DC Circuit Analysis and Source Transformation Study Notes

DC Electric Circuits and Circuit Elements

• Definition of an Electric Circuit: An electric circuit is an interconnection of electrical devices containing at least one closed path through which electric current may flow.

• Basic Circuit Concepts:

  • Direct Current (DC) circuits operate with non-time-varying currents and voltages.

  • Circuits consist of active elements that deliver energy and passive elements that absorb or store energy.

Active and Passive Circuit Elements

• Active Elements:

  • Definition: Circuit elements capable of generating or supplying electrical energy to a network.

  • Classification by Source Type:

  • Voltage Sources

  • Current Sources

  • Classification by Control Mechanism:

  • Independent Sources: Maintain a specified voltage or current regardless of other variables in the circuit.

  • Dependent (Controlled) Sources: The source value is determined or controlled by a voltage or current existing elsewhere in the circuit.

• Passive Elements:

  • Definition: Circuit elements that absorb, dissipate, or store electrical power or energy.

  • Specific Sinks:

  • Resistor: Dissipates electrical energy in the form of heat.

  • Inductor: Stores electrical energy within a magnetic field.

  • Capacitor: Stores electrical energy within an electric field.

  • Structural and Behavior Classifications:

  • Linear: Operating characteristics satisfy the principles of superposition and homogeneity.

  • Bilateral: Conduction properties are identical in either current flow direction.

  • Lumped: Physical dimensions are small relative to the operational wavelength, allowing parameter representation at discrete points.

Electrical Sources: Ideal and Practical

• Voltage Sources:

  • Ideal Voltage Source: Provides a continuous terminal voltage independent of the load current drawn from it. Possesses zero internal resistance ($R_S = 0\,\Omega$).

  • Practical Voltage Source: Modelled as an ideal voltage source ($V_S$) in series with an internal source resistance ($R_S$). Terminal voltage decreases as load current increases due to internal voltage drop.

• Current Sources:

  • Ideal Current Source: Supplies a constant current regardless of the voltage across its terminals. Possesses infinite internal resistance ($R_S = \infty\,\Omega$).

  • Practical Current Source: Modelled as an ideal current source ($I_S$) connected in parallel with an internal source resistance ($R_S$). Terminal current decreases as terminal voltage increases due to current diversion through internal resistance.

Resistance, Conductance, and Power Dissipation

• Resistor and Resistance ($R$):

  • Physical Relationship Formula:

$R = \frac{\rho \times l}{A}$

• Parameters:

  • $\rho$: Resistivity of the material

  • $l$: Length of the conductor

  • $A$: Cross-sectional area of the conductor

• Unit of Measurement: Ohms ($\Omega$

  • Conductance ($G$):

• Definition: The reciprocal of electrical resistance, measuring the ease with which current flows.

• Formula:

$G = \frac{1}{R}$

• Unit of Measurement: Siemens ($\text{S}$

  • Power Associated with Resistors ($P$):

• Energy conversion rate from electrical energy to thermal dissipation.

• Governing Formulas:

$P = V \times I$

$P = I^2 \times R$

$P = \frac{V^2}{R}$

• Unit of Measurement: Watts ($\text{W}$

Network Reduction Technique: Source Transformation

• Fundamental Concept:

  • Source transformation is a circuit reduction technique in which a practical voltage source is transformed into an equivalent practical current source, or vice versa, without affecting terminal characteristics.

• Equivalence Formulations:

  • A practical voltage source consisting of $V_S$ in series with $R_S$ converts to a practical current source consisting of $I_S$ in parallel with $R_S$.

  • Relationship Formula:

$I_S = \frac{V_S}{R_S}$

$V_S = I_S \times R_S$

• Resistance Equality: The internal resistance $R_S$ remains identical in magnitude for both representations.

Worked Example 1: Load Current and Power Calculation

• Problem Statement: Determine the current through and the power consumed by the $15\,\Omega$ resistor in the given circuit.

• Given Circuit Parameters:

  • Left branch: Independent current source $I_{S1} = 20\,A$ in parallel with resistor $R_1 = 100\,\Omega$

  • Series resistor $R_2 = 6\,\Omega$

  • Shunt resistor $R_3 = 6\,\Omega$

  • Series resistor $R_4 = 6\,\Omega$

  • Voltage branch: Independent voltage source $V_{S2} = 50\,V$ in series with resistor $R_5 = 0.5\,\Omega$

  • Target load resistor $R_L = 15\,\Omega$

• Step-by-Step Transformation Procedure:

  • Step 1: Transform the parallel combination of $20\,A$ and $100\,\Omega$ into an equivalent voltage source:

$V_1 = 20\,A \times 100\,\Omega = 2000\,V$

- Resistor $100\,\Omega$ is now in series with $2000\,V$.

• Step 2: Combine series resistors $100\,\Omega$ and $6\,\Omega$:

$R_{s1} = 100\,\Omega + 6\,\Omega = 106\,\Omega$

• Step 3: Transform $2000\,V$ in series with $106\,\Omega$ back to a current source:

$I_{S2} = \frac{2000\,V}{106\,\Omega} = 18.8679\,A$

- Resistor $106\,\Omega$ is in parallel with $18.8679\,A$

• Step 4: Combine parallel resistors $106\,\Omega$ and $6\,\Omega$:

$R_{p1} = \frac{106 \times 6}{106 + 6} = \frac{636}{112} = 5.6786\,\Omega$

• Step 5: Transform $18.8679\,A$ in parallel with $5.6786\,\Omega$ into a voltage source:

$V_2 = 18.8679\,A \times 5.6786\,\Omega = 107.1429\,V$

- Internal series resistance is $5.6786\,\Omega$

• Step 6: Add second series resistor $6\,\Omega$:

$R_{s2} = 5.6786\,\Omega + 6\,\Omega = 11.6786\,\Omega$

• Step 7: Transform $107.1429\,V$ in series with $11.6786\,\Omega$ to a current source:

$I_{S3} = \frac{107.1429\,V}{11.6786\,\Omega} = 9.1743\,A$

• Step 8: Transform the right branch voltage source $50\,V$ in series with $0.5\,\Omega$ into a current source:

$I_{S4} = \frac{50\,V}{0.5\,\Omega} = 100\,A$

- Parallel resistance is $0.5\,\Omega$

• Step 9: Combine total parallel current sources and total parallel resistance connected to the $15\,\Omega$ load:

$I_{total} = 9.1743\,A + 100\,A = 109.1743\,A$

$R_{eq, parallel} = \left(\frac{1}{11.6786} + \frac{1}{0.5} + \frac{1}{15}\right)^{-1} = 0.46462\,\Omega$

$V_{load} = 109.1743\,A \times 0.46462\,\Omega = 50.724\,V$

• Step 10: Compute load current and dissipated power:

$I_{15\,\Omega} = \frac{50.724\,V}{15\,\Omega} = 3.381\,A$

$P_{15\,\Omega} = (3.381\,A)^2 \times 15\,\Omega = 171.497\,W$

• Final Calculated Values:

  • $I_{15\,\Omega} = 3.381\,A$

  • $P_{15\,\Omega} = 171.497\,W$

Worked Example 2: Practical Source Reduction and Load Voltage

• Problem Statement:

  • (a) Simplify the circuit shown into a single practical voltage source connected across the load resistor $R_L$

  • (b) Calculate the voltage drop across the load resistance $R_L = 10\,\Omega$

• Given Circuit Parameters:

  • Voltage source 1: $24\,V$ with internal resistance $0.1\,\Omega$

  • Voltage source 2: $24\,V$ with internal resistance $0.15\,\Omega$

  • Parallel shunt resistor: $100\,\Omega$

  • Series resistor: $6\,\Omega$

  • Parallel resistor branch: $40\,\Omega$ in parallel with $60\,\Omega$

  • Load resistor $R_L = 10\,\Omega$

• Step-by-Step Reduction Procedure:

  • Step 1: Convert parallel voltage sources into current sources:

$I_1 = \frac{24\,V}{0.1\,\Omega} = 240\,A$

$I_2 = \frac{24\,V}{0.15\,\Omega} = 160\,A$

• Step 2: Combine parallel current sources and internal parallel resistances:

$I_p = 240\,A + 160\,A = 400\,A$

$R_{p1} = \frac{0.1 \times 0.15}{0.1 + 0.15} = \frac{0.015}{0.25} = 0.06\,\Omega$

• Step 3: Convert current source $400\,A$ with parallel resistance $0.06\,\Omega$ back to a voltage source:

$V_{p1} = 400\,A \times 0.06\,\Omega = 24\,V$

- Series resistance is $0.06\,\Omega$

• Step 4: Include parallel $100\,\Omega$ shunt resistor:

$R_{p2} = \frac{0.06 \times 100}{0.06 + 100} = 0.05996\,\Omega$

$V_{p2} = 400\,A \times 0.05996\,\Omega = 24\,V$

• Step 5: Simplify parallel load-side resistors $40\,\Omega$ and $60\,\Omega$:

$R_{parallel} = \frac{40 \times 60}{40 + 60} = \frac{2400}{100} = 24\,\Omega$

• Step 6: Determine total equivalent internal source resistance $R_S$:

$R_S = 0.05996\,\Omega + 6\,\Omega + 24\,\Omega = 30.05996\,\Omega \approx 30.06\,\Omega$

$V_S = 24\,V$

• Step 7: Calculate voltage drop across load $R_L = 10\,\Omega$ using potential divider formula:

$V_{R_L} = V_S \times \frac{R_L}{R_S + R_L} = 24\,V \times \frac{10\,\Omega}{30.06\,\Omega + 10\,\Omega} = 5.99\,V$

• Final Calculated Values:

  • Practical Voltage Source: $V_S = 24\,V$

  • Source Resistance: $R_S = 30.06\,\Omega$

  • Load Voltage Drop: $V_{R_L} = 5.99\,V$

Self-Practice Problem: Power Dissipation in a Circuit

• Problem Statement: In the circuit shown, find the amount of power dissipated in the $3\,\Omega$ resistor.

• Circuit Analysis & Derivation:

  • Formulate nodal equation at the top node above the $3\,\Omega$ resistor ($V_1$):

$\frac{V_1 - 24}{6} + \frac{V_1 - 0}{3} + \frac{V_1 - 6}{6} + \frac{V_1 - 12}{6} = 0$

• Expand and solve for node voltage $V_1$:

$(V_1 - 24) + 2(V_1) + (V_1 - 6) + (V_1 - 12) = 0$

$5 V_1 - 42 = 0$

$5 V_1 = 42$

$V_1 = 8.4\,V$

• Calculate power dissipated in $3\,\Omega$ resistor:

$P_{3\,\Omega} = \frac{V_1^2}{3\,\Omega} = \frac{(8.4\,V)^2}{3\,\Omega} = \frac{70.56}{3} = 23.52\,W$

• Final Calculated Value:

  • Dissipated Power $P_{3\,\Omega} = 23.52\,W$

Summary of DC Circuit Concepts

• Electric Circuit Core Definition: Interconnection of electrical components establishing at least one closed current path.

• Circuit Elements Classification:

  • Active Elements: Deliver power (Independent and Dependent Voltage/Current Sources).

  • Passive Elements: Absorb or store energy (Resistors, Inductors, Capacitors).

  • Structural Attributes: Linear, Bilateral, Lumped.

• Source Characteristics:

  • Ideal Voltage Source: $R_S = 0\,\Omega$

  • Practical Voltage Source: $V_S$ in series with $R_S$

  • Ideal Current Source: $R_S = \infty\,\Omega$

  • Practical Current Source: $I_S$ in parallel with $R_S$

• Sinks and Parameters:

  • Resistance: $R = \frac{\rho \times l}{A}$ in ohms ($\Omega$

  • Conductance: $G = \frac{1}{R}$ in siemens ($\text{S}$

  • Power: $P = V \times I = I^2 \times R = \frac{V^2}{R}$ in watts ($\text{W}$

• Source Transformation Core Rule:

  • Practical voltage source ($V_S, R_S$) converts to practical current source ($I_S, R_S$) with $I_S = \frac{V_S}{R_S}$, keeping resistance value $R_S$ identical.