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Chapter 3 — Methods of Analysis

Adapted from C. K. Alexander and M. N. O. Sadiku, Fundamentals of Electric Circuits.

Learning Objectives

After studying this chapter, you should be able to:

3.1 Introduction

Having understood the fundamental laws of circuit theory (Ohm’s law and Kirchhoff’s laws), we are now prepared to apply them to develop two powerful general techniques for circuit analysis:

These two methods are general—they work on any planar linear circuit—and they produce the minimum number of simultaneous equations needed to solve the circuit. After presenting each method, we show how to solve the resulting equations using Cramer’s rule.

3.2 Nodal Analysis

Nodal analysis determines the node voltages in a circuit. Once the node voltages are known, every element’s voltage and current can be found by Ohm’s law.

Steps to determine node voltages

  1. Select a node as the reference (ground) node. Assign voltages \(v_1, v_2, \ldots, v_{n-1}\) to the remaining \(n-1\) nodes. Each node voltage is referenced with respect to the reference node.
  2. Apply KCL to each of the non-reference nodes. Use Ohm’s law to express the branch currents in terms of the node voltages.
  3. Solve the resulting simultaneous equations to obtain the unknown node voltages.

The reference node is commonly called the ground since it is assumed to have zero potential. We use the earth-ground symbol throughout.

Figure 3.1 Ground symbols.

Key principle: current flows from higher to lower potential

Since a resistor is passive, by the passive sign convention current must always flow from a higher potential to a lower potential. For a resistor \(R\) connected between node \(a\) (voltage \(V_a\)) and node \(b\) (voltage \(V_b\)), the current from \(a\) to \(b\) is

\[ \boxed{i_{a\to b} = \frac{V_a - V_b}{R}} \tag{3.4} \]

This is the central idea that lets us express every branch current in terms of node voltages.

Illustration

Consider a circuit with a reference node 0 and two non-reference nodes (voltages \(v_1\), \(v_2\)), resistors \(R_1, R_2, R_3\), and sources. Applying KCL at node 1 and node 2 and using Eq. (3.4) yields two simultaneous equations in \(v_1\) and \(v_2\). In conductance form (\(G_k = 1/R_k\)):

\[ G_{11} v_1 - G_{12} v_2 = I_1 \] \[ -G_{21} v_1 + G_{22} v_2 = I_2 \tag{3.7-3.8} \]

where \(G_{11}\) is the sum of conductances connected to node 1, \(G_{12} = G_{21}\) is the (negative of the) conductance shared between nodes 1 and 2, and \(I_1, I_2\) are the source currents injected into nodes 1 and 2. In matrix form:

\[ \begin{bmatrix} G_{11} & -G_{12} \\ -G_{21} & G_{22} \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \begin{bmatrix} I_1 \\ I_2 \end{bmatrix} \tag{3.9} \]

These are solved by substitution, elimination, Cramer’s rule, or matrix inversion (see Appendix A).

Example 3.1 — Find the node voltages in the circuit of Fig. 3.2. In Fig. 3.2, a 2-node circuit with node 1 connected via \(5\ \Omega\) to a \(10\)-V source-referenced node and via \(2\ \Omega\) to node 2, and node 2 connected via \(1\ \Omega\) to ground; a \(6\)-A source feeds node 1.

Figure 3.2 Circuit for Example 3.1.

Solution:

Take the bottom node as reference (ground, \(0\) V). Node voltages are \(v_1\) and \(v_2\). The node connected to the +terminal of the \(10\)-V source is at \(10\) V.

At node 1, KCL (sum of currents leaving = 0), using \(i = (v_1 - v_{\text{other}})/R\):

\[ \frac{v_1 - 10}{5} + \frac{v_1 - v_2}{2} + \frac{v_1 - 0}{\infty} = -6 \;\;(\text{the 6-A source pushes current into node 1}). \]

Simplifying (currents leaving + injected):

\[ \frac{v_1 - 10}{5} + \frac{v_1 - v_2}{2} = 6. \]

At node 2:

\[ \frac{v_2 - v_1}{2} + \frac{v_2 - 0}{1} = 0. \]

Clearing denominators and solving the two equations:

From the node-2 equation: \(v_2 - v_1 + 2v_2 = 0 \Rightarrow v_1 = 3v_2\).

Substitute into the node-1 equation (multiply by 10): \(2(v_1 - 10) + 5(v_1 - v_2) = 60 \Rightarrow 2v_1 - 20 + 5v_1 - 5v_2 = 60 \Rightarrow 7v_1 - 5v_2 = 80\).

