Circuit Analysis using KCL and KVL, Equivalent Circuits, and Realistic Sources

Complete Circuit Analysis Example Using KCL and KVL

Circuit Setup and Parameter Definition

• The circuit consists of two ideal sources and four resistors:

  • A $50\,\text{V}$ independent voltage source on the left branch.

  • A $2\,\text{A}$ independent current source on the right branch.

  • A $1\,\text{Ohm}$ resistor ($R_1$) located on the top horizontal branch.

  • A $2\,\text{Ohm}$ resistor ($R_2$) located on the central vertical branch.

  • A $4\,\text{Ohm}$ resistor ($R_3$) located on the bottom horizontal branch of the left loop.

  • A $10\,\text{Ohm}$ resistor ($R_4$) located on the right branch in series with the current source.

• Key structural components of the circuit include:

  • Extraordinary node $A$ located at the top center.

  • Extraordinary node $B$ located at the bottom center.

  • Loop 1 ($L_1$) defined as the left loop.

  • Loop 2 ($L_2$) defined as the right loop.

• Unknown branch currents and voltage variables:

  • $i_1$: Current flowing clockwise through the left loop (top and bottom branches).

  • $i_2$: Current flowing downward through the middle branch from node $A$ to node $B$.

  • Current coming from the right side towards node $A$ is known and equal to $2\,\text{A}$.

  • Voltage variables across resistors follow the passive sign convention: $v_1$ across $1\,\text{Ohm}$, $v_2$ across $2\,\text{Ohm}$, $v_3$ across $4\,\text{Ohm}$, and $v_4$ across $10\,\text{Ohm}$.

  • $V_C$: Unknown voltage across the $2\,\text{A}$ current source ($+$ terminal on top, $-$ terminal on bottom).

System of Equations Derivation (Part A)

Step 1: Apply Kirchhoff's Current Law (KCL) at Node A

• KCL states that the sum of currents entering a node must equal the sum of currents leaving the node:   $i_1 + 2 = i_2$

• Rearranging this into a standard linear equation yields Equation 1:   $i_1 - i_2 = -2$

Step 2: Apply Kirchhoff's Voltage Law (KVL) on Loop 1 ($L_1$)

• Starting at the bottom-left corner and traversing Loop 1 clockwise:

  • Rise in potential through the voltage source: $+50\,\text{V}$

  • Drop in potential through the $1\,\text{Ohm}$ resistor: $-v_1$

  • Drop in potential through the $2\,\text{Ohm}$ resistor: $-v_2$

  • Drop in potential through the $4\,\text{Ohm}$ resistor (moving right to left): $-v_3$

• KVL equation for Loop 1:   $50 - v_1 - v_2 - v_3 = 0$

Step 3: Express Voltages in Terms of Currents Using Ohm's Law

• Applying Ohm's Law ($V = I \times R$) while respecting passive sign convention:

  • $v_1 = i_1 \times 1\,\text{Ohm} = i_1$

  • $v_2 = i_2 \times 2\,\text{Ohm} = 2i_2$

  • $v_3 = i_1 \times 4\,\text{Ohm} = 4i_1$

• Substituting these expressions back into the Loop 1 KVL equation:   $50 - i_1 - 2i_2 - 4i_1 = 0$   $50 - 5i_1 - 2i_2 = 0$

• Rearranging yields Equation 2:   $5i_1 + 2i_2 = 50$

Step 4: Simultaneous Equation Solution

• Multiply Equation 1 by $2$:   $2(i_1 - i_2) = 2(-2)$   $2i_1 - 2i_2 = -4$

• Add this modified equation to Equation 2:   $(2i_1 - 2i_2) + (5i_1 + 2i_2) = -4 + 50$   $7i_1 = 46$   $i_1 = \frac{46}{7}\,\text{A}$

• Substitute $i_1 = \frac{46}{7}\,\text{A}$ back into Equation 1 to solve for $i_2$:   $i_2 = i_1 + 2$   $i_2 = \frac{46}{7} + 2$   $i_2 = \frac{46}{7} + \frac{14}{7} = \frac{60}{7}\,\text{A}$

Step 5: Solve for Voltage $V_C$ Across Current Source Using Loop 2 ($L_2$)

• Apply KVL around Loop 2 ($L_2$), starting at node $B$ and traversing clockwise:

  • Moving up through the $2\,\text{Ohm}$ resistor from $-$ to $+$ terminal: $+v_2$

  • Moving down through the $2\,\text{A}$ current source from $+$ to $-$ terminal: $-V_C$

  • Moving right to left through the $10\,\text{Ohm}$ resistor from $-$ to $+$ terminal: $+v_4$

• KVL Equation for Loop 2:   $v_2 - V_C + v_4 = 0$

• Compute individual voltage values:

  • $v_2 = i_2 \times 2\,\text{Ohm} = \left(\frac{60}{7}\,\text{A}\right) \times 2\,\text{Ohm} = \frac{120}{7}\,\text{V}$

  • $v_4 = 2\,\text{A} \times 10\,\text{Ohm} = 20\,\text{V}$

• Substitute voltage values into KVL equation:   $\frac{120}{7} - V_C + 20 = 0$   $V_C = \frac{120}{7} + 20$   $V_C = \frac{120}{7} + \frac{140}{7} = \frac{260}{7}\,\text{V}$

Power Calculations and Conservation of Energy (Part B)

Passive Sign Convention (PSC) Rules

• Power $P = V \times I_{\text{entering}+}$, where $I_{\text{entering}+}$ is the current entering the positive voltage terminal.

• If $P > 0$, power is absorbed/dissipated by the element.

• If $P < 0$, power is delivered/supplied by the element.

• Resistors always absorb power ($P = I^2 R = \frac{V^2}{R} > 0$).

Individual Element Power Calculations

• Top $1\,\text{Ohm}$ Resistor:   $P_{1\Omega} = i_1^2 \times 1\,\text{Ohm} = \left(\frac{46}{7}\,\text{A}\right)^2 \times 1\,\text{Ohm} = \frac{2116}{49}\,\text{W} \approx 43.2\,\text{W}$   This power is dissipated.

• Center $2\,\text{Ohm}$ Resistor:   $P_{2\Omega} = i_2^2 \times 2\,\text{Ohm} = \left(\frac{60}{7}\,\text{A}\right)^2 \times 2\,\text{Ohm} = \frac{7200}{49}\,\text{W} \approx 146.9\,\text{W}$   This power is dissipated.

• Bottom $4\,\text{Ohm}$ Resistor:   $P_{4\Omega} = i_1^2 \times 4\,\text{Ohm} = \left(\frac{46}{7}\,\text{A}\right)^2 \times 4\,\text{Ohm} = \frac{8464}{49}\,\text{W} \approx 172.7\,\text{W}$   This power is dissipated.

• Right $10\,\text{Ohm}$ Resistor:   $P_{10\Omega} = (2\,\text{A})^2 \times 10\,\text{Ohm} = 40\,\text{W}$   This power is dissipated.

• Left $50\,\text{V}$ Voltage Source:   Current $i_1$ leaves the positive terminal, so the current entering the positive terminal is $-i_1 = -\frac{46}{7}\,\text{A}$.   $P_{50\text{V}} = 50\,\text{V} \times \left(-\frac{46}{7}\,\text{A}\right) = -\frac{2300}{7}\,\text{W} \approx -328.6\,\text{W}$   This power is supplied.

• Right $2\,\text{A}$ Current Source:   Voltage across source is $V_C = \frac{260}{7}\,\text{V}$ ($+$ on top). Current $2\,\text{A}$ leaves the positive terminal, so the entering current is $-2\,\text{A}$.   $P_{2\text{A}} = \left(\frac{260}{7}\,\text{V}\right) \times (-2\,\text{A}) = -\frac{520}{7}\,\text{W} \approx -74.3\,\text{W}$   This power is supplied.

Conservation of Power Verification

• Summing all powers dissipated and supplied:   $\sum P = P_{1\Omega} + P_{2\Omega} + P_{4\Omega} + P_{10\Omega} + P_{50\text{V}} + P_{2\text{A}}$   $\sum P = 43.2\,\text{W} + 146.9\,\text{W} + 172.7\,\text{W} + 40\,\text{W} - 328.6\,\text{W} - 74.3\,\text{W} = 0\,\text{W}$

• Total power dissipated ($402.8\,\text{W}$) equals total power supplied ($402.9\,\text{W}$) within rounding approximations of recurring fractions, verifying Tellegen's theorem / power conservation.

Equivalent Circuits and Source Transformations

Source Transformation

• Source transformation converts a voltage source in series with a resistance into an equivalent current source in parallel with a resistance, and vice versa.

• Given terminals $A$ and $B$:

  • A voltage source $V_S$ in series with resistor $R_S$ transforms to:

  • A current source $I_S$ in parallel with resistor $R_P$.

• Transformation Equations:   $I_S = \frac{V_S}{R_S}$   $R_P = R_S$

• Terminal Sign and Orientation Rules:

  • The current $I_S$ is injected into terminal $A$ corresponding to the terminal connected to the positive terminal of $V_S$, and extracted from terminal $B$.

Sources Connected in Series and Parallel

Voltage Sources in Series

• For $n$ voltage sources connected in series between terminals $A$ and $B$, the total equivalent voltage source $V_{\text{eq}}$ is the algebraic sum of all series voltage sources:   $V_{\text{eq}} = \sum_{i=1}^{n} V_{Si}$

• Algebraic Sign Rules:

  • Polarity must be strictly tracked relative to the terminals $A$ and $B$.

  • Example with three series voltage sources ($V_{S1}, V_{S2}, V_{S3}$):

  • Let $V_{\text{eq}}$ be defined with positive terminal at $B$ and negative terminal at $A$.

  • If $V_{S1}$ has $+$ toward $B$ and $-$ toward $A$: sign is positive ($+V_{S1}$).

  • If $V_{S2}$ has $-$ toward $B$ and $+$ toward $A$: sign is negative ($-V_{S2}$).

  • If $V_{S3}$ has $+$ toward $B$ and $-$ toward $A$: sign is positive ($+V_{S3}$).

  • Resulting equivalent equation:       $V_{\text{eq}} = V_{S1} - V_{S2} + V_{S3}$

  • This expression derives directly from applying KVL from terminal $A$ to terminal $B$

Current Sources in Parallel

• For $n$ current sources connected in parallel between terminals $A$ and $B$, the total equivalent current source $I_{\text{eq}}$ is the algebraic sum:   $I_{\text{eq}} = \sum_{i=1}^{n} I_{Si}$

• Algebraic Sign Rules:

  • Example with three parallel current sources ($I_{S1}, I_{S2}, I_{S3}$):

  • Define $I_{\text{eq}}$ pointing such that current leaves terminal $A$ and enters terminal $B$

  • Sources leaving $A$ and entering $B$ take a positive sign.

  • Sources entering $A$ and leaving $B$ take a negative sign.

  • Equivalent current equation:       $I_{\text{eq}} = I_{S1} - I_{S2} + I_{S3}$

Circuit Constraints and Realistic Sources

Constraints on Ideal Source Configurations

Parallel Ideal Voltage Sources

• Connecting two distinct ideal voltage sources $V_{S1}$ and $V_{S2}$ in parallel is physically invalid unless $V_{S1} = V_{S2}$.

• Applying KVL around the closed loop formed by two parallel voltage sources gives:   $V_{S1} - V_{S2} = 0 \implies V_{S1} = V_{S2}$

• If $V_{S1} \neq V_{S2}$, Kirchhoff's Voltage Law is violated.

Series Ideal Current Sources

• Connecting two distinct ideal current sources $I_{S1}$ and $I_{S2}$ in series is physically invalid unless $I_{S1} = I_{S2}$.

• Applying KCL at the intermediate node connecting the two series current sources yields:   $I_{S1} = I_{S2}$

• If $I_{S1} \neq I_{S2}$, Kirchhoff's Current Law is violated.

Realistic Source Models

Realistic Voltage Sources

• Ideal voltage sources maintain fixed voltage regardless of current, which is an idealization.

• Real-world sources (such as AA flashlight batteries or laboratory power supplies) are modeled as realistic voltage sources.

• Circuit Model: An ideal voltage source $V_S$ in series with an internal resistance $R_S$.

• Practical Value: For a good voltage source, $R_S$ is very small (e.g., $10\,\text{mA}$ listed as a real-world scale value).

• Terminal Current-Voltage ($I-V$) Relation:

  • Applying KVL for terminal voltage $V$ across terminals $A, B$ and injected current $I$:     $V_S - I R_S = V$

  • Expressing $I$ as a function of independent variable $V$:     $I = -\frac{V}{R_S} + \frac{V_S}{R_S}$

• Graphical Behavior:

  • Represents a straight line on an $I-V$ plot.

  • Intersects the voltage axis ($I = 0$) at $V = V_S$

  • Magnitude of slope is equal to $\frac{1}{R_S}$.

  • An ideal voltage source has $R_S = 0$, yielding an infinite slope (vertical line at $V = V_S$).

  • To approximate an ideal voltage source, $R_S$ must be as small as possible.

• Parallel Compatibility: Two realistic voltage sources ($V_{S1}, R_{S1}$ and $V_{S2}, R_{S2}$) can be connected in parallel because internal series resistors absorb voltage differences, allowing KVL to be satisfied.

Realistic Current Sources

• Circuit Model: An ideal current source $I_S$ in parallel with an internal parallel resistance $R_P$.

• Terminal Current Supplied Equation (via KCL at top node $A$):   $I = I_S - \frac{V}{R_P}$

• Parameter Requirements for High Quality:

  • To approach ideal current source behavior ($I \approx I_S$), the term $\frac{V}{R_P}$ must be minimized.

  • Thus, $R_P$ must be as large as possible ($R_P \rightarrow \infty$).

• Practical Values:

  • A practical, realistic lab-grade current source has $R_P$ on the order of Gigaohms ($10^9\,\text{Ohm}$ to $10^{10}\,\text{Ohm}$).