Super-node Analysis | Network Analysis

Tutorial 6 — Super-node Analysis

When a voltage source connects two non-reference nodes, the two nodes are enclosed in a supernode: one KCL equation is written around the whole supernode, and a constraint equation relates the two node voltages through the source. These worked problems cover supernodes with independent and dependent sources, and the technique for recovering the current through a voltage source after the node voltages are known.

Network Analysis · Electric Circuit Analysis · 8 solved problems

Demonstrative VideoWalkthrough

Problem 1Dependent-Source Supernode

Using nodal analysis, determine \(i_\phi\).

Solution

Node 1 is fixed by the 50 V source, so \(v_1 = 50\,\text{V}\). The dependent source \(10\,i_\phi\) sits between nodes 2 and 3, so those two form a supernode. Writing one KCL around the supernode:

\[ \frac{v_2 - 50}{5} + \frac{v_2}{50} + \frac{v_3}{100} = 4 \]

The source provides the constraint, with the controlling current \(i_\phi = \dfrac{v_2 - v_1}{5}\):

\[ v_3 = v_2 + 10\,i_\phi = v_2 + 2(v_2 - 50) = 3 v_2 - 100 \]

Substituting into the supernode equation:

\[ \frac{v_2 - 50}{5} + \frac{v_2}{50} + \frac{3 v_2 - 100}{100} = 4 \;\Rightarrow\; 25 v_2 = 1500 \;\Rightarrow\; v_2 = 60\ \text{V},\quad v_3 = 80\ \text{V} \]

Therefore:

\[ i_\phi = \frac{v_2 - v_1}{5} = \frac{60 - 50}{5} = 2\ \text{A} \]

Answer\(i_\phi = 2\ \text{A}\)

Problem 2Dependent-Source Nodal

Using nodal analysis, determine \(i_\phi\).

Solution

KCL at the two nodes, with the dependent source \(8 i_\phi\) and the controlling current \(i_\phi = (V_1 - V_2)/5\):

\[ \begin{aligned} \frac{V_1 - 20}{2} + \frac{V_1}{20} + \frac{V_1 - V_2}{5} &= 0\\ \frac{V_2 - V_1}{5} + \frac{V_2}{10} + \frac{V_2 - 8 i_\phi}{2} &= 0\\ i_\phi &= \frac{V_1 - V_2}{5} \end{aligned} \]

The first equation gives \(15 V_1 - 4 V_2 = 200\); substituting \(i_\phi\) into the second reduces it to \(V_2 = 0.625\,V_1\). Solving the pair:

\[ V_1 = 16\ \text{V},\qquad V_2 = 10\ \text{V} \]

Hence:

\[ i_\phi = \frac{V_1 - V_2}{5} = \frac{16 - 10}{5} = 1.2\ \text{A} \]

Answer\(i_\phi = 1.2\ \text{A}\)

Problem 3BJT Bias (Symbolic)

Determine \(i_B\).

Solution

Node a is at \(V_{CC}\). The source \(V_0\) sits between nodes b and c, giving the supernode constraint \(v_b - v_c = V_0\). Applying KCL at node b:

\[ \frac{v_b}{R_2} + \frac{v_b - V_{CC}}{R_1} + i_B = 0 \]

At node c the emitter current is \((1+\beta) i_B\) (base current plus the dependent collector current \(\beta i_B\)), so:

\[ (1+\beta) i_B = \frac{v_c}{R_E} \;\Rightarrow\; v_c = (1+\beta) i_B R_E,\qquad v_b = v_c + V_0 \]

Substituting \(v_b\) into the node-b equation and solving for \(i_B\) gives the closed form:

\[ i_B = \frac{\dfrac{V_{CC} - V_0}{R_1} - \dfrac{V_0}{R_2}}{\,1 + (1+\beta)\,R_E\left(\dfrac{1}{R_1} + \dfrac{1}{R_2}\right)} \]

No component values are given, so the result is symbolic. The same answer follows whether node b and node c are treated separately (as above) or merged into a single supernode b–c.

Answer\(i_B = \dfrac{(V_{CC}-V_0)/R_1 - V_0/R_2}{1 + (1+\beta)R_E(1/R_1 + 1/R_2)}\)

Problem 4Current Through a Source

Determine the current flowing through the 6 V source.

Solution

The 6 V source between nodes 1 and 2 makes them a supernode. The supernode KCL (the 0.1 A source feeds node 2) and the constraint are:

\[ \frac{V_1 - 24}{150} + \frac{V_1}{100} + \frac{V_2}{75} - 0.1 = 0,\qquad V_1 + 6 - V_2 = 0 \]

Substituting \(V_2 = V_1 + 6\) gives \(5 V_1 + 4 V_2 = 78\), so:

\[ V_1 = 6\ \text{V},\qquad V_2 = 12\ \text{V} \]

The current through the source cannot be read from the supernode equation; it comes from KCL at one node alone. At node 1 (current \(I\) leaving toward node 2 through the source):

\[ \frac{V_1 - 24}{150} + \frac{V_1}{100} + I = 0 \;\Rightarrow\; -0.12 + 0.06 + I = 0 \;\Rightarrow\; I = 0.06\ \text{A} \]

Check at node 2: \(I + 0.1 = V_2/75 \Rightarrow 0.06 + 0.1 = 0.16 = 12/75\). ✓

Answer\(I_{6\text{V}} = 0.06\ \text{A} = 60\ \text{mA}\)

Problem 5Two Voltage Sources

Determine \(V_a\) and \(V_b\). All resistors are 1 kΩ; \(V_{S1} = 4\,\text{V}\), \(V_{S2} = 8\,\text{V}\), \(I_S = 4\,\text{mA}\).

Solution

Writing KCL at nodes a and b (the branch through \(R_4\) carries \(V_{S2} + v_a - v_b\)):

\[ \begin{aligned} \text{a:}\quad & \frac{V_{S1} - v_a}{R_2} = \frac{v_a}{R_1} + \frac{V_{S2} + v_a - v_b}{R_4}\\ \text{b:}\quad & \frac{V_{S1} - v_b}{R_3} + \frac{V_{S2} + v_a - v_b}{R_4} + I_S = 0 \end{aligned} \]

With all resistances equal to 1 kΩ and the given source values, these reduce to:

\[ 3 v_a - v_b = 12,\qquad -v_a + 2 v_b = 16 \]

Solving:

\[ V_a = 8\ \text{V},\qquad V_b = 12\ \text{V} \]

Answer\(V_a = 8\ \text{V},\quad V_b = 12\ \text{V}\)

Problem 6Supernode + Internal Resistor

Determine \(V_1\) and \(V_2\).

Solution

The 4 V source ties nodes 1 and 2 into a supernode. The 2 Ω resistor bridges the same two nodes, so it is internal to the supernode and drops out of the KCL. The supernode equation (5 A in, 2 A out) is:

\[ 5 = \frac{v_1}{6} + \frac{v_2}{3} + 2 \;\Rightarrow\; v_1 + 2 v_2 = 18 \]

The source constraint:

\[ v_1 + 4 - v_2 = 0 \;\Rightarrow\; v_2 = v_1 + 4 \]

Solving the two together:

\[ v_1 + 2(v_1 + 4) = 18 \;\Rightarrow\; 3 v_1 = 10 \;\Rightarrow\; v_1 = 3.33\ \text{V},\quad v_2 = 7.33\ \text{V} \]

Answer\(V_1 = 3.33\ \text{V},\quad V_2 = 7.33\ \text{V}\)

Additional Practice Problems

Two fully-worked supernode examples with self-contained schematics: recovering the current through the connecting source, and a supernode whose branch terminates on another voltage source.

Problem 7Find Current Through Source

A 6 A source feeds node A, which connects to node B through a 6 V source (\(V_A - V_B = 6\)). Node A has a 2 Ω resistor to ground and node B a 4 Ω resistor to ground. Find \(V_A\), \(V_B\) and the current through the 6 V source.

Solution

The 6 V source between A and B forms a supernode fed by the 6 A source:

\[ 6 = \frac{V_A}{2} + \frac{V_B}{4},\qquad V_A - V_B = 6 \]

Substituting \(V_A = V_B + 6\):

\[ 6 = \frac{V_B + 6}{2} + \frac{V_B}{4} = \frac{3}{4}V_B + 3 \;\Rightarrow\; V_B = 4\ \text{V},\quad V_A = 10\ \text{V} \]

The current through the 6 V source follows from KCL at node B alone (only the source branch and the 4 Ω meet there):

\[ I_{6\text{V}} = \frac{V_B}{4} = \frac{4}{4} = 1\ \text{A} \]

Answer\(V_A = 10\ \text{V},\ V_B = 4\ \text{V},\ I_{6\text{V}} = 1\ \text{A}\)

Problem 8Branch to a Voltage Source

A 2 A source feeds node A (2 Ω to ground). A 4 V source connects A to B (\(V_B - V_A = 4\)), and node B reaches a fixed 10 V source through a 2 Ω resistor. Find \(V_A\) and \(V_B\).

Solution

Nodes A and B form a supernode (4 V source between them). The branch from B carries current \((V_B - 10)/2\) toward the 10 V source, so the supernode KCL and constraint are:

\[ 2 = \frac{V_A}{2} + \frac{V_B - 10}{2},\qquad V_B = V_A + 4 \]

Substituting:

\[ 2 = \frac{V_A}{2} + \frac{V_A - 6}{2} = V_A - 3 \;\Rightarrow\; V_A = 5\ \text{V},\quad V_B = 9\ \text{V} \]

Answer\(V_A = 5\ \text{V},\quad V_B = 9\ \text{V}\)