With \(v_1 = 3v_2\): \(7(3v_2) - 5v_2 = 80 \Rightarrow 16v_2 = 80 \Rightarrow v_2 = 5\) V, and \(v_1 = 15\) V.

\[ \boxed{v_1 = 15\ \text{V},\quad v_2 = 5\ \text{V}} \]

(The exact numbers depend on the specific resistances and source value in your figure; the procedure—KCL at each node with \(i=(V_a-V_b)/R\), then solve—is general.) [equation reconstructed]

Practice Problem 3.1 — Find \(v_1\) and \(v_2\) in the circuit of Fig. 3.4 using nodal analysis. [values reconstructed]

Figure 3.4 Circuit for Practice Problem 3.1.

Solution:

We need to determine the node voltages \(v_1\) and \(v_2\) in Fig. 3.4 using Kirchhoff’s Current Law (KCL). From the circuit:
• A \(2\,\Omega\) resistor connects node 1 to ground.
• A \(7\,\Omega\) resistor connects node 2 to ground.
• A \(6\,\Omega\) resistor connects nodes 1 and 2.
• The \(3\,\text{A}\) current source injects current into node 1.
• The \(12\,\text{A}\) current source directs current from node 2 toward ground.
The bottom conductor is the reference node (ground), so the unknown node voltages are \(v_1\) and \(v_2\). The bottom conductor is the reference node (ground), so the unknown node voltages are \(v_1\) and \(v_2\). Step 1: Apply KCL at node 1 The current leaving node 1 through the \(2\,\Omega\) resistor is \[ i_1=\frac{v_1}{2} \] The current leaving node 1 toward node 2 through the \(6\,\Omega\) resistor is \[ i_{12}=\frac{v_1-v_2}{6} \] Since the \(3\,\text{A}\) source injects current into node 1, KCL gives \[ \frac{v_1}{2}+\frac{v_1-v_2}{6}=3 \] Multiplying both sides by 6: \[ 3v_1+v_1-v_2=18 \] Therefore, \[ \boxed{4v_1-v_2=18} \tag{1} \] Step 2: Apply KCL at node 2 The current leaving node 2 through the \(7\,\Omega\) resistor is \[ i_2=\frac{v_2}{7} \] The current leaving node 2 toward node 1 through the \(6\,\Omega\) resistor is \[ i_{21}=\frac{v_2-v_1}{6} \] The \(12\,\text{A}\) current source also directs current out of node 2. Taking currents leaving node 2 as positive, KCL gives \[ \frac{v_2-v_1}{6}+\frac{v_2}{7}+12=0 \] Multiplying both sides by 42: \[ 7(v_2-v_1)+6v_2+504=0 \] Expanding and rearranging: \[ -7v_1+13v_2=-504 \] Thus, \[ \boxed{-7v_1+13v_2=-504} \tag{2} \] Step 3: Solve the simultaneous equations From Equation (1), \[ 4v_1-v_2=18 \] Rearranging, \[ \boxed{v_2=4v_1-18} \tag{3} \] Substitute Equation (3) into Equation (2): \[ -7v_1+13(4v_1-18)=-504 \] Expanding: \[ -7v_1+52v_1-234=-504 \] \[ 45v_1=-270 \] Therefore, \[ \boxed{v_1=-6\text{ V}} \] Substitute \(v_1=-6\,\text{V}\) into Equation (3): \[ v_2=4(-6)-18 \] \[ v_2=-24-18 \] Therefore, \[ \boxed{v_2=-42\text{ V}} \] Final answer Node 1 voltage \[v_1=-6\text{ V}\] Node 2 voltage \[v_2=-42\text{ V}\]

3.3 Nodal Analysis with Voltage Sources

When a voltage source is present, two cases arise:

A supernode is formed by enclosing a (dependent or independent) voltage source connected between two non-reference nodes and any elements connected in parallel with it.

Figure 3.7 A supernode formed by nodes 2 and 3.

Properties of a supernode

We analyze a circuit with supernodes using the same three steps as ordinary nodal analysis, treating each supernode specially: apply KCL to the supernode as a whole (currents in = currents out across its boundary), and add the KVL constraint from the internal voltage source.

Example 3.3 — For the circuit of Fig. 3.9, find the node voltages. [values reconstructed: a \(2\)-V source between nodes 1 and 2 with a \(10\ \Omega\) resistor across the supernode, node 1 fed by a current source and connected to ground via a resistor, node 2 connected to ground via a resistor]

Figure 3.9 Circuit for Example 3.3.

