Nodal Analysis | Network Analysis

Tutorial 5 — Nodal Analysis

Worked problems on the node-voltage method: writing KCL at each non-reference node, handling voltage sources with the supernode technique, and incorporating dependent sources. Each problem is followed by a complete solution with the final result highlighted.

Network Analysis · Electric Circuit Analysis · 12 solved problems

Demonstrative VideoWalkthrough

Problem 1Nodal + Short Circuit

If \(R = 0\) and the current \(i_R = 10\,\text{A}\), find the value of \(E\).

Solution

Since \(R = 0\) is a short circuit, the top node sits at the reference potential (0 V). Applying nodal analysis at node 1:

\[ \frac{V_1 - E}{4} + \frac{V_1}{4} + \frac{V_1}{2} = 0 \;\Rightarrow\; E = 4 V_1 \]

At node 2:

\[ \frac{V_2 - 0}{2} + \frac{V_2}{2} = 4 \;\Rightarrow\; V_2 = 4\ \text{V} \]

KCL for \(i_R\), then substitute \(i_R = 10\,\text{A}\) and \(V_2 = 4\,\text{V}\):

\[ i_R = \frac{0 - V_1}{2} + \frac{0 - V_2}{2} \;\Rightarrow\; 10 = -\frac{V_1}{2} - 2 \;\Rightarrow\; V_1 = -24\ \text{V} \]

Finally:

\[ E = 4 V_1 = 4(-24) = -96\ \text{V} \]

Answer\(E = -96\ \text{V}\)

Problem 2Basic Nodal

Using nodal analysis, find the currents in the resistors.

Solution

KCL at the two nodes:

\[ \begin{aligned} \text{Node 1:}\quad & -5 + \frac{V_1}{1} + \frac{V_1 - V_2}{1} = 0 \;\Rightarrow\; 2 V_1 - V_2 = 5\\ \text{Node 2:}\quad & -2 + \frac{V_2}{2} + \frac{V_2 - V_1}{1} = 0 \;\Rightarrow\; -V_1 + \tfrac{3}{2} V_2 = 2 \end{aligned} \]

Solving the pair:

\[ V_1 = 4.75\ \text{V},\qquad V_2 = 4.5\ \text{V} \]

The resistor currents then follow from Ohm's law \(I = V/R\):

\[ I_{1\Omega} = \frac{V_1}{1} = 4.75\ \text{A},\quad I_{\text{mid}} = \frac{V_1 - V_2}{1} = 0.25\ \text{A},\quad I_{2\Omega} = \frac{V_2}{2} = 2.25\ \text{A} \]

Correction: the previously listed values \(V_1 = 0.25\,\text{V},\ V_2 = -4.5\,\text{V}\) satisfy the node-1 equation but not node-2; the consistent solution is \(V_1 = 4.75\,\text{V},\ V_2 = 4.5\,\text{V}\).

Answer\(V_1 = 4.75\ \text{V},\ V_2 = 4.5\ \text{V}\)

Problem 3Nodal + Dependent Source

Find the values of \(V_1\) and \(V_2\).

Solution

Applying KCL at the node, with the dependent current source \(2I\) and \(I = V/4\):

\[ -5 + \frac{V}{4} + \frac{V}{4} + 2I = 0,\qquad I = \frac{V}{4} \]

Substituting gives \(-5 + \tfrac{V}{4} + \tfrac{V}{4} + \tfrac{V}{2} = 0 \Rightarrow V = 5\,\text{V}\), so:

\[ V_1 = 5\ \text{V},\qquad V_2 = 25\ \text{V} \]

Answer\(V_1 = 5\ \text{V},\ V_2 = 25\ \text{V}\)

Problem 4Supernode

Find the node voltages.

