Electric Circuit Analysis Study Guide

Circuit Analysis Principles and Network Equivalents

• Capacitance and Resistance Network Simplifications

  • Calculation of Equivalent Capacitance in Series-Parallel Configurations

  • Step 1: Combine parallel capacitors by summing their individual capacitance values:

      $C_{\text{parallel}} = C_1 + C_2 = 23\,\mu\text{F} + 35\,\mu\text{F} = 58\,\mu\text{F}$

- Step 2: Combine the parallel result with the remaining series capacitor $C_3 = 40\,\mu\text{F}$ using the reciprocal sum formula:

      $\frac{1}{C_{\text{total}}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$

      $\frac{1}{C_{\text{total}}} = \frac{1}{58\,\mu\text{F}} + \frac{1}{40\,\mu\text{F}} = \frac{40 + 58}{58 \times 40} = \frac{98}{2320} = \frac{49}{1160}$

- Step 3: Solve for total capacitance $C_{\text{total}}$:

      $C_{\text{total}} = \frac{1160}{49} = 23.673\dots\,\mu\text{F} \approx 24\,\mu\text{F}\text{ (2 s.f.)}$

• Star to Delta ($Y - \Delta$) Resistance Conversion

  • For three star-connected resistors ($R_1$, $R_2$, and $R_3$) meeting at a common central node, the equivalent delta-connected resistors ($R_A$, $R_B$, and $R_C$) formed between terminals A, B, and C are given by:

      $R_A = \frac{R_1 R_2 + R_1 R_3 + R_2 R_3}{R_1}$

      $R_B = \frac{R_1 R_2 + R_1 R_3 + R_2 R_3}{R_2}$

      $R_C = \frac{R_1 R_2 + R_1 R_3 + R_2 R_3}{R_3}$

• Equivalent Resistance between Terminals X and Y

  • A network composed of bridge elements rated $r$ and $2r$ simplifies into four parallel branches between terminals X and Y, each having an effective resistance of $2r$

  • Summing the conductance of the four parallel branches yields:

      $\frac{1}{R_{xy}} = \frac{1}{2r} + \frac{1}{2r} + \frac{1}{2r} + \frac{1}{2r} = \frac{4}{2r}$

      $R_{xy} = \frac{2r}{4} = \frac{r}{2}\,\Omega$

Fundamental Circuit Theorems and Power Calculations

• Thevenin's Theorem

  • Definition: Any linear bidirectional electric circuit, regardless of complexity, can be replaced across its load terminals by an equivalent circuit consisting of a single independent voltage source ($V_{\text{Th}}$) connected in series with a single resistor ($R_{\text{Th}}$).

  • Practical Thevenin Calculation Example:

  • Objective: Calculate the current flowing through a $10\,\Omega$ load resistor connected across terminals A and B of a dual-source network containing a $10\,\text{V}$ source in series with $4\,\Omega$ and a $12\,\text{V}$ source in series with $6\,\Omega$

  • Open-Circuit Voltage ($V_{\text{Th}}$) Calculation:

      $I = \frac{12\,\text{V} - 10\,\text{V}}{4\,\Omega + 6\,\Omega} = \frac{2\,\text{V}}{10\,\Omega} = 0.2\,\text{A}$

      $V_{\text{Th}} = V_{AB} = 6 I - 12 = 6(0.2) - 12 = 1.2 - 12 = -10.8\,\text{V}$

      (Terminal B has negative polarity and terminal A has positive polarity relative to the loop direction).

- Equivalent Resistance ($R_{\text{Th}}$) Calculation with independent sources deactivated (short-circuited):

      $R_{\text{Th}} = R_{AB} = 4\,\Omega \parallel 6\,\Omega = \frac{4 \times 6}{4 + 6} = \frac{24}{10} = 2.4\,\Omega$

- Load Current ($I_L$) Determination:

      $I_L = \frac{V_{\text{Th}}}{R_{\text{Th}} + R_L} = \frac{10.8\,\text{V}}{2.4\,\Omega + 10\,\Omega} = \frac{10.8}{12.4} = 0.871\,\text{A}$

• Superposition Theorem

  • Statement: In any linear bidirectional circuit containing multiple independent sources, the total response (current or voltage) across any element equals the algebraic sum of individual responses produced by each independent source acting alone, with all other independent voltage sources short-circuited and independent current sources open-circuited.

  • Limitations:

  • Non-linear circuits: Cannot be applied to circuits containing non-linear elements such as diodes and transistors.

  • Power calculations: Cannot be used directly to calculate power dissipation because power is proportional to the square of current/voltage ($P = I^2 R = \frac{V^2}{R}$), making power a non-linear quantity.

  • Imbalanced bridge networks: Inapplicable to imbalanced bridge circuits.

  • Single-source requirements: Requires a minimum of two independent sources to be applied.

  • Comprehensive Multi-Source Superposition Problem:

  • Objective: Find current flowing through a $25\,\Omega$ resistor in a network containing a $62.5\,\text{V}$ source and a $12.5\,\text{V}$ source.

  • Case 1 ($62.5\,\text{V}$ source acting alone, $12.5\,\text{V}$ source shorted):

      $50 I_1 - 37.5 I_2 + 0 I_3 = 0$

      $-37.5 I_1 + 50 I_2 - 12.5 I_3 = 62.5$

      $0 I_1 - 12.5 I_2 + 37.5 I_3 = 0$

      Solving mesh currents yields current in $25\,\Omega$ resistor:

      $I_{25\Omega (62.5\text{V})} = 1.1765\,\text{A}$

- Case 2 ($12.5\,\text{V}$ source acting alone, $62.5\,\text{V}$ source shorted):

      $50 I_1 - 37.5 I_2 + 0 I_3 = -12.5$

      $-37.5 I_1 + 50 I_2 - 12.5 I_3 = 0$

      $0 I_1 - 12.5 I_2 + 37.5 I_3 = 12.5$

      Solving mesh currents yields current in $25\,\Omega$ resistor:

      $I_{25\Omega (12.5\text{V})} = 0.2353\,\text{A}$

- Superposition Summation:

      $I_{25\Omega} = I_{25\Omega (62.5\text{V})} + I_{25\Omega (12.5\text{V})} = 1.1765\,\text{A} + 0.2353\,\text{A} = 1.4118\,\text{A}$

• Power Dissipation and Power Balance Calculations

  • Power Dissipation in Voltage Reduction:

  • Problem: Calculate power consumed when $100\,\text{V}$ is applied across a bulb rated at $200\,\text{V}, 100\,\text{W}$

  • Bulb filament resistance $R$ calculation:

      $P = \frac{V^2}{R} \implies R = \frac{V_{\text{rated}}^2}{P_{\text{rated}}} = \frac{200^2}{100} = \frac{40000}{100} = 400\,\Omega$

- Actual power consumed at $100\,\text{V}$:

      $P_{\text{actual}} = \frac{V_{\text{applied}}^2}{R} = \frac{100^2}{400} = \frac{10000}{400} = 25\,\text{W}$

• Total Power Absorbed in a Multi-Branch Network Junction:

  • Branch 1 ($100\,\text{V}$ source with $10\,\text{A}$ entering positive terminal):

      $P_{100(\text{A})} = 100\,\text{V} \times 10\,\text{A} = 1000\,\text{W}$

- Branch 2 ($15\,\text{V}$ source with $2\,\text{A}$ leaving positive terminal):

      $P_{15(\text{D})} = 15\,\text{V} \times 2\,\text{A} = 30\,\text{W} \implies P_{15(\text{A})} = -30\,\text{W}$

- Branch 3 ($80\,\text{V}$ source with $8\,\text{A}$ leaving positive terminal):

      $P_{80(\text{D})} = 80\,\text{V} \times 8\,\text{A} = 640\,\text{W} \implies P_{80(\text{A})} = -640\,\text{W}$

- Total Power Absorbed (T.P.A.):

      $\text{T.P.A.} = P_{100(\text{A})} + P_{15(\text{A})} + P_{80(\text{A})} = 1000\,\text{W} - 30\,\text{W} - 640\,\text{W} = 330\,\text{W}$

• Maximum Power Transfer Theorem

  • Condition: Maximum power is transferred from a source to a load when the load resistance ($R_L$) equals the internal source resistance or Thevenin resistance ($R_{\text{Th}}$) as seen from the load terminals ($R_L = R_{\text{Th}}$).

