Basic Electrical Engineering - Circuit Analysis and Network Theorems Study Notes

Fundamental Definitions and Concepts in Circuit Theory

• Active Element: An active element is a circuit element that is capable of delivering energy or power to the rest of the circuit. It can generate electrical energy. Examples include batteries, generators, and dependent or independent voltage and current sources.

• Passive Element: A passive element is a circuit element that only absorbs or stores energy; it cannot generate or supply net energy to the circuit. Examples include resistors, inductors, and capacitors.

• Node: A node is a point in a circuit where two or more circuit elements (branches) meet or are joined together.

• Mesh: A mesh is a loop that does not contain any other loop within it. It represents the smallest, independent closed loop in a planar circuit.

• Loop: A loop is any closed path in a circuit formed by traversing circuit elements such that the starting node is reached again without passing through any other node more than once.

• Kirchhoff's Current Law (KCL):

  • KCL states that the algebraic sum of all currents meeting at a node is zero, which means the total current entering a node is equal to the total current leaving that node:     $\sum I_{\text{in}} = \sum I_{\text{out}}$
  • Physical Principle: Kirchhoff's Current Law is a direct consequence of the conservation of electric charge.

• Kirchhoff's Voltage Law (KVL):

  • KVL states that the algebraic sum of all the voltages (potential rises and drops) around any closed loop in a circuit is zero:     $\sum V = 0$
  • Physical Principle: Kirchhoff's Voltage Law is a direct consequence of the conservation of energy.

• Current Divider Rule:

  • When a total current $I$ is divided between two or more resistors connected in parallel, the current through any single branch is proportional to the opposite branch resistance divided by the sum of the parallel resistances.
  • For two resistors $R_1$ and $R_2$ connected in parallel carrying a total current $I$:     $I_1 = I \times \frac{R_2}{R_1 + R_2}$$I_2 = I \times \frac{R_1}{R_1 + R_2}$

• Source Conversion:

  • A practical voltage source consisting of an ideal voltage source $V$ in series with an internal resistance $R_s$ can be converted into an equivalent practical current source consisting of an ideal current source $I = \frac{V}{R_s}$ in parallel with the same resistance $R_s$, and vice versa.
  • Under this conversion, the terminal behavior of the circuit remains identical.

Network Theorems

• Superposition Theorem:

  • Statement: In any linear, bilateral network containing more than one independent source, the current through (or voltage across) any element is equal to the algebraic sum of the currents (or voltages) produced by each independent source acting alone, with all other independent sources deactivated.
  • Source Deactivation Rules:
  • Voltage sources are deactivated by replacing them with a short circuit (retaining their internal resistance if specified).
  • Current sources are deactivated by replacing them with an open circuit.

• Thevenin's Theorem:

  • Statement: Any linear, bilateral two-terminal (one-port) network, however complex, can be replaced at its terminals by an equivalent circuit consisting of a single voltage source $E_{\text{th}}$ (the Thevenin voltage) in series with a single resistance $R_{\text{th}}$ (the Thevenin resistance).
  • $E_{\text{th}}$ Determination: $E_{\text{th}}$ is the open-circuit voltage measured across the designated load terminals.
  • $R_{\text{th}}$ Determination: $R_{\text{th}}$ is the equivalent resistance seen looking back into the network from the opened terminals with all independent sources deactivated (voltage sources shorted and current sources opened).

Numericals on Nodal Analysis (KCL Applications)

• Nodal Analysis Example 1:

  • Circuit Configuration: Contains passive resistors of values $1\,\Omega$, $0.5\,\Omega$, $1\,\Omega$, $2\,\Omega$, and $1\,\Omega$ with two independent voltage sources of $15\,V$ and $20\,V$
  • Image Illustration:     
  • Node Equations:     $4V_A - 2V_B = 15$$4V_A - 7V_B = -40$
  • Intermediate Nodal Voltages:     $V_A = 9.25\,V$$V_B = 11\,V$
  • Calculated Branch Currents:     $I_1 = 5.75\,A$$I_2 = 9\,A$

• Nodal Analysis Example 2:

  • Circuit Configuration: Two batteries connected in parallel to a $10\,\Omega$ load resistor. Battery A has an electromotive force (emf) of $12\,V$ and internal resistance of $1\,\Omega$. Battery B has an emf of $10\,V$ and internal resistance of $1\,\Omega$
  • Image Illustration:     
  • Nodal Voltages:     $V_X = V_Y = 10\,V$
  • Currents Supplied and Load Current:
  • Current supplied by battery A: $1\,A$
  • Current supplied by battery B: $0\,A$
  • Load current $I_3$ through the $10\,\Omega$ resistor: $1\,A$

• Nodal Analysis Example 3:

  • Circuit Configuration: Three-node network with nodes A, B, and C, containing voltage sources of $15\,V$ and $30\,V$, and resistors of $4\,\Omega$, $5\,\Omega$, $10\,\Omega$, $6\,\Omega$, and $4\,\Omega$
  • Image Illustration:     
  • Method: Solved by formulating KCL node equations at nodes A and B to obtain nodal voltages $V_A$ and $V_B$
  • Current Formula through the $10\,\Omega$ resistor:     $I = \frac{V_A - V_B}{10}$

• Nodal Analysis Example 4:

  • Circuit Configuration: Circuit containing a $6\,V$ fixed voltage source, resistors of $2\,\Omega$, $3\,\Omega$, and $12\,\Omega$, and a $4\,A$ independent current source across nodes A, B, and C
  • Image Illustration:     
  • Node Voltages & Equations:
  • $V_A = 6\,V$ (fixed directly by the $6\,V$ source connected to node A)
  • Node B KCL Equation:       $\frac{V_B - V_A}{2} + \frac{V_B}{12} = 4$
  • Resulting Node Voltage:     $V_B = 12\,V$

• Nodal Analysis Example 5:

  • Circuit Configuration: Four-node network containing two current sources ($10\,A$ and $2\,A$) and six resistors ($5\,\Omega$, $3\,\Omega$, $1\,\Omega$, $2\,\Omega$, $5\,\Omega$, $4\,\Omega$)
  • Image Illustration:     
  • System of Simultaneous Node Equations:     $31V_A - 10V_B - 6V_C = 300$$5V_A - 23V_B + 15V_C = 0$$4V_A + 20V_B - 29V_C = 40$

Numericals on Superposition Theorem

• Superposition Example 1:

  • Circuit Configuration: Network containing a $20\,V$ source, a $32\,V$ source, and resistors of values $14.5\,\Omega$, $9\,\Omega$, $10\,\Omega$, $0.5\,\Omega$, and $1\,\Omega$
  • Image Illustration:     
  • Step 1: Considering $20\,V$ source alone ($32\,V$ source shorted):     $I_1' = 1\,A$$I_2' = 0.5\,A$$I_3' = 0.5\,A$
  • Step 2: Considering $32\,V$ source alone ($20\,V$ source shorted):     $I_1'' = 0.8\,A$$I_3'' = 1.2\,A$
  • Final Combined Branch Currents:
  • $I_1 = 0.2\,A$ (direction: $A \rightarrow B$)
  • $I_2 = -1.5\,A$ (direction: $C \rightarrow B$)
  • $I_3 = 1.7\,A$ (direction: $B \rightarrow E$)

• Superposition Example 2:

  • Circuit Configuration: Circuit containing a $20\,V$ voltage source, a $2\,A$ current source, and resistors of $2\,\Omega$, $4\,\Omega$, and $8\,\Omega$. Current $I_1 = 0.4\,A$ is given in one branch
  • Image Illustration:     
  • Step 1: Considering $20\,V$ source alone ($2\,A$ current source open-circuited):     $I_1 = \frac{20}{2 + 8} = 2\,A \quad (\text{direction: } A \rightarrow B)$
  • Step 2: Considering $2\,A$ current source alone ($20\,V$ voltage source short-circuited):     $I_2 = 2 \times \frac{2}{2 + 8} = 0.4\,A \quad (\text{direction: } B \rightarrow A)$
  • Net Current through the $8\,\Omega$ resistor:     $I = I_1 - I_2 = 2 - 0.4 = 1.6\,A \quad (\text{direction: } A \rightarrow B)$