Solution:

Find the Node Voltages Using Supernode Analysis The circuit contains a voltage source between nodes 1 and 2, so we use the supernode method to determine \(v_1\) and \(v_2\).
Given:
• \(R_1=2\Omega\), connected from node 1 to ground.
• \(R_2=4\Omega\), connected from node 2 to ground.
• A \(2 \text{V}\) voltage source between nodes 1 and 2, with its positive terminal at node 2.
• A \(2\,\text{A}\) current source directed upward into node 1.
• A \(7\,\text{A}\) current source directed downward from node 2 to ground.
• A \(10\Omega\) resistor connected across the supernode, between nodes 1 and 2.
Step 1: Write the voltage constraint The voltage source has its negative terminal at node 1 and positive terminal at node 2. Therefore, \[v_2-v_1=2\] Hence, \[\boxed{v_2=v_1+2} \tag{1}\] Step 2: Apply KCL to the supernode Enclose nodes 1 and 2, including the voltage source and the \(10\Omega\) resistor between them, within one supernode. The \(10\Omega\) resistor is internal to the supernode. Its current does not appear in the supernode KCL equation because it flows between two nodes inside the supernode. Taking currents leaving the supernode as positive:
• Current through the \(2\Omega\) resistor: \(\dfrac{v_1}{2}\).
• Current through the \(4\Omega\) resistor: \(\dfrac{v_2}{4}\).
• The \(7\text{A}\) source carries current out of the supernode.
• The \(2\,\text{A}\) source carries current into the supernode, so it is negative in this convention.
Applying KCL, \[ \frac{v_1}{2}+\frac{v_2}{4}+7-2=0\] Simplifying, \[ \frac{v_1}{2}+\frac{v_2}{4}=-5\] Multiply both sides by 4: \[\boxed{2v_1+v_2=-20} \tag{2}\] Step 3: Solve the simultaneous equations From Equation (1), \[v_2=v_1+2\] Substitute this into Equation (2): \[2v_1+(v_1+2)=-20\] \[3v_1+2=-20\] \[3v_1=-22\] Therefore, \[\boxed{v_1=-\frac{22}{3}=-7.33\text{ V}}\] Now substitute \(v_1=-\dfrac{22}{3}\) into Equation (1): \[v_2=-\frac{22}{3}+2\] \[v_2=-\frac{22}{3}+\frac{6}{3}\] Thus, \[\boxed{v_2=-\frac{16}{3}=-5.33\text{ V}}\] Final answer Node 1 voltage \[v_1=-7.33\text{ V}\] Node 2 voltage \[v_2=-5.33\text{ V}\] Step 4: Verify the answer Check the voltage-source constraint: \[v_2-v_1=-5.33-(-7.33)=2.00\text{ V}\] This agrees with the \(2\text{V}\) source polarity. Check the supernode KCL equation: \[\frac{-7.33}{2}+\frac{-5.33}{4}+7-2 \approx 0\] The small discrepancy is due to rounding; the exact values give zero. Therefore, the node voltages are \(v_1=-7.33\,\text{V}\) and \(v_2=-5.33\,\text{V}\).
Practice Problem 3.3 — Find \(v\) and \(i\) in the circuit of Fig. 3.11.

Figure 3.11 Circuit for Practice Problem 3.3.

Solution:

The 14-V source is connected directly between the reference node and node 1, so that node voltage is known at once (Case 1):

\[ v_1 = 14\ \text{V}. \]

The 6-V source between nodes 2 and 3 (its + terminal at node 3) makes nodes 2 and 3 a supernode, with the constraint

\[ v_3 - v_2 = 6 \quad\Longrightarrow\quad v_3 = v_2 + 6. \tag{1} \]

Apply KCL at the supernode as a whole: the 4-\(\Omega\) resistor carries current from node 2 toward node 1, while the 3-\(\Omega\), 2-\(\Omega\), and 6-\(\Omega\) resistors carry current from the supernode down to ground:

\[ \frac{v_2 - v_1}{4} + \frac{v_2}{3} + \frac{v_3}{2} + \frac{v_3}{6} = 0. \]

Substituting \(v_1 = 14\) and \(v_3 = v_2 + 6\) and multiplying through by 12:

\[ 3(v_2 - 14) + 4v_2 + 6(v_2 + 6) + 2(v_2 + 6) = 0, \]

\[ 3v_2 - 42 + 4v_2 + 6v_2 + 36 + 2v_2 + 12 = 0 \;\Longrightarrow\; 15v_2 = -6 \;\Longrightarrow\; v_2 = -0.4\ \text{V}. \]

The required voltage \(v\) is the voltage across the 3-\(\Omega\) resistor (+ at node 2), so

\[ v = v_2 = -0.4\ \text{V} = -400\ \text{mV}. \]

Then \(v_3 = v_2 + 6 = 5.6\) V, and the required current \(i\) is the current downward through the 2-\(\Omega\) resistor:

\[ i = \frac{v_3}{2} = \frac{5.6}{2} = 2.8\ \text{A}. \]

\[ \boxed{v = -400\ \text{mV}, \qquad i = 2.8\ \text{A}} \]

Example 3.4 — Find the node voltages in the circuit of Fig. 3.12.

Figure 3.12 Circuit for Example 3.4.

Solution:

Nodes 1 and 2 form a supernode (the 20-V source between them); so do nodes 3 and 4 (the dependent source \(3v_x\) between them). We apply KCL at each supernode. At supernode 1-2,

\[ i_3 + 10 = i_1 + i_2. \]

Expressing this in terms of the node voltages,

\[ \frac{v_3 - v_2}{6} + 10 = \frac{v_1 - v_4}{3} + \frac{v_1}{2} \quad\Longrightarrow\quad 5v_1 + v_2 - v_3 - 2v_4 = 60. \tag{3.4.1} \]

At supernode 3-4,

\[ i_1 = i_3 + i_4 + i_5 \quad\Longrightarrow\quad \frac{v_1 - v_4}{3} = \frac{v_3 - v_2}{6} + \frac{v_4}{1} + \frac{v_3}{4}, \]

\[ 4v_1 + 2v_2 - 5v_3 - 16v_4 = 0. \tag{3.4.2} \]

We now apply KVL around the loops containing the voltage sources. For loop 1 (the 20-V source between nodes 1 and 2, + at node 1),

\[ -v_1 + 20 + v_2 = 0 \quad\Longrightarrow\quad v_1 - v_2 = 20 \quad\Longrightarrow\quad v_2 = v_1 - 20. \tag{3.4.3} \]

For loop 2 (the dependent source between nodes 3 and 4, + at node 3), with \(v_x = v_1 - v_4\) (the voltage across the 3-\(\Omega\) resistor),

\[ -v_3 + 3v_x + v_4 = 0 \quad\Longrightarrow\quad v_3 - v_4 = 3(v_1 - v_4) \quad\Longrightarrow\quad 3v_1 - v_3 - 2v_4 = 0. \tag{3.4.4} \]

Substituting \(v_2 = v_1 - 20\) from Eq. (3.4.3) into Eqs. (3.4.1) and (3.4.2) gives, respectively,

\[ 6v_1 - v_3 - 2v_4 = 80, \tag{3.4.6} \]

and

\[ 6v_1 - 5v_3 - 16v_4 = 40. \tag{3.4.7} \]

Equations (3.4.4), (3.4.6), and (3.4.7) can be cast in matrix form as

\[ \begin{bmatrix} 3 & -1 & -2 \\ 6 & -1 & -2 \\ 6 & -5 & -16 \end{bmatrix} \begin{bmatrix} v_1 \\ v_3 \\ v_4 \end{bmatrix} = \begin{bmatrix} 0 \\ 80 \\ 40 \end{bmatrix}. \]

Using Cramer’s rule gives

\[ \Delta = -18, \qquad \Delta_1 = -480, \qquad \Delta_3 = -3120, \qquad \Delta_4 = 840. \]

Thus we arrive at the node voltages

\[ v_1 = \frac{\Delta_1}{\Delta} = \frac{-480}{-18} = 26.67\ \text{V}, \qquad v_3 = \frac{\Delta_3}{\Delta} = \frac{-3120}{-18} = 173.33\ \text{V}, \]

\[ v_4 = \frac{\Delta_4}{\Delta} = \frac{840}{-18} = -46.67\ \text{V}, \qquad v_2 = v_1 - 20 = 6.667\ \text{V}. \]

\[ \boxed{v_1 = 26.67\ \text{V},\quad v_2 = 6.667\ \text{V},\quad v_3 = 173.33\ \text{V},\quad v_4 = -46.67\ \text{V}} \]

Practice Problem 3.4 — Find \(v_1\) and \(v_2\) in the circuit of Fig. 3.14 using nodal analysis. Answer: \(v_1 = 7.608\) V, \(v_2 = -17.39\) V, \(v_3 = 1.6305\) V.

Figure 3.14 Circuit for Practice Problem 3.4.

Solution:

The 25-V source between nodes 1 and 2 (+ at node 1) and the dependent source \(5i\) between nodes 2 and 3 (+ at node 3) both connect non-reference nodes, so nodes 1, 2, and 3 together form a single (three-node) supernode.

The two sources inside the supernode provide the constraint equations

\[ v_1 - v_2 = 25, \qquad v_3 - v_2 = 5i, \]

where the controlling current \(i\) is the current downward through the 2-\(\Omega\) resistor, \(i = v_1/2\). Note that the 6-\(\Omega\) resistor is connected between nodes 1 and 3, both of which lie inside the supernode, so its current is internal and does not appear in the supernode equation. Applying KCL at the supernode,

\[ \frac{v_1}{2} + \frac{v_2}{4} + \frac{v_3}{3} = 0. \]

From the constraints,

\[ v_1 = v_2 + 25, \qquad v_3 = v_2 + 5i = v_2 + \frac{5v_1}{2} = v_2 + 2.5(v_2 + 25) = 3.5v_2 + 62.5. \]

Substituting into the supernode equation and multiplying through by 12:

\[ 6(v_2 + 25) + 3v_2 + 4(3.5v_2 + 62.5) = 0 \]

\[ 6v_2 + 150 + 3v_2 + 14v_2 + 250 = 0 \;\Longrightarrow\; 23v_2 = -400 \;\Longrightarrow\; v_2 = -17.39\ \text{V}. \]

Then

\[ v_1 = v_2 + 25 = 7.608\ \text{V}, \qquad v_3 = 3.5v_2 + 62.5 = 1.6305\ \text{V}, \qquad i = \frac{v_1}{2} = 3.804\ \text{A}. \]

\[ \boxed{v_1 = 7.608\ \text{V},\quad v_2 = -17.39\ \text{V},\quad v_3 = 1.6305\ \text{V}} \]

3.4 Mesh Analysis

Using mesh currents instead of element currents as circuit variables reduces the number of simultaneous equations. Recall that a loop is a closed path with no node passed more than once. A mesh is a loop that does not contain any other loop within it.

Figure 3.17 Meshes in a planar circuit.

In Fig. 3.17, paths abefa and bcdeb are meshes, but path abcdefa is not a mesh (it contains the other two). The current through a mesh is the mesh current. In mesh analysis we apply KVL to find the mesh currents.

Nodal analysis applies KCL to find unknown voltages; mesh analysis applies KVL to find unknown currents. For a circuit with \(n\) nodes, \(b\) branches, and \(\ell\) independent loops/meshes, \(\ell = b - n + 1\); hence \(\ell\) simultaneous equations are required for mesh analysis.

Steps to determine mesh currents

  1. Assign mesh currents \(i_1, i_2, \ldots, i_n\) to the \(n\) meshes (conventionally clockwise).
  2. Apply KVL to each of the \(n\) meshes. Use Ohm’s law to express the voltages in terms of mesh currents.
  3. Solve the resulting \(n\) simultaneous equations to get the mesh currents.

Inspection shortcut

For mesh 1, the coefficient of \(i_1\) is the sum of resistances in mesh 1, and the coefficient of \(i_2\) is the negative of the resistance common to meshes 1 and 2. This pattern lets you write the mesh equations by inspection. Branch currents are algebraic sums of the mesh currents: a branch on the boundary of mesh 1 alone carries \(i_1\); a shared branch carries \(i_1 - i_2\) (depending on directions).

Example 3.5 — For the circuit of Fig. 3.18, find the branch currents \(i_1\) and \(i_2\) using mesh analysis. [values reconstructed: two-mesh circuit, mesh 1 has a \(12\)-V source and resistors \(2\ \Omega, 4\ \Omega\); mesh 2 has a \(-6\)-V source (or \(6\) V opposite polarity) and resistors \(4\ \Omega, 6\ \Omega\); \(4\ \Omega\) is shared]

Figure 3.18 Circuit for Example 3.5.

Solution:

Apply KVL to mesh 1 (clockwise, sum of voltage drops = sum of rises):

\[ 2 i_1 + 4(i_1 - i_2) = 12 \;\Rightarrow\; 6 i_1 - 4 i_2 = 12. \tag{3.5.1} \]

For mesh 2:

\[ 4(i_2 - i_1) + 6 i_2 = -6 \;\Rightarrow\; -4 i_1 + 10 i_2 = -6. \tag{3.5.2} \]

Method 1 (substitution): From Eq. (3.5.1), \(i_1 = \frac{12 + 4 i_2}{6} = 2 + \frac{2}{3}i_2\). Substitute into Eq. (3.5.2):

\[ -4\!\left(2 + \tfrac{2}{3}i_2\right) + 10 i_2 = -6 \;\Rightarrow\; -8 - \tfrac{8}{3}i_2 + 10 i_2 = -6 \;\Rightarrow\; \tfrac{22}{3}i_2 = 2 \;\Rightarrow\; i_2 = \tfrac{6}{22} = 0.273\ \text{A}. \]

Then \(i_1 = 2 + \tfrac{2}{3}(0.273) = 2.182\) A.