Solution

Combining nodes \(v_1\) and \(v_2\) into a supernode, the KCL equation is:

\[ -2 + \frac{v_1}{2} + \frac{v_2}{4} + 7 = 0 \]

The supernode voltage-constraint equation (from the source between the nodes):

\[ v_2 - v_1 = 2 \]

Solving:

\[ v_1 = -7.333\ \text{V},\qquad v_2 = -5.333\ \text{V} \]

Note: the 10 Ω resistor makes no difference here because it is connected directly across the supernode.

Answer\(v_1 = -7.333\ \text{V},\ v_2 = -5.333\ \text{V}\)

Problem 5Dependent Source Power

Find the power of the dependent current source.

Solution

Apply KCL at node A:

\[ \frac{V_A - 20}{5} + \frac{V_A}{5} = 4 \;\Rightarrow\; V_A = 20\ \text{V} \]

The dependent source carries 4 A with \(V_A = 20\,\text{V}\) across it, so the power delivered is:

\[ P = V_A \times 4 = 20 \times 4 = 80\ \text{W} \]

Answer\(P = 80\ \text{W}\)

Problem 6Basic Nodal

Find the current through the 5 Ω resistor.

Solution

KCL at the two nodes:

\[ \begin{aligned} \text{Node 1:}\quad & -2 + \frac{V_1}{2} + \frac{V_1 - V_2}{5} = 0\\ \text{Node 2:}\quad & -\frac{V_1 - V_2}{5} + 4 + \frac{V_2}{4} = 0 \end{aligned} \]

Solving:

\[ V_1 = 0.36\ \text{V},\qquad V_2 = -8.73\ \text{V} \]

The current through the 5 Ω resistor (the quantity asked):

\[ I_{5\Omega} = \frac{V_1 - V_2}{5} = \frac{0.36 - (-8.73)}{5} = \frac{9.09}{5} \approx 1.82\ \text{A} \]

Answer\(I_{5\Omega} \approx 1.82\ \text{A}\)

Problem 7Nodal + VCCS

Find \(V_1\).

Solution

Applying KCL at the node and collecting the conductance terms (the \(0.2\,V_1\) dependent term, the 10 Ω branch and the 1 Ω branch) against the net source current:

\[ 0.2 V_1 + \frac{V_1}{10} + \frac{V_1}{1} = (10 + 2)\ \text{mA} \;\Rightarrow\; 1.3\,V_1 = 12\ \text{mA} \;\Rightarrow\; V_1 = 9.23\ \text{mV} \]

Answer\(V_1 = 9.23\ \text{mV}\)

Problem 8Single-Node Nodal

Find the current through the 5 mΩ resistor.

Solution

Applying KCL at the single node (three milliohm branches to ground, fed by three current sources):

\[ \frac{V}{4\,\text{m}\Omega} + \frac{V}{3\,\text{m}\Omega} + \frac{V}{5\,\text{m}\Omega} = (4 + 10 + 5)\ \text{mA} = 19\ \text{mA} \]

The total conductance is \(250 + 333.3 + 200 = 783.3\,\text{S}\), so:

\[ V = \frac{19\ \text{mA}}{783.3\ \text{S}} = 24.26\ \mu\text{V} \]

Current through the 5 mΩ resistor:

\[ I = \frac{V}{5\,\text{m}\Omega} = \frac{24.26 \times 10^{-6}}{5 \times 10^{-3}} = 4.85\ \text{mA} \]

Correction: the previously listed \(V = 11.54\,\mu\text{V},\ I = 2.31\,\text{mA}\) do not satisfy the node equation; with the three sources totalling 19 mA across 783.3 S the node voltage is \(24.26\,\mu\text{V}\) and \(I_{5m\Omega} = 4.85\,\text{mA}\).

Answer\(V \approx 24.26\ \mu\text{V},\quad I_{5m\Omega} \approx 4.85\ \text{mA}\)

Problem 9Dependent Source Power

Find the power supplied by the dependent source.

Solution

KCL at the two nodes, with the controlling relation \(i_1 = v_1/2\):

\[ \begin{aligned} \text{Node } v_1:\quad & -15 + \frac{v_1 - v_2}{1} + \frac{v_1}{2} = 0\\ \text{Node } v_2:\quad & -3 i_1 + \frac{v_2}{3} + \frac{v_2 - v_1}{1} = 0,\qquad i_1 = \frac{v_1}{2} \end{aligned} \]

Solving:

\[ v_1 = -40\ \text{V},\quad v_2 = -75\ \text{V},\quad i_1 = -20\ \text{A} \]

The dependent source \(3 i_1\) with \(v_2\) across it supplies:

\[ P = (3 i_1)(v_2) = (3 \times -20)(-75) = 4500\ \text{W} = 4.5\ \text{kW} \]

Answer\(P = 4.5\ \text{kW}\)

Problem 10Supernode (3 Nodes)

Determine the value of \(v_1\).

Solution

KCL at node \(v_1\):

\[ -(-8) - (-3) + \frac{v_1 - v_2}{3} + \frac{v_1 - v_3}{4} = 0 \]

Supernode KCL across \(v_2\)–\(v_3\), and the constraint equation:

\[ -3 + \frac{v_2 - v_1}{3} + \frac{v_2}{1} + \frac{v_3}{5} - 25 + \frac{v_3 - v_1}{4} = 0,\qquad v_3 - v_2 = 22 \]

Solving the system:

\[ v_1 = 1.071\ \text{V},\quad v_2 = 10.5\ \text{V},\quad v_3 = 32.5\ \text{V} \]

Answer\(v_1 = 1.071\ \text{V}\)

Additional Practice Problems

Two foundational problems with self-contained schematics: a basic two-node circuit solved directly by KCL, and a supernode formed by a voltage source connecting two non-reference nodes.

Problem 11Basic Two-Node

For the circuit below, a 4 A source feeds node 1 and a 1 A source feeds node 2. All three resistors are 2 Ω. Find the node voltages \(V_1, V_2\) and the current through the middle resistor.

Solution

KCL at each node (current in = current out):

\[ \begin{aligned} \text{Node 1:}\quad & 4 = \frac{V_1}{2} + \frac{V_1 - V_2}{2}\\ \text{Node 2:}\quad & 1 = \frac{V_2}{2} + \frac{V_2 - V_1}{2} \end{aligned} \]

These simplify to \(2V_1 - V_2 = 8\) and \(-V_1 + 2V_2 = 2\). Solving:

\[ V_1 = 6\ \text{V},\qquad V_2 = 4\ \text{V} \]

Current through the middle 2 Ω resistor:

\[ I_{\text{mid}} = \frac{V_1 - V_2}{2} = \frac{6 - 4}{2} = 1\ \text{A} \]

Answer\(V_1 = 6\ \text{V},\ V_2 = 4\ \text{V},\ I_{\text{mid}} = 1\ \text{A}\)

Problem 12Supernode (Worked)

A 6 V source connects nodes 1 and 2 (so \(V_2 - V_1 = 6\)), forming a supernode fed by a 5 A source; each node has a 2 Ω resistor to ground. Find \(V_1\) and \(V_2\).

Solution

Because a voltage source sits between two non-reference nodes, enclose both in a supernode and write one KCL around it (the 5 A source feeds the two ground resistors):

\[ 5 = \frac{V_1}{2} + \frac{V_2}{2} \]

The source between the nodes supplies the constraint:

\[ V_2 - V_1 = 6 \;\Rightarrow\; V_2 = V_1 + 6 \]

Substituting:

\[ 5 = \frac{V_1}{2} + \frac{V_1 + 6}{2} = V_1 + 3 \;\Rightarrow\; V_1 = 2\ \text{V},\quad V_2 = 8\ \text{V} \]

Answer\(V_1 = 2\ \text{V},\quad V_2 = 8\ \text{V}\)