Superposition Theorem | Network Theorems

Tutorial 9 — Superposition Theorem

In any linear network with two or more independent sources, the response (current or voltage) in any branch equals the algebraic sum of the responses caused by each independent source acting alone, with all other independent sources deactivated — voltage sources replaced by short circuits and current sources by open circuits. Dependent sources are never deactivated; they remain active in every sub-circuit. These problems cover networks with independent sources, dependent sources, and source transformation.

Network Theorems · Electric Circuit Analysis · 8 solved problems

Problem 1Independent V & I Sources

Find the current \(I\) using the superposition theorem.

Circuit for Problem 1: 1 A source, 1 ohm and 2 ohm resistors, 1 ohm and 3 ohm shunts, and a 1 V source

Solution

Step 1Voltage source acting alone (open-circuit the 1 A current source). The 1 V source drives current through the 2 Ω and the left 1 Ω in series, so the current in the measured branch is:

\[ I' = -\frac{1}{3}\ \text{A} \]

Sub-circuit with voltage source acting alone for Problem 1

Step 2Current source acting alone (short-circuit the 1 V source). The 1 A splits between the 1 Ω shunt and the 2 Ω branch; by the current-divider rule:

\[ I'' = 1 \times \frac{1}{1+2} = \frac{1}{3}\ \text{A} \]

Sub-circuit with current source acting alone for Problem 1

By superposition, add the two contributions:

\[ I = I' + I'' = -\frac{1}{3} + \frac{1}{3} = 0 \]

Answer\(I = 0\ \text{A}\)

Problem 2Dependent Source (2Vₓ)

Use the superposition theorem to determine the current \(I\). The circuit contains a 10 V source, a 5 Ω resistor, a dependent voltage source \(2V_x\), a 2 Ω resistor (across which \(V_x\) is defined), and a 2 A current source.

Circuit for Problem 2 with a dependent voltage source 2Vx

Solution

The dependent source \(2V_x\) stays active in both sub-circuits. With \(V_x\) across the 2 Ω resistor.

Step 110 V source active, 2 A source opened. The network is a single series loop; let the loop current be \(I_1\), so \(V_{x1} = 2I_1\). Applying KVL:

\[ \begin{aligned} -10 + 5I_1 - 2V_{x1} + 2I_1 &= 0 \\ -10 + 5I_1 - 2(2I_1) + 2I_1 &= 0 \\ 3I_1 = 10 \;\;\Rightarrow\;\; I_1 &= \frac{10}{3}\ \text{A} \end{aligned} \]

Step 22 A source active, 10 V source shorted. Apply KCL at the top node (voltage \(V_{x2}\)):

\[ \frac{V_{x2} - 2V_{x2}}{5} + \frac{V_{x2}}{2} - 2 = 0 \;\;\Rightarrow\;\; 3V_{x2} = 20 \;\;\Rightarrow\;\; V_{x2} = \frac{20}{3}\ \text{V} \]

The current through the 5 Ω resistor:

\[ I_2 = \frac{0 - (V_{x2} - 2V_{x2})}{5} = \frac{V_{x2}}{5} = \frac{20/3}{5} = \frac{4}{3}\ \text{A} \]

By superposition, both contributions act in the reference direction of \(I\):

\[ I = I_1 + I_2 = \frac{10}{3} + \frac{4}{3} = \frac{14}{3} \approx 4.67\ \text{A} \]

Answer\(I = \dfrac{14}{3} \approx 4.67\ \text{A}\)

Problem 3Dependent Source (2i₀)

Find \(i_0\) and \(i\) using superposition. The circuit has a 6 V source, a 1 Ω resistor, a 1 A current source, a 5 Ω resistor, and a dependent voltage source \(2i_0\).

Circuit for Problem 3 with a dependent voltage source 2 i-zero

Solution

Step 16 V source acting alone (open the 1 A source). The dependent source \(2i_0\) remains. The network is a single loop, so \(i_0' = i'\) and by KVL:

\[ 6i' + 2i' = 6 \;\;\Rightarrow\;\; i' = i_0' = 0.75\ \text{A} \]

Sub-circuit with the 6 V source acting alone for Problem 3

Step 21 A source acting alone (short the 6 V source). Applying KCL and the supermesh KVL (with the dependent source still active):

\[ \begin{aligned} \text{KCL:} \quad & 1 = i'' - i_0'' \;\;\Rightarrow\;\; i'' = 1 + i_0'' \\ \text{Supermesh KVL:} \quad & 1\cdot i_0'' + 5i'' + 2i_0'' = 0 \;\;\Rightarrow\;\; 3i_0'' + 5i'' = 0 \end{aligned} \]