Method 2 (Cramer’s rule): Cast in matrix form:

\[ \begin{bmatrix} 6 & -4 \\ -4 & 10 \end{bmatrix} \begin{bmatrix} i_1 \\ i_2 \end{bmatrix} = \begin{bmatrix} 12 \\ -6 \end{bmatrix}. \]

Determinant \(\Delta = (6)(10) - (-4)(-4) = 60 - 16 = 44\).

\[ i_1 = \frac{\begin{vmatrix}12 & -4 \\ -6 & 10\end{vmatrix}}{\Delta} = \frac{120 - 24}{44} = \frac{96}{44} = 2.182\ \text{A}, \]

\[ i_2 = \frac{\begin{vmatrix}6 & 12 \\ -4 & -6\end{vmatrix}}{\Delta} = \frac{-36 + 48}{44} = \frac{12}{44} = 0.273\ \text{A}. \]

\[ \boxed{i_1 \approx 2.18\ \text{A},\quad i_2 \approx 0.273\ \text{A}} \]

(Exact values depend on the specific source polarities/resistances in your figure; the mesh-current method and Cramer’s rule procedure are general.) [equation reconstructed]

Practice Problem 3.5 — Calculate the mesh currents \(i_1\) and \(i_2\) of the circuit of Fig. 3.19. Answer: \(i_1 = 2.5\) A, \(i_2 = 0\) A.

Figure 3.19 Circuit for Practice Problem 3.5.

Solution:

Assign clockwise mesh currents \(i_1\) (left mesh, containing the 45-V source, the 2-\(\Omega\) resistor on top and the 4-\(\Omega\) resistor on the bottom) and \(i_2\) (right mesh, containing the 30-V source, the 9-\(\Omega\) resistor on top and the 3-\(\Omega\) resistor on the bottom). The 12-\(\Omega\) resistor is shared by the two meshes.

For mesh 1, write the KVL equation by inspection: the coefficient of \(i_1\) is the sum of the resistances in mesh 1, the coefficient of \(i_2\) is minus the shared resistance, and the 45-V source enters positively (it is met from − to + traveling clockwise around the mesh):

\[ (2 + 4 + 12)i_1 - 12i_2 = 45 \quad\Longrightarrow\quad 18i_1 - 12i_2 = 45. \tag{1} \]

For mesh 2, the 30-V source is met from + to − traveling clockwise, so it enters with a negative sign:

\[ -12i_1 + (3 + 9 + 12)i_2 = -30 \quad\Longrightarrow\quad -12i_1 + 24i_2 = -30. \tag{2} \]

Solving, multiply Eq. (1) by 2 and add Eq. (2):

\[ 36i_1 - 24i_2 = 90, \qquad 24i_1 = 60 \;\Longrightarrow\; i_1 = 2.5\ \text{A}. \]

From Eq. (2),

\[ 24i_2 = 12i_1 - 30 = 30 - 30 = 0 \;\Longrightarrow\; i_2 = 0\ \text{A}. \]

\[ \boxed{i_1 = 2.5\ \text{A}, \qquad i_2 = 0\ \text{A}} \]

(Check: the 45-V source delivers \(45 \times 2.5 = 112.5\) W, and the resistors absorb \(2.5^2(2 + 4 + 12) = 112.5\) W. ✓)

Example 3.6 — Use mesh analysis to find the current \(I_o\) in the circuit of Fig. 3.20.

Figure 3.20 Circuit for Example 3.6.

Solution:

The circuit has three meshes; assign clockwise mesh currents \(i_1\), \(i_2\), and \(i_3\). The desired current \(I_o\) is the branch current through the 10-\(\Omega\) resistor at node A, so

\[ I_o = i_1 - i_2. \]

We apply KVL to the three meshes in turn. For mesh 1 (the 24-V source is met from − to + going clockwise),

\[ -24 + 10(i_1 - i_2) + 12(i_1 - i_3) = 0 \]

or

\[ 11i_1 - 5i_2 - 6i_3 = 12. \tag{3.6.1} \]

For mesh 2 (which contains the 24-\(\Omega\) resistor and shares the 10-\(\Omega\) and 4-\(\Omega\) resistors with meshes 1 and 3),

\[ 24i_2 + 4(i_2 - i_3) + 10(i_2 - i_1) = 0 \]

or

\[ -5i_1 + 19i_2 - 2i_3 = 0. \tag{3.6.2} \]

For mesh 3, the dependent voltage source \(4I_o\) appears directly in the KVL equation:

\[ 4I_o + 12(i_3 - i_1) + 4(i_3 - i_2) = 0. \]

But at node A, \(I_o = i_1 - i_2\), so that

\[ 4(i_1 - i_2) + 12(i_3 - i_1) + 4(i_3 - i_2) = 0 \]

or

\[ -i_1 - i_2 + 2i_3 = 0. \tag{3.6.3} \]

In matrix form, Eqs. (3.6.1) to (3.6.3) become

\[ \begin{bmatrix} 11 & -5 & -6 \\ -5 & 19 & -2 \\ -1 & -1 & 2 \end{bmatrix} \begin{bmatrix} i_1 \\ i_2 \\ i_3 \end{bmatrix} = \begin{bmatrix} 12 \\ 0 \\ 0 \end{bmatrix}. \]

We calculate the mesh currents using Cramer’s rule:

\[ \Delta = 192, \qquad \Delta_1 = 432, \qquad \Delta_2 = 144, \qquad \Delta_3 = 288, \]

\[ i_1 = \frac{\Delta_1}{\Delta} = \frac{432}{192} = 2.25\ \text{A}, \qquad i_2 = \frac{\Delta_2}{\Delta} = \frac{144}{192} = 0.75\ \text{A}, \qquad i_3 = \frac{\Delta_3}{\Delta} = \frac{288}{192} = 1.5\ \text{A}. \]

Thus,

\[ I_o = i_1 - i_2 = 2.25 - 0.75 = 1.5\ \text{A}. \]

\[ \boxed{i_1 = 2.25\ \text{A},\quad i_2 = 0.75\ \text{A},\quad i_3 = 1.5\ \text{A},\quad I_o = 1.5\ \text{A}} \]

3.5 Mesh Analysis with Current Sources (Supermesh)

When a current source lies on the boundary between two meshes, the current through it is known but its voltage is unknown—so a direct KVL around either mesh would introduce that unknown voltage. The solution is the supermesh: we form a larger loop that excludes the current source (and any elements in parallel with it), apply KVL to the supermesh, and add a constraint equation relating the two mesh currents to the current source.

Figure 3.21 A supermesh excludes the current source.

Steps for a supermesh

  1. Assign mesh currents as usual.
  2. Form a supermesh by excluding the current source (and elements in parallel with it) and apply KVL around the perimeter of the combined region.
  3. Add the constraint: the difference of the two mesh currents equals the current source value (with sign depending on direction).
  4. Solve the resulting equations.

Example 3.7 — For the circuit of Fig. 3.23, find the mesh currents.

Figure 3.23 Circuit for Example 3.7.

Solution:

The 6-A current source (in series with the 2-\(\Omega\) resistor) lies on the boundary shared by mesh 1 (which contains the 20-V source and the 6-\(\Omega\) resistor) and mesh 2 (which contains the 10-\(\Omega\) and 4-\(\Omega\) resistors). Since the voltage across the current source is not known in advance, we exclude the current source and the 2-\(\Omega\) resistor in series with it, and apply KVL to the resulting supermesh:

\[ -20 + 6i_1 + 10i_2 + 4i_2 = 0 \]

or

\[ 6i_1 + 14i_2 = 20. \tag{3.7.1} \]

The current source provides the constraint equation. Applying KCL at node 0 (the bottom of the shared branch), where the 6-A arrow is directed upward,

\[ i_2 - i_1 = 6 \]

or

\[ i_2 = i_1 + 6. \tag{3.7.2} \]

Substituting Eq. (3.7.2) into Eq. (3.7.1) gives

\[ 6i_1 + 14(i_1 + 6) = 20 \;\Longrightarrow\; 20i_1 = -64 \;\Longrightarrow\; i_1 = -3.2\ \text{A}, \]

and hence

\[ i_2 = i_1 + 6 = 2.8\ \text{A}. \]

\[ \boxed{i_1 = -3.2\ \text{A}, \qquad i_2 = 2.8\ \text{A}} \]

Note that the 2-\(\Omega\) resistor in series with the current source does not appear in Eq. (3.7.1), because it is excluded from the supermesh along with the source.

Practice Problem 3.7 — Use mesh analysis to determine the mesh currents \(i_1\), \(i_2\), and \(i_3\) in the circuit of Fig. 3.25.

Figure 3.25 Circuit for Practice Problem 3.7.

Solution:

The 4-A current source (in series with the 1-\(\Omega\) resistor) lies on the boundary shared by mesh 1 (which contains the 8-V source and shares the 2-\(\Omega\) resistor with mesh 3) and mesh 2 (which contains the 8-\(\Omega\) resistor and shares the 4-\(\Omega\) resistor with mesh 3). Exclude the current source and the 1-\(\Omega\) resistor in series with it, and apply KVL to the larger supermesh made up of meshes 1 and 2:

\[ -8 + 2(i_1 - i_3) + 4(i_2 - i_3) + 8i_2 = 0, \]

where \(2(i_1 - i_3)\) is the voltage across the 2-\(\Omega\) resistor shared with mesh 3 and \(4(i_2 - i_3)\) is the voltage across the 4-\(\Omega\) resistor shared with mesh 3. Simplifying,

\[ 2i_1 + 12i_2 - 6i_3 = 8. \tag{1} \]

The current source provides the constraint: its arrow points downward through the shared branch, and with both mesh currents taken clockwise,

\[ i_1 - i_2 = 4. \tag{2} \]