  • Practical Circuit Application:

  • Ladder network parameters: $V_{\text{Th}} = -7.142\,\text{V}$, $R_{\text{Th}} = 1.7143\,\Omega$

  • Optimum load resistance for maximum power: $R_L = R_{\text{Th}} = 1.7143\,\Omega$

  • Load current $I_L$ under matched conditions:

      $I_L = \frac{V_{\text{Th}}}{2 R_{\text{Th}}} = \frac{7.142\,\text{V}}{2 \times 1.7143\,\Omega} = 2.083\,\text{A}$

- Maximum power delivered to the load ($P_L$):

      $P_L = I_L^2 R_L = (2.083)^2 \times 1.7143 = 7.438\,\text{W}$

Nodal Analysis and Network Formulation

• Kirchhoff's Laws

  • Kirchhoff's Voltage Law (KVL): The algebraic sum of all potential differences around any closed loop in a circuit is zero ($\sum V = 0$). Alternatively, the sum of voltage rises equals the sum of voltage drops.

  • Kirchhoff's Current Law (KCL): The algebraic sum of all electric currents entering and exiting a node is zero ($\sum I = 0$).

  • KVL Analysis of a Two-Mesh Network with a $3.3\,\text{k}\Omega$ Resistor:

  • Mesh 1 equation ($I_1$):

      $-8 + 2 I_1 + 3.3 I_1 + 3.3 I_2 = 0 \implies 5.3 I_1 - 3.3 I_2 = 8$

- Mesh 2 equation ($I_2$):

      $4.7 I_2 + 5 + 3.3 I_2 - 3.3 I_1 = 0 \implies -3.3 I_1 + 8.0 I_2 = -5$

- Solving simultaneous mesh currents:

      $I_1 = 1.5\,\text{mA}$

      $I_2 = -0.0031\,\text{mA}$

- Net current $I_R$ and voltage $V_R$ across the $3.3\,\text{k}\Omega$ resistor:

      $I_R = I_1 - I_2 = 1.5 - (-0.0031) = 1.5031\,\text{mA}$

      $V_R = I_R \times R = 1.5031\,\text{mA} \times 3.3\,\text{k}\Omega = 4.96\,\text{V} \approx 5\,\text{V}$

• Three-Node Nodal Matrix Problem

  • Circuit configuration: Node 1 ($V_1$), Node 2 ($V_2$), and Node 3 ($V_3$) with source transformation applied to convert a $10\,\text{V}$ voltage source with $10\,\Omega$ series resistor into a $1\,\text{A}$ parallel current source.

  • Node 1 KCL Formulation:

    $\frac{V_1}{10} + \frac{V_1}{5} + \frac{V_1}{3} + \frac{V_1}{3} - \frac{V_2}{3} - \frac{V_2}{3} = 1$

    $V_1 \left( \frac{1}{10} + \frac{1}{5} + \frac{1}{3} + \frac{1}{3} \right) - V_2 \left( \frac{2}{3} \right) = 1$

    $0.967 V_1 - 0.667 V_2 = 1$

• Node 2 KCL Formulation:

    $\frac{V_2}{3} + \frac{V_2}{3} - \frac{V_1}{3} - \frac{V_1}{3} + \frac{V_2}{2} - \frac{V_3}{2} = 5$

    $-0.667 V_1 + 1.167 V_2 - 0.5 V_3 = 5$

• Node 3 KCL Formulation:

    $\frac{V_3}{6} + \frac{V_3}{1} + \frac{V_3}{2} - \frac{V_2}{2} = 0$

    $0 V_1 - 0.5 V_2 + 1.667 V_3 = 0$

• Calculated Node Voltages:

    $V_1 = 8.08\,\text{V}$

    $V_2 = 10.21\,\text{V}$

    $V_3 = 3.06\,\text{V}$

• Single Non-Reference Node Voltage Determination

  • Circuit parameters: Node $V_1$ connected to a $50\,\text{V}$ source via $(3 + 6)\,\Omega$, a $25\,\text{V}$ source via $(2 + 8)\,\Omega$, and a $5\,\Omega$ resistor to reference ground.

  • KCL Equation at Node $V_1$:

    $\left[ \frac{V_1 - 50}{3 + 6} + \frac{V_1 - 25}{2 + 8} + \frac{V_1}{5} \right] = 0$

    $\left[ \frac{V_1 - 50}{9} + \frac{V_1 - 25}{10} + \frac{V_1}{5} \right] = 0$

    $[0.1111 + 0.1000 + 0.2000] V_1 - 5.556 - 2.5 = 0$

    $[0.4111] V_1 = 8.056 \implies V_1 = 19.59\,\text{V}$

• Branch Current Calculations:

  • Current through $(2 + 8)\,\Omega$ branch ($I_5$):

      $I_5 = \frac{V_1 - 25}{2 + 8} = \frac{19.59 - 25}{10} = -0.541\,\text{A}$

- Current through $(3 + 6)\,\Omega$ branch ($I_3$):

      $I_3 = \frac{V_1 - 50}{3 + 6} = \frac{19.59 - 50}{9} = -3.37\,\text{A}$

- Current through $5\,\Omega$ vertical resistor:

      $I_{5\Omega} = \frac{V_1}{5} = \frac{19.59}{5} = 3.918\,\text{A}$

- Voltage across $5\,\Omega$ resistor:

      $V_{5\Omega} = I_{5\Omega} \times 5 = 3.918 \times 5 = 19.59\,\text{V}$

Mesh Analysis and Network Transformations

• Bridge Network Equivalent Resistance via Delta-Star Conversion

  • Network composed of five $9\,\Omega$ resistors connected between terminals A and B.

  • Star-to-Delta Conversion of Star Point N (where branch resistances equal $R = 9\,\Omega$):

  • Equivalent delta branches equal $3 R = 3 \times 9 = 27\,\Omega$

  • Circuit simplification:

  • Parallel combination of $27\,\Omega$ and $9\,\Omega$:

      $\frac{27 \times 9}{27 + 9} = \frac{243}{36} = 6.75\,\Omega$

- Series and parallel branch reductions result in:

      $R_{AB} = 27\,\Omega \parallel 13.5\,\Omega = \frac{27 \times 13.5}{27 + 13.5} = \frac{364.5}{40.5} = 9\,\Omega$

• Mesh Analysis of a Three-Mesh Resistive Circuit

  • Circuit parameters: Mesh 1 ($I_1$), Mesh 2 ($I_2$), and Mesh 3 ($I_3$) with dual $100\,\text{V}$ sources.

  • KVL Equations:

  • Mesh 1: $-100 + 15 I_1 + 20 I_1 - 20 I_2 = 0 \implies 35 I_1 - 20 I_2 = 100 \implies 7 I_1 - 4 I_2 = 20$

  • Mesh 2: $-20 I_1 + 20 I_2 + 5 I_2 + 30 I_2 - 30 I_3 = 0 \implies -20 I_1 + 55 I_2 - 30 I_3 = 0 \implies 4 I_1 - 11 I_2 + 6 I_3 = 0$

  • Mesh 3: $30 I_3 - 30 I_2 + 5 I_3 + 100 = 0 \implies -30 I_2 + 35 I_3 = -100 \implies 30 I_2 - 35 I_3 = 100$

  • Matrix Formulation:

    $\begin{pmatrix} 7 & -4 & 0 \\ 4 & -11 & 6 \\ 0 & 30 & -35 \end{pmatrix} \begin{pmatrix} I_1 \\ I_2 \\ I_3 \end{pmatrix} = \begin{pmatrix} 20 \\ 0 \\ 100 \end{pmatrix}$

• Calculated Mesh Currents:

    $I_1 = 1.94\,\text{A}$

    $I_2 = -1.6\,\text{A}$

    $I_3 = -4.22\,\text{A}$

• Current through $30\,\Omega$ shared resistor:

    $I_{30\Omega} = I_2 - I_3 = -1.6 - (-4.22) = 2.62\,\text{A}$

• Bridge Network Terminal Resistance ($R_{PQ}$) and Source Current

  • Circuit parameters: Terminal P with $0.75\,\Omega$ series resistor connected to node A of a bridge (edges $R_{AB} = 5\,\Omega$, $R_{AC} = 2\,\Omega$, $R_{BC} = 3\,\Omega$, $R_{BD} = 1.5\,\Omega$, $R_{CD} = 0.4\,\Omega$) and terminal Q connected to node D.

  • Delta-to-Star Conversion of Delta ABC ($R_{\Delta} = 5 + 3 + 2 = 10\,\Omega$):

    $R_A = \frac{5 \times 2}{10} = 1\,\Omega$

    $R_B = \frac{5 \times 3}{10} = 1.5\,\Omega$

    $R_C = \frac{3 \times 2}{10} = 0.6\,\Omega$

• Network Path Reduction:

  • Left path to node D: $R_B + R_{BD} = 1.5 + 1.5 = 3\,\Omega$

  • Right path to node D: $R_C + R_{CD} = 0.6 + 0.4 = 1\,\Omega$

  • Parallel combination of reduced paths:

      $\frac{1}{R_{\text{parallel}}} = \frac{1}{3} + \frac{1}{1} = \frac{4}{3} \implies R_{\text{parallel}} = 0.75\,\Omega$

• Total Resistance $R_{PQ}$:

    $R_{PQ} = 0.75 + R_A + R_{\text{parallel}} = 0.75 + 1 + 0.75 = 2.50\,\Omega$

• Current Supplied by a $10\,\text{V}$ Battery across Terminals PQ:

    $I = \frac{V}{R_{PQ}} = \frac{10\,\text{V}}{2.5\,\Omega} = 4\,\text{A}$

• Mesh Analysis with Source Transformation

  • Objective: Calculate current $I_x$ in a multi-loop circuit.

  • Source Transformation Step: Convert an $8\,\text{A}$ current source in parallel with a $10\,\Omega$ resistor into an equivalent $80\,\text{V}$ voltage source ($V = 8\,\text{A} \times 10\,\Omega = 80\,\text{V}$) in series with a $10\,\Omega$ resistor.

  • KVL for Mesh 1 (loop abcda):

    $8 i_1 + 4(i_1 - i_2) - 100 = 0 \implies 12 i_1 - 4 i_2 = 100$

• KVL for Mesh 2 (loop cefdc):

    $2 i_2 + 3(i_2 - i_3) + 4(i_2 - i_1) = 0 \implies -4 i_1 + 9 i_2 - 3 i_3 = 0$

• KVL for Mesh 3 (loop eghfe):

    $10 i_3 + 5 i_3 + 3(i_3 - i_2) + 80 = 0 \implies 0 i_1 - 3 i_2 + 18 i_3 = -80$

• Determinant ($\Delta$) Matrix Solution:

    $\Delta = \begin{vmatrix} 12 & -4 & 0 \\ -4 & 9 & -3 \\ 0 & -3 & 18 \end{vmatrix} = 12(162 - 9) + 4(-72 - 0) + 0 = 1836 - 288 = 1548$

• Solving Mesh Currents via Cramer's Rule:

  • Determinant for $i_1$:

      $\Delta_1 = \begin{vmatrix} 100 & -4 & 0 \\ 0 & 9 & -3 \\ -80 & -3 & 18 \end{vmatrix} = 100(162 - 9) + 4(0 - 240) = 15300 - 960 = 14340$

      $i_1 = \frac{14340}{1548} = 9.26\,\text{A}$

- Determinant for $i_2$ (Current $I_x$):

      $\Delta_2 = \begin{vmatrix} 12 & 100 & 0 \\ -4 & 0 & -3 \\ 0 & -80 & 18 \end{vmatrix} = 12(0 - 240) - 100(-72 - 0) = -2880 + 7200 = 4320$

      $i_2 = I_x = \frac{4320}{1548} = 2.79\,\text{A}$

- Determinant for $i_3$:

      $\Delta_3 = \begin{vmatrix} 12 & -4 & 100 \\ -4 & 9 & 0 \\ 0 & -3 & -80 \end{vmatrix} = 12(-720 - 0) + 4(320 - 0) + 100(12 + 0) = -8640 + 1280 + 1200 = -6160$

      $i_3 = \frac{-6160}{1548} = -3.97\,\text{A}$

• Final Result: Current $I_x = i_2 = 2.79\,\text{A}$