2.2 Series, Parallel, and Series-Parallel Circuits and Power Calculations

Key Takeaways

  • In a series circuit, current is identical through all components, total resistance is the sum of all individual resistances, and applied voltage divides proportionally across loads.
  • In a parallel circuit, voltage is identical across all branches, total current equals the sum of individual branch currents, and total equivalent resistance is always less than the lowest branch resistance.
  • HVAC safety controls (pressure switches, limit switches, freeze stats) are always wired in series with the control load, while primary power loads (compressors, condenser fans, blower motors) are wired in parallel across line voltage.
  • Real Power (Watts) performs actual work; Apparent Power (Volt-Amps) is the total supplied power; Reactive Power (VARs) sustains electromagnetic fields in inductive motor windings.
  • Power Factor (PF = Real Power / Apparent Power = cos θ) measures electrical efficiency; run capacitors provide leading reactive power to correct lagging power factors toward unity.
Last updated: August 2026

2.2 Series, Parallel, and Series-Parallel Circuits and Power Calculations

Every electrical schematic in an air conditioner, heat pump, refrigeration rack, or gas furnace is composed of basic circuit arrangements: series, parallel, or complex combination (series-parallel) networks. Understanding how voltage, current, resistance, and power behave across these configurations is essential for diagnosing open circuits, component failures, and electrical power quality issues.


1. Series Circuit Principles and Laws

A series circuit provides only one single, continuous path for electron flow. All components are connected end-to-end like links in a chain.

The Three Fundamental Laws of Series Circuits

  1. Current is Constant Throughout: Because there is only one path, the exact same rate of electron flow passes through every component: Itotal=I1=I2=I3==InI_{\text{total}} = I_1 = I_2 = I_3 = \dots = I_n
  2. Total Resistance is Additive: Each added resistive load increases the total opposition to electron flow in the circuit: Rtotal=R1+R2+R3++RnR_{\text{total}} = R_1 + R_2 + R_3 + \dots + R_n
  3. Voltage Divides Across Loads (Kirchhoff's Voltage Law - KVL): The total applied source voltage equals the sum of the individual voltage drops across each series load, with voltage dividing in direct proportion to each component's resistance ($E_n = I \times R_n$): Etotal=E1+E2+E3++EnE_{\text{total}} = E_1 + E_2 + E_3 + \dots + E_n

Component Faults in Series Circuits

  • Open Circuit Fault: If any single switch or component opens in a series string, total current immediately drops to zero ($I = 0\text{ A}$). The entire circuit stops operating. The full source voltage appears across the open contact.
  • Short Circuit Fault: If one component in a series string shorts out ($R = 0\ \Omega$), total circuit resistance decreases ($R_{\text{total}} \downarrow$), causing total current to increase ($I_{\text{total}} \uparrow$). The remaining components experience higher voltage drops and higher current, which can lead to thermal overload.

HVAC Application: The Series Safety String

In HVAC systems, safety and operating limit controls are always wired in series with the load they protect (e.g., compressor contactor coil, gas valve). These safety devices include:

  • High-pressure cutout switch (opens on excessive head pressure)
  • Low-pressure cutout / loss-of-charge switch (opens on low suction pressure)
  • Compressor internal thermal overload / discharge line thermostat
  • High limit switch (opens on furnace heat exchanger overheating)
  • Flame rollout safety switch (opens on burner draft failure)
  • Condensate overflow float switch
  • Freeze protection thermostat (freeze stat on water chillers or evaporator coils)

Why? If any single safety control detects an unsafe operating condition and opens its contacts, the single path for control current is severed, immediately de-energizing the contactor coil or gas valve and protecting equipment and building occupants.

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HVAC Control Circuit Architecture: Series Safety Chain Feeding Parallel Operational Loads

2. Parallel Circuit Principles and Laws

A parallel circuit contains two or more distinct branches connected across the same common voltage source, providing multiple independent paths for electron flow.

The Three Fundamental Laws of Parallel Circuits

  1. Voltage is Constant Across All Branches: Every parallel branch connects directly across the full line voltage: Etotal=E1=E2=E3==EnE_{\text{total}} = E_1 = E_2 = E_3 = \dots = E_n
  2. Current is Additive (Kirchhoff's Current Law - KCL): Total current entering a common junction equals the sum of the individual branch currents flowing out of that junction: Itotal=I1+I2+I3++InI_{\text{total}} = I_1 + I_2 + I_3 + \dots + I_n
  3. Total Equivalent Resistance Decreases: Adding parallel branches creates additional paths for current flow, reducing the overall equivalent resistance of the network. Total resistance is always less than the resistance of the smallest individual branch: 1Rtotal=1R1+1R2+1R3++1Rn\frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots + \frac{1}{R_n}

Mathematical Shortcuts for Parallel Resistors

  • Product Over Sum Formula (for exactly two parallel resistors): Rtotal=R1×R2R1+R2R_{\text{total}} = \frac{R_1 \times R_2}{R_1 + R_2}
  • Equal Value Resistors in Parallel ($N$ identical resistors of value $R$): Rtotal=RNR_{\text{total}} = \frac{R}{N} Example: Three $30\ \Omega$ crankcase heaters wired in parallel produce $R_{\text{total}} = 30 / 3 = 10\ \Omega$.

Component Faults in Parallel Circuits

  • Open Branch Fault: If one branch opens (e.g., a burned-out crankcase heater or outdoor fan motor), only that specific branch ceases to function. All other parallel branches continue to receive full supply voltage and operate normally.
  • Short Circuit Fault: If a dead short ($R = 0\ \Omega$) occurs in any single parallel branch, the entire power supply rail is shorted directly to neutral or ground. Total resistance drops to near zero, creating massive fault current that instantly opens the branch overcurrent protective device (fuse or circuit breaker).

HVAC Application: Parallel Line-Voltage Power Loads

All primary operating loads in residential and commercial HVAC units—such as the compressor motor, outdoor condenser fan motor, indoor blower motor, crankcase heater, and electric heat elements—are wired in parallel across the main supply voltage. This ensures every motor and heater receives full design operating voltage (e.g., 240 VAC) regardless of which other components are cycling on or off.

3. Series-Parallel (Combination) Circuits

Practical HVAC machinery combines series and parallel networks into series-parallel combination circuits. In these circuits, switches and safety controls are wired in series with parallel load banks, or complex relay logic routes power through interconnected branches.

Analysis Method: Circuit Reduction

To analyze a combination circuit:

  1. Identify purely parallel sub-branches and calculate their single equivalent resistance ($R_p$).
  2. Replace the parallel branches with their equivalent resistance ($R_p$) to form a simplified series circuit.
  3. Sum all remaining series resistances ($R_{\text{total}} = R_s + R_p$).
  4. Use Ohm's Law to determine total current ($I_{\text{total}} = E / R_{\text{total}}$).
  5. Calculate the voltage drop across the series section ($V_s = I_{\text{total}} \times R_s$).
  6. Subtract the series voltage drop from total voltage to find the remaining voltage across the parallel branch bank ($V_p = E_{\text{total}} - V_s$).
  7. Calculate individual branch currents ($I_n = V_p / R_n$).

4. Worked Calculation Examples

Example 1: Parallel Electric Heating Element Bank

Problem: An electric furnace contains three resistive heating elements wired in parallel across a 240 VAC power supply. The hot resistances are $R_1 = 20\ \Omega$, $R_2 = 24\ \Omega$, and $R_3 = 30\ \Omega$.

  1. Calculate the total equivalent resistance ($R_{\text{total}}$) of the heating bank.
  2. Calculate the individual branch currents ($I_1, I_2, I_3$).
  3. Calculate the total line current ($I_{\text{total}}$).
  4. Calculate the total power consumed in kW and total heat output in BTU/hr.

Solution Steps:

  • Step 1: Calculate Total Equivalent Resistance: 1Rtotal=120+124+130=0.050+0.04167+0.03333=0.125 Ω1\frac{1}{R_{\text{total}}} = \frac{1}{20} + \frac{1}{24} + \frac{1}{30} = 0.050 + 0.04167 + 0.03333 = 0.125\ \Omega^{-1} Rtotal=10.125=8.0 ΩR_{\text{total}} = \frac{1}{0.125} = 8.0\ \Omega
  • Step 2: Calculate Individual Branch Currents: I1=240 V20 Ω=12.0 A,I2=240 V24 Ω=10.0 A,I3=240 V30 Ω=8.0 AI_1 = \frac{240\text{ V}}{20\ \Omega} = 12.0\text{ A}, \quad I_2 = \frac{240\text{ V}}{24\ \Omega} = 10.0\text{ A}, \quad I_3 = \frac{240\text{ V}}{30\ \Omega} = 8.0\text{ A}
  • Step 3: Calculate Total Line Current: Itotal=I1+I2+I3=12.0 A+10.0 A+8.0 A=30.0 AI_{\text{total}} = I_1 + I_2 + I_3 = 12.0\text{ A} + 10.0\text{ A} + 8.0\text{ A} = 30.0\text{ A} Verification via Ohm's Law: $I_{\text{total}} = E / R_{\text{total}} = 240\text{ V} / 8.0\ \Omega = 30.0\text{ A}$ (Matches).
  • Step 4: Calculate Power and Heat Output: Ptotal=E×Itotal=240 V×30.0 A=7,200 Watts=7.2 kWP_{\text{total}} = E \times I_{\text{total}} = 240\text{ V} \times 30.0\text{ A} = 7,200\text{ Watts} = 7.2\text{ kW} Qheat=7.2 kW×3,412.14 BTU/kW=24,567.4 BTU/hrQ_{\text{heat}} = 7.2\text{ kW} \times 3,412.14\text{ BTU/kW} = 24,567.4\text{ BTU/hr}

5. AC Power Components: Watts, Volt-Amperes, VARs, and Power Factor

In AC circuits containing inductive loads (compressor motors, fan motors, contactor coils) or capacitive loads (run capacitors, start capacitors), voltage and current sine waves do not peak at the exact same instant. This phase displacement creates three distinct types of electrical power:

The Three Components of AC Power

  1. Real Power (True Power - $P$):
    • Measured in Watts (W) or Kilowatts (kW).
    • The actual rate at which electrical energy is converted into usable mechanical work, heat, or light by resistive components.
    • Formula: $P = E \times I \times \cos(\theta)$
  2. Apparent Power ($S$):
    • Measured in Volt-Amperes (VA) or Kilovolt-Amperes (kVA).
    • The total "raw" power supplied to the circuit by the utility or control transformer, calculated as the simple scalar product of total RMS voltage and total RMS current.
    • Formula: $S = E \times I$
  3. Reactive Power ($Q$):
    • Measured in Volt-Amperes Reactive (VAR) or kVAR.
    • The non-working power that oscillates back and forth between the power source and the magnetic fields of inductors (motor windings, transformers) or the electric fields of capacitors.
    • Formula: $Q = E \times I \times \sin(\theta)$

The Power Triangle and Pythagