1.3 Circuit Analysis: Series, Parallel & Combination Circuits

1.3 Circuit Analysis: Series, Parallel & Combination Circuits

Fundamental Circuit Topologies and Series Configurations

Electrical systems route current through networks classified as series, parallel, or combination circuits. A series circuit provides a single continuous pathway for electron flow. Because charge cannot accumulate along an unbroken loop, current is constant across every component:

• Current Law: $I_{\text{total}} = I_1 = I_2 = I_3 = \dots = I_n$
• Total Resistance: Opposition to current flow is cumulative, meaning total resistance equals the sum of individual resistances: $R_{\text{total}} = R_1 + R_2 + R_3 + \dots + R_n$
• Voltage Division: Applied voltage distributes across components proportionally to individual resistance: $E_{\text{total}} = E_1 + E_2 + E_3 + \dots + E_n$ $E_x = E_{\text{total}} \times \left(\frac{R_x}{R_{\text{total}}}\right)$

In field installations, loose terminations or corroded splices act as high-resistance series elements, causing unwanted voltage drop and dangerous localized thermal buildup.

Parallel Circuit Configurations and Branch Laws

A parallel circuit connects multiple devices across identical common nodes, providing separate conductive pathways for current:

• Voltage Law: Because all branches connect across the same terminals, full source voltage appears across every branch: $E_{\text{total}} = E_1 = E_2 = E_3 = \dots = E_n$
• Current Law: Total current equals the sum of individual branch currents: $I_{\text{total}} = I_1 + I_2 + I_3 + \dots + I_n$
• Equivalent Resistance: Adding parallel branches creates additional paths, reducing overall opposition to current flow. Equivalent circuit resistance ($R_{\text{eq}}$) is always less than the lowest branch resistance:
  • Reciprocal Formula: $\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \dots + \frac{1}{R_n} \implies R_{\text{eq}} = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2} + \dots + \frac{1}{R_n}}$
  • Product-Over-Sum Formula (two branches): $R_{\text{eq}} = \frac{R_1 \times R_2}{R_1 + R_2}$
  • Equal Resistance Formula (for $n$ identical resistors $R$): $R_{\text{eq}} = \frac{R}{n}$

Kirchhoff's Laws in Electrical Circuit Analysis

Complex networks follow Gustav Kirchhoff's two governing theorems:

• Kirchhoff's Current Law (KCL / Node Rule): The algebraic sum of currents entering and exiting an electrical node equals zero ($\sum I_{\text{in}} = \sum I_{\text{out}}$). Current entering a junction must balance across departing conductor paths.
• Kirchhoff's Voltage Law (KVL / Loop Rule): The algebraic sum of voltages around any closed conductive loop equals zero ($\sum E_{\text{source}} - \sum V_{\text{drops}} = 0$). KVL is applied in field diagnostics to verify loop drops and ground-fault paths.

Combination (Series-Parallel) Network Reduction Technique

Combination circuits require systematic five-step reduction:

1. Identify Isolated Clusters: Locate purely parallel or series resistor groups furthest from the supply.
2. Calculate Equivalent Values: Resolve isolated clusters into single equivalent resistances.
3. Simplify Schematic: Repeat reductions until collapsing into a single equivalent resistance ($R_{\text{total}}$).
4. Determine Source Current: Calculate source current using Ohm's Law ($I_{\text{total}} = E_{\text{source}} / R_{\text{total}}$).
5. Expand Backward: Trace backward through the circuit to solve individual component voltage drops and branch currents.

Conductor Resistance and Circular Mil Area

Every electrical conductor possesses internal resistance that creates voltage drop over distance. Resistance depends on material resistivity, length, and cross-sectional area:

$R = \frac{K \times L}{\text{CM}}$

• $K$: Specific resistivity in ohms per circular mil-foot at $75^\circ\text{C}$:
  • Copper: $K = 12.9,\Omega\cdot\text{cmil/ft}$
  • Aluminum: $K = 21.2,\Omega\cdot\text{cmil/ft}$
• $L$: One-way length of the conductor run in feet.
• $\text{CM}$: Conductor cross-sectional area in circular mils ($1\text{ mil} = 0.001\text{ inch}$; $\text{CM} = d^2$). Exact areas are listed in NEC Chapter 9, Table 8.

NEC Voltage Drop Standards and Formulas

Excessive voltage drop causes motor overheating, relay chatter, and lumen loss. NEC Informational Notes 210.19(A) Note 4 and 215.2(A)(1) Note 2 recommend sizing conductors so voltage drop does not exceed:

• 3% on branch circuits (or feeders) at the farthest connected outlet.
• 5% total overall voltage drop combining feeder and branch conductors to the final load.

Standard Journeyman calculation formulas:

• Single-Phase Circuits: $VD = \frac{2 \times K \times I \times D}{\text{CM}}$
• Three-Phase Balanced Circuits: $VD = \frac{\sqrt{3} \times K \times I \times D}{\text{CM}} = \frac{1.732 \times K \times I \times D}{\text{CM}}$
• Minimum Conductor Sizing for Maximum Allowable Voltage Drop:
  • Single-Phase: $\text{CM} = \frac{2 \times K \times I \times D}{VD_{\text{allowed}}}$
  • Three-Phase: $\text{CM} = \frac{1.732 \times K \times I \times D}{VD_{\text{allowed}}}$

Step-by-Step Worked Calculations

Worked Calculation 1: Combination Circuit Reduction

A 120-volt source supplies a circuit containing a $10,\Omega$ series resistor ($R_1$) connected in series with a parallel pair consisting of $R_2 = 30,\Omega$ and $R_3 = 60,\Omega$.

1. Equivalent parallel resistance: $R_{23} = \frac{30 \times 60}{30 + 60} = \frac{1800}{90} = 20\,\Omega$
2. Total circuit resistance: $R_{\text{total}} = R_1 + R_{23} = 10 + 20 = 30\,\Omega$
3. Total circuit current: $I_{\text{total}} = \frac{120\text{ V}}{30\,\Omega} = 4.0\text{ Amperes}$
4. Component voltage drops: $E_1 = 4.0\text{ A} \times 10\,\Omega = 40\text{ Volts}$ $E_{23} = 120\text{ V} - 40\text{ V} = 80\text{ Volts}$
5. Branch currents: $I_2 = \frac{80\text{ V}}{30\,\Omega} = 2.67\text{ A}; \quad I_3 = \frac{80\text{ V}}{60\,\Omega} = 1.33\text{ A}$ Verifying KCL: $2.67\text{ A} + 1.33\text{ A} = 4.0\text{ Amperes}$.

Worked Calculation 2: Feeder Sizing for Voltage Drop Compliance

A 240-volt single-phase feeder supplies a continuous 45-ampere load over 200 feet using copper conductors ($K = 12.9$). Limit voltage drop to 3%.

1. Allowable voltage drop: $VD_{\text{allowed}} = 240\text{ V} \times 0.03 = 7.2\text{ Volts}$
2. Required circular mils: $\text{CM} = \frac{2 \times 12.9 \times 45 \times 200}{7.2} = \frac{232,200}{7.2} = 32,250\text{ cmil}$
3. NEC Chapter 9, Table 8 Evaluation:
   • #6 AWG copper = $26,240\text{ cmil}$ ($VD = 8.85\text{V}$ or $3.69%$, exceeds limit)
   • #4 AWG copper = $41,740\text{ cmil}$ ($VD = 5.56\text{V}$ or $2.32%$, compliant)

Conclusion: Install #4 AWG copper conductors to meet the 3% threshold.