Mesh 3 contains no current source, so its KVL equation is written directly (by inspection: the sum of its resistances is \(2 + 4 + 2 = 8\ \Omega\), sharing the 2-\(\Omega\) resistor with mesh 1 and the 4-\(\Omega\) resistor with mesh 2):

\[ -2i_1 - 4i_2 + 8i_3 = 0. \tag{3} \]

From Eq. (2), \(i_1 = i_2 + 4\). Substituting into Eq. (1):

\[ 2(i_2 + 4) + 12i_2 - 6i_3 = 8 \;\Longrightarrow\; 14i_2 = 6i_3 \;\Longrightarrow\; i_3 = \frac{7}{3}i_2. \]

Substituting both relations into Eq. (3):

\[ -2(i_2 + 4) - 4i_2 + 8\left(\frac{7}{3}i_2\right) = 0 \;\Longrightarrow\; \frac{38}{3}i_2 = 8 \;\Longrightarrow\; i_2 = \frac{12}{19} = 0.6316\ \text{A}. \]

Hence

\[ i_1 = i_2 + 4 = 4.632\ \text{A}, \qquad i_3 = \frac{7}{3}i_2 = \frac{28}{19} = 1.4736\ \text{A}. \]

\[ \boxed{i_1 = 4.632\ \text{A}, \qquad i_2 = 631.6\ \text{mA}, \qquad i_3 = 1.4736\ \text{A}} \]

3.6 Cramer’s Rule

When nodal or mesh analysis produces \(n\) simultaneous linear equations, cast them in matrix form

\[ \mathbf{A}\mathbf{x} = \mathbf{b}, \]

where \(\mathbf{A}\) is the \(n\times n\) coefficient matrix, \(\mathbf{x}\) the unknown vector, and \(\mathbf{b}\) the source vector. Cramer’s rule gives each unknown as

\[ \boxed{x_k = \frac{\Delta_k}{\Delta}} \]

where \(\Delta = \det(\mathbf{A})\) and \(\Delta_k\) is the determinant of \(\mathbf{A}\) with its \(k\)-th column replaced by \(\mathbf{b}\). (A full treatment with \(2\times2\) and \(3\times3\) formulas is in Appendix A.) The same equations can be solved by matrix inversion, \(\mathbf{x} = \mathbf{A}^{-1}\mathbf{b}\), or with software such as MATLAB.

Example 3.8 — Solve the nodal equations of Example 3.4 by Cramer’s rule. [values reconstructed]

Solution:

Cast the three (reduced) equations in matrix form, compute \(\Delta\) and each \(\Delta_k\), then \(v_k = \Delta_k/\Delta\). (Reproduce with the specific values in your figure.) [equation reconstructed]

Practice Problem 3.8 — Solve the mesh equations of Practice Problem 3.6 by Cramer’s rule.

Solution:

Form the coefficient matrix from the mesh equations, compute \(\Delta = \det(\mathbf{A})\) and each \(\Delta_k\), and obtain \(i_k = \Delta_k/\Delta\). (Apply with the specific values in your figure.) [equation reconstructed]

3.7 Homework Problems

Homework Set 3.1 (from EEE141-HW-3)

North South University, Department of ECE, EEE141: Electrical Circuits I, HW-3, Fall 2026. Instructor: Prof. Miftahur Rahman, Ph.D.

Problem 3.1. Obtain the node voltages \(v_1, v_2,\) and \(v_3\) in the circuit of Fig. 1.

Figure for Problem 3.1

Hint: Pick the bottom node as reference, assign \(v_1, v_2, v_3\), apply KCL at each non-reference node using \(i = (V_a - V_b)/R\), and solve the three simultaneous equations (Cramer’s rule helps).

Problem 3.2. Find the node voltages for the circuit in Fig. 2.

Figure for Problem 3.2

Hint: Watch for any voltage source between two non-reference nodes—if so, form a supernode: write KCL on the supernode boundary and add the KVL constraint \(v_2 - v_1 = V_s\), then solve.

Problem 3.3. Use mesh analysis to obtain \(i_1, i_2,\) and \(i_3\) in the circuit in Fig. 3.

Figure for Problem 3.3

Hint: Assign clockwise mesh currents, write KVL in each mesh by inspection (diagonal = sum of mesh resistances, off-diagonal = −shared resistance), and solve the three equations.

Problem 3.4. Apply mesh analysis to the circuit in Fig. 4 and obtain \(v_o\).

Figure for Problem 3.4

Hint: If a current source lies on a mesh boundary, form a supermesh (exclude the current source), write the supermesh KVL plus the current-source constraint, solve for mesh currents, then \(v_o = i_{\text{branch}} R\).

Key Points