• Superposition Example 3:

  • Circuit Configuration: Network with a $24\,V$ voltage source, a $4\,A$ current source, and resistors of $6\,\Omega$, $2\,\Omega$, and $5\,\Omega$
  • Image Illustration:     
  • Step 1: Considering $24\,V$ source alone ($4\,A$ source open-circuited):     $I_1 = \frac{24}{6 + 5} = 2.18\,A \quad (\text{direction: } A \rightarrow B)$
  • Step 2: Considering $4\,A$ source alone ($24\,V$ source short-circuited):     $I_2 = \left(\frac{6}{5 + 6}\right) \times 4 = 2.18\,A \quad (\text{direction: } B \rightarrow A)$
  • Net Current $I$ through the $5\,\Omega$ branch:     $I = 2.18 - 2.18 = 0\,A$

• Superposition Example 4:

  • Circuit Configuration: Circuit with $75\,V$ and $64\,V$ voltage sources and resistors of $5\,\Omega$, $5\,\Omega$, $4\,\Omega$, $20\,\Omega$, and $12\,\Omega$
  • Image Illustration:     
  • Equivalent Resistance Calculation (with $75\,V$ source alone, $64\,V$ source shorted):     $R_{\text{eq}}' = \left[\left\{(4 \parallel 12) + 5\right\} \parallel 20\right] + 5 = \frac{75}{7}\,\Omega$

• Superposition Example 5:

  • Circuit Configuration: Circuit with a $20\,V$ voltage source, a $5\,A$ current source, and resistors of $5\,\Omega$, $10\,\Omega$, and $3\,\Omega$. Find current through the $3\,\Omega$ resistor
  • Image Illustration:     
  • Step 1: Considering $20\,V$ source alone ($5\,A$ source open-circuited):     $I_1 = \frac{20}{5 + 3} = \frac{20}{8} = 2.5\,A$
  • Step 2: Considering $5\,A$ current source alone ($20\,V$ source short-circuited):     $I_2 = 5 \times \frac{5}{5 + 3} = 5 \times \frac{5}{8} = 3.125\,A$
  • Net Current $I$ through the $3\,\Omega$ resistor:     $I = I_1 + I_2 = 2.5 + 3.125 = 5.625\,A$

Numericals on Thevenin's Theorem

• Thevenin Example 1:

  • Circuit Problem: Determine the current through the $6\,\Omega$ resistor connected across terminals A–B using Thevenin's theorem. Circuit contains $6\,V$ and $15\,V$ sources and resistors of $6\,\Omega$, $4\,\Omega$, $3\,\Omega$, and $6\,\Omega$
  • Image Illustration:     
  • Thevenin Parameters:     $R_{\text{th}} = 4 + (3 \parallel 6) = 6\,\Omega$$E_{\text{th}} = 12\,V$
  • Load Current Calculation (for $R_L = 6\,\Omega$):     $I_L = \frac{E_{\text{th}}}{R_{\text{th}} + R_L} = \frac{12}{6 + 6} = 1\,A$

• Thevenin Example 2:

  • Circuit Problem: Ladder network with resistors $18\,\Omega$, $18\,\Omega$, $9\,\Omega$, $9\,\Omega$, $3\,\Omega$ and a $54\,V$ source. Determine current through $R_L$ for $R_L = 3\,\Omega$, $6\,\Omega$, and $9\,\Omega$
  • Image Illustration:     
  • Thevenin Equivalent Parameters:     $R_{\text{th}} = 9\,\Omega$$E_{\text{th}} = 9\,V$
  • Load Current Results for Various Values of $R_L$:
  • For $R_L = 3\,\Omega$:       $I(3\,\Omega) = \frac{9}{9 + 3} = 0.75\,A$
  • For $R_L = 6\,\Omega$:       $I(6\,\Omega) = \frac{9}{9 + 6} = 0.6\,A$
  • For $R_L = 9\,\Omega$:       $I(9\,\Omega) = \frac{9}{9 + 9} = 0.5\,A$

• Thevenin Example 3:

  • Circuit Problem: Find the branch current $I_2$ through $R_2$ for $R_2 = 5\,\Omega$, $15\,\Omega$, and $50\,\Omega$ using Thevenin's theorem. Circuit contains sources $140\,V$ and $85\,V$, and resistors $R_1 = 30\,\Omega$, $R_3 = 70\,\Omega$
  • Image Illustration:     
  • Thevenin Equivalent Parameters:     $R_{\text{th}} = 30 \parallel 70 = 21\,\Omega$$E_{\text{th}} = 123.5\,V$
  • Calculated Branch Current Results:
  • For $R_2 = 5\,\Omega$:       $I(5\,\Omega) = \frac{123.5}{21 + 5} = 4.75\,A$
  • For $R_2 = 15\,\Omega$:       $I(15\,\Omega) = \frac{123.5}{21 + 15} = 3.43\,A$
  • For $R_2 = 50\,\Omega$:       $I(50\,\Omega) = \frac{123.5}{21 + 50} = 1.74\,A$

• Thevenin Example 4:

  • Circuit Problem: Determine the current and voltage in the $2\,\Omega$ resistor using Thevenin's theorem. Circuit contains a $7\,A$ current source, a $12\,V$ voltage source, and resistors of $2\,\Omega$, $5\,\Omega$, $6\,\Omega$, and $10\,\Omega$
  • Image Illustration:     
  • Intermediate Voltage Calculations & Thevenin Parameters:     $V_{AC} = 35\,V$$V_{BC} = 7.5\,V$$E_{\text{th}} = V_{AC} - V_{BC} = 35 - 7.5 = 27.5\,V$$R_{\text{th}} = 5 + (10 \parallel 6) = 8.75\,\Omega$
  • Load Current Result:     $I_L = \frac{27.5}{8.75 + 2} = 2.56\,A$

• Thevenin Example 5:

  • Circuit Problem: Obtain the Thevenin equivalent circuit at terminals AB for the given network containing $80\,V$ and $20\,V$ voltage sources and resistors of $5\,\Omega$, $8\,\Omega$, $3\,\Omega$, and $4\,\Omega$
  • Image Illustration:     
  • Mesh / Branch Currents:     $I_1 = \frac{1280}{119}\,A$$I_2 = \frac{500}{119}\,A$
  • Thevenin Equivalent Parameters:     $R_{\text{th}} = 3 \parallel \left[8 + (4 \parallel 5)\right] = \frac{276}{119} = 2.32\,\Omega$$E_{\text{th}} = 80 - 5I_1 - 8I_2 = -\frac{880}{119} = -7.395\,V$

• Thevenin Example 6:

  • Circuit Problem: Determine the Thevenin equivalent circuit across terminals AB for the network given with $50\,V$ and $25\,V$ voltage sources and resistors of $10\,\Omega$ and $5\,\Omega$
  • Image Illustration:     
  • Step 1: Open-circuit Current $I$ in the main loop:     $I = \frac{50 - 25}{10 + 5} = \frac{25}{15} = 1.67\,A$
  • Step 2: Open-circuit Voltage $V_{AB}$ ($E_{\text{th}}$):     $V_{AB} = 50 - 10 \times I = 50 - 10 \times 1.67 = 33.35\,V$     Alternatively:     $V_{AB} = 5 \times I + 25 = 5 \times 1.67 + 25 = 33.35\,V$     Therefore:     $E_{\text{th}} = V_{\text{th}} = 33.35\,V$
  • Step 3: Thevenin Resistance $R_{\text{th}}$:     $R_{\text{th}} = 10 \parallel 5 = \frac{10 \times 5}{10 + 5} = \frac{50}{15} = 3.33\,\Omega$
  • Step 4: Final Equivalent Circuit:
  • A Thevenin voltage source $E_{\text{th}} = 33.35\,V$ in series with a Thevenin resistance $R_{\text{th}} = 3.33\,\Omega$ connected across terminals A and B.