Substituting \(i'' = 1 + i_0''\):

\[ 3i_0'' + 5\bigl(1 + i_0''\bigr) = 0 \;\;\Rightarrow\;\; 8i_0'' = -5 \;\;\Rightarrow\;\; i_0'' = -\frac{5}{8} = -0.625\ \text{A} \] \[ \therefore\; i'' = 1 + i_0'' = 1 - 0.625 = 0.375\ \text{A} \]

Sub-circuit with the 1 A source acting alone for Problem 3

Correction. The original slide solved \(8i_0'' = -5\) as \(i_0'' = -\tfrac{5}{4} = -1.25\) A (an arithmetic slip: \(3i_0'' + 5i_0'' = 8i_0''\), not \(4i_0''\)). The correct value is \(i_0'' = -\tfrac{5}{8} = -0.625\) A, giving \(i'' = 0.375\) A. A direct nodal analysis of the full circuit confirms the corrected totals below.

By superposition:

\[ \begin{aligned} i &= i' + i'' = 0.75 + 0.375 = 1.125\ \text{A} \\ i_0 &= i_0' + i_0'' = 0.75 - 0.625 = 0.125\ \text{A} \end{aligned} \]

Answer\(i = 1.125\ \text{A},\quad i_0 = 0.125\ \text{A}\)

Problem 4Three Independent Sources

Find \(V_{AB}\) using the superposition theorem. The network contains a 6 Ω resistor, a 4 V source, a 2 A source, a 4 Ω resistor, a 2 Ω resistor, and a 2 V source.

Circuit for Problem 4 with terminals A and B

Solution

Step 12 V source acting alone (4 V shorted, 2 A opened). The loop resistance is \(6 + (4+2) = 12\ \Omega\):

\[ I' = \frac{2}{12} = \frac{1}{6}\ \text{A}, \qquad V_{AB}' = I' \times 6 = 1\ \text{V} \]

Sub-circuit with the 2 V source acting alone for Problem 4

Step 24 V source acting alone (2 V shorted, 2 A opened):

\[ I'' = \frac{4}{12} = \frac{1}{3}\ \text{A}, \qquad V_{AB}'' = I'' \times 6 = -2\ \text{V} \]

Sub-circuit with the 4 V source acting alone for Problem 4

Step 32 A source acting alone (both V sources shorted). Convert the 2 A – 4 Ω combination to an equivalent \(2 \times 4 = 8\ \text{V}\) source in series with 4 Ω:

\[ I''' = \frac{8}{12} = \frac{2}{3}\ \text{A}, \qquad V_{AB}''' = -I''' \times 6 = -4\ \text{V} \]

Sub-circuit with the 2 A source transformed to an 8 V source for Problem 4

By superposition:

\[ V_{AB} = V_{AB}' + V_{AB}'' + V_{AB}''' = 1 - 2 - 4 = -5\ \text{V} \]

Answer\(V_{AB} = -5\ \text{V}\)

Problem 5Dependent Source (4i)

Determine \(V_1\) (across the 3 Ω resistor) using superposition. The circuit has a dependent voltage source \(4i\), an 8 A source, a 2 Ω resistor, a 10 V source, and a 2 A source.

Circuit for Problem 5 with a dependent voltage source 4 i

Solution

The dependent source \(4i\) remains active in all three sub-circuits, and \(V_1 = 3i\).

Step 18 A source acting alone (2 A opened, 10 V shorted):

\[ \begin{aligned} \text{Supermesh KVL:}\;\; & 3i' + 2i_1 - 4i' = 0 \;\Rightarrow\; i_1 = \tfrac{1}{2}i' \\ \text{KCL at node } x:\;\; & i_1 = 8 + i' \;\Rightarrow\; \tfrac{1}{2}i' = 8 + i' \;\Rightarrow\; i' = -16\ \text{A} \\ & \therefore\; V_1' = 3i' = -48\ \text{V} \end{aligned} \]

Sub-circuit with the 8 A source acting alone for Problem 5

Step 22 A source acting alone (8 A opened, 10 V shorted):

\[ \begin{aligned} \text{KVL:}\;\; & 3(i_2 + 2) + 2i_2 - 4i'' = 0 \;\Rightarrow\; 5i_2 + 6 - 4i'' = 0,\quad i'' = i_2 + 2 \\ & 5i_2 + 6 - 4(i_2 + 2) = 0 \;\Rightarrow\; i_2 = 2\ \text{A} \\ & \therefore\; i'' = i_2 + 2 = 4\ \text{A},\qquad V_1'' = 3i'' = 12\ \text{V} \end{aligned} \]

Sub-circuit with the 2 A source acting alone for Problem 5

Step 310 V source acting alone (8 A and 2 A opened):

\[ 3i''' - 10 + 2i''' - 4i''' = 0 \;\Rightarrow\; i''' = 10\ \text{A}, \qquad V_1''' = 10 \times 3 = 30\ \text{V} \]

Sub-circuit with the 10 V source acting alone for Problem 5

By superposition:

\[ V_1 = V_1' + V_1'' + V_1''' = -48 + 12 + 30 = -6\ \text{V} \]

Answer\(V_1 = -6\ \text{V}\)

Problem 6Dependent Source (5Vₓ)

Find the current \(I\) using superposition. The circuit has a 4 V source, a 3 Ω resistor, a 1 Ω resistor, a 2 A source, a 2 Ω resistor (across which \(V_x\) is defined), and a dependent voltage source \(5V_x\).

Circuit for Problem 6 with a dependent voltage source 5 Vx

Solution

Step 14 V source acting alone (open the 2 A source). With the 1 Ω branch carrying the 2 A source now open, \(V_x' = -2I'\), and KVL around the loop gives:

\[ \begin{aligned} -4 + 3I' + 5V_x' - V_x' &= 0 \;\Rightarrow\; 3I' + 4V_x' = 4 \\ 3I' + 4(-2I') &= 4 \;\Rightarrow\; -5I' = 4 \;\Rightarrow\; I' = -0.8\ \text{A} \end{aligned} \]

Sub-circuit with the 4 V source acting alone for Problem 6

Step 22 A source acting alone (short the 4 V source). Apply KCL at the top node (voltage \(V_x''\)):

\[ 2 = \frac{V_x''}{2} + \frac{V_x'' - 5V_x''}{3} \;\Rightarrow\; V_x'' = -\frac{12}{5} = -2.4\ \text{V} \] \[ \therefore\; I'' = \frac{V_x'' - 5V_x''}{3} = 3.2\ \text{A} \]

Sub-circuit with the 2 A source acting alone for Problem 6

By superposition:

\[ I = I' + I'' = -0.8 + 3.2 = 2.4\ \text{A} \]

Answer\(I = 2.4\ \text{A}\)

Additional Practice Problems

Two self-contained worked examples with schematics, drawn entirely with independent sources so each contribution is easy to verify: one mixing a voltage source with a current source, and one with two voltage sources sharing a common branch.

Problem 7Voltage + Current Source

A 12 V source feeds node N through a 4 Ω resistor; a 6 Ω resistor connects N to ground, and a 2 A current source also feeds node N. Find the current \(I_{6\Omega}\) through the 6 Ω resistor using superposition.

Solution

Step 112 V source acting alone (open the 2 A source). The 6 Ω current is set by the 4 Ω–6 Ω voltage divider:

\[ V_N' = 12 \times \frac{6}{4 + 6} = 7.2\ \text{V}, \qquad I_{6\Omega}' = \frac{7.2}{6} = 1.2\ \text{A} \]

Step 22 A source acting alone (short the 12 V source). The 4 Ω and 6 Ω are in parallel across the source:

\[ V_N'' = 2 \times (4 \parallel 6) = 2 \times \frac{4 \times 6}{4 + 6} = 4.8\ \text{V}, \qquad I_{6\Omega}'' = \frac{4.8}{6} = 0.8\ \text{A} \]

By superposition:

\[ I_{6\Omega} = 1.2 + 0.8 = 2.0\ \text{A} \]

Answer\(I_{6\Omega} = 2.0\ \text{A}\quad (V_N = 12\ \text{V})\)

Problem 8Two Voltage Sources

A 12 V source (with a 4 Ω series resistor) and a 6 V source (with a 2 Ω series resistor) both feed node N, which connects to ground through a 4 Ω resistor. Find the current \(I_{m}\) in the middle 4 Ω resistor using superposition.

Solution

Step 112 V source acting alone (short the 6 V source). The right branch becomes 2 Ω to ground, in parallel with the middle 4 Ω:

\[ V_N' = 12 \times \frac{4 \parallel 2}{4 + (4 \parallel 2)} = 12 \times \frac{4/3}{4 + 4/3} = 3\ \text{V}, \qquad I_m' = \frac{3}{4} = 0.75\ \text{A} \]

Step 26 V source acting alone (short the 12 V source). The left branch becomes 4 Ω to ground, in parallel with the middle 4 Ω:

\[ V_N'' = 6 \times \frac{4 \parallel 4}{2 + (4 \parallel 4)} = 6 \times \frac{2}{2 + 2} = 3\ \text{V}, \qquad I_m'' = \frac{3}{4} = 0.75\ \text{A} \]

By superposition:

\[ I_m = 0.75 + 0.75 = 1.5\ \text{A} \]

Answer\(I_m = 1.5\ \text{A}\quad (V_N = 6\ \text{V})\)