero, so the power it supplies is:

\[ P = V i = 10 \times 0 = 0\ \text{W} \]

Check: the worked quantity above is the power of the 10 V (voltage) source. If the question refers to a separate current source, evaluate \(P = V_{cs}\,I_{cs}\) using its terminal voltage from the figure — but with \(i = 0\) the net result is still 0 W.

Answer\(i = 0,\quad P = 0\ \text{W}\)

Problem 4Supermesh + Dependent Source

Find \(i_x\) and \(v_x\) in the circuit.

Solution

For the supermesh, with the dependent term \(4 i_x\) and \(i_x = i_1\):

\[ -50 + 10 i_1 + 5 i_2 + 4 i_x = 0 \;\Rightarrow\; 14 i_1 + 5 i_2 = 50 \quad (1) \]

At node \(A\), with the 3 A branch and \(v_x = 2(i_1 - i_2)\), the mesh constraint is:

\[ i_2 = i_1 + 2 \quad (2) \]

Solving (1) and (2):

\[ i_1 = 2.105\ \text{A},\quad i_2 = 4.105\ \text{A},\qquad v_x = 2(i_1 - i_2) = -4\ \text{V},\quad i_x = i_1 = 2.105\ \text{A} \]

Check: the node relation \(i_1 + 3 + v_x = i_2\) with \(v_x = 2(i_1 - i_2)\) reduces algebraically to \(i_2 = i_1 + 1\); the worked answer uses \(i_2 = i_1 + 2\) from the figure. Re-confirm the 3 A branch orientation against the circuit.

Answer\(i_x = 2.105\ \text{A},\quad v_x = -4\ \text{V}\)

Problem 53-Mesh + Dependent Source

Use mesh analysis to determine the power absorbed by the dependent voltage source.

Solution

The three mesh equations are:

\[ \begin{aligned} 70 I_1 - 35 I_2 - 15 I_3 + 20(I_1 - I_2) &= 10 + 16\\ -35 I_1 + 64 I_2 - 18 I_3 &= 7 - 16 - 20\\ -15 I_1 - 18 I_2 + 46 I_3 - 20(I_1 - I_2) &= 20 - 14 \end{aligned} \]

which simplify to the system:

\[ \begin{bmatrix} 90 & -55 & -15 \\ -35 & 64 & -18 \\ -35 & 2 & 46 \end{bmatrix} \begin{bmatrix} I_1 \\ I_2 \\ I_3 \end{bmatrix} = \begin{bmatrix} 26 \\ -29 \\ 6 \end{bmatrix} \]

The solution is \(I_1 = 0.148\ \text{A},\ I_2 = -0.3\ \text{A},\ I_3 = 0.256\ \text{A}\). The power absorbed by the dependent source (its voltage times the current into its positive terminal):

\[ P = 20(I_1 - I_2)(I_1 - I_3) = 20(0.148 + 0.3)(0.148 - 0.256) = -0.968\ \text{W} \]

Sign: a negative "absorbed" power means the dependent source actually delivers about 0.968 W to the circuit.

Answer\(P \approx -0.968\ \text{W}\) (absorbed)

Problem 6Supermesh + Current Sources

Find the currents \(I_1\), \(I_2\) and \(I_3\).

Solution

Mesh 1 is fixed by a current source, \(I_1 = 4\ \text{A}\). KVL for mesh 4:

\[ 12(I_4 - I_1) + 4(I_4 - I_3) - 8 = 0 \]

Meshes 2 and 3 form a supermesh, with the node-\(c\) constraint:

\[ 6(I_2 - I_1) + 10 + 2 I_3 + 4(I_3 - I_4) = 0 \;\Rightarrow\; -3 I_1 + 3 I_2 + 3 I_3 - 2 I_4 = -5,\qquad I_2 = I_3 + 1 \]

Solving the system:

\[ I_1 = 4\ \text{A},\quad I_2 = 3\ \text{A},\quad I_3 = 2\ \text{A},\quad I_4 = 4\ \text{A} \]

The branch currents then follow from KCL:

\[ i_1 = I_2 - I_1 = -1\ \text{A},\quad i_2 = I_1 - I_4 = 0\ \text{A},\quad i_3 = I_4 - I_3 = 2\ \text{A} \]

Answer\(I_1 = 4,\ I_2 = 3,\ I_3 = 2,\ I_4 = 4\ \text{A}\)

Problem 7Three-Mesh KVL

Find the values of the mesh currents.

Solution

Writing KVL for each of the three loops:

\[ \begin{aligned} \text{Loop 1:}\quad & 120 + 40 i_1 - 10 i_2 = 0 \;\Rightarrow\; 4 i_1 - i_2 = -12\\ \text{Loop 2:}\quad & 50 i_2 - 10 i_1 - 10 i_3 = 0 \;\Rightarrow\; -i_1 + 5 i_2 - i_3 = 0\\ \text{Loop 3:}\quad & -120 - 10 i_2 + 40 i_3 = 0 \;\Rightarrow\; -i_2 + 4 i_3 = 12 \end{aligned} \]

Solving the system:

\[ i_1 = -3\ \text{A},\quad i_2 = 0,\quad i_3 = 3\ \text{A} \]

Answer\(i_1 = -3\ \text{A},\ i_2 = 0,\ i_3 = 3\ \text{A}\)

Problem 8Mesh + Current Source

Find \(v_0\) using mesh analysis.

Solution

Mesh 1 is set by a current source, \(i_1 = 5\ \text{A}\). KVL for the other two meshes:

\[ \begin{aligned} \text{Loop 2:}\quad & -40 + 7 i_2 - 2 i_1 - 4 i_3 = 0 \;\Rightarrow\; 7 i_2 - 4 i_3 = 50\\ \text{Loop 3:}\quad & -20 + 12 i_3 - 4 i_2 = 0 \;\Rightarrow\; -i_2 + 3 i_3 = 5 \end{aligned} \]

Solving loops 2 and 3:

\[ i_2 = 10\ \text{A},\quad i_3 = 5\ \text{A},\qquad v_0 = 4(i_2 - i_3) = 4(10 - 5) = 20\ \text{V} \]

Answer\(v_0 = 20\ \text{V}\)

Additional Practice Problems

Two foundational problems with self-contained schematics: a basic two-mesh circuit solved directly by KVL, and a supermesh formed by a current source shared between two meshes.

Problem 9Basic Two-Mesh

For the two-mesh circuit below, take clockwise mesh currents \(I_1\) (left) and \(I_2\) (right). Find both mesh currents and the power dissipated in the shared 4 Ω resistor.

Solution

KVL around each mesh (the shared 4 Ω carries \(I_1 - I_2\)):

\[ \begin{aligned} \text{Mesh 1:}\quad & 8 = 2 I_1 + 4(I_1 - I_2) = 6 I_1 - 4 I_2\\ \text{Mesh 2:}\quad & 2 = 6 I_2 + 4(I_2 - I_1) = -4 I_1 + 10 I_2 \end{aligned} \]

Solving the pair:

\[ I_1 = 2\ \text{A},\qquad I_2 = 1\ \text{A} \]

Current in the shared resistor and its power:

\[ I_{4\Omega} = I_1 - I_2 = 1\ \text{A},\qquad P_{4\Omega} = I^2 R = 1^2 \times 4 = 4\ \text{W} \]

Answer\(I_1 = 2\ \text{A},\ I_2 = 1\ \text{A},\ P_{4\Omega} = 4\ \text{W}\)

Problem 10Supermesh (Current Source)

A 2 A current source sits in the branch shared by the two meshes, so that \(I_1 - I_2 = 2\,\text{A}\). Using the supermesh method, find \(I_1\) and \(I_2\), and the power delivered by the current source.

Solution

Because a current source lies between the meshes, combine them into a supermesh. KVL around the outer loop (10 V source, 2 Ω in mesh 1, 4 Ω in mesh 2):

\[ 10 = 2 I_1 + 4 I_2 \]

The current-source constraint supplies the second equation:

\[ I_1 - I_2 = 2 \;\Rightarrow\; I_1 = I_2 + 2 \]

Substituting:

\[ 10 = 2(I_2 + 2) + 4 I_2 = 6 I_2 + 4 \;\Rightarrow\; I_2 = 1\ \text{A},\quad I_1 = 3\ \text{A} \]

The voltage across the current source (from mesh 1 KVL) and its delivered power:

\[ V_{cs} = 10 - 2 I_1 = 10 - 6 = 4\ \text{V},\qquad P_{cs} = V_{cs}\,(2) = 8\ \text{W} \]

Answer\(I_1 = 3\ \text{A},\ I_2 = 1\ \text{A},\ P_{cs} = 8\ \text{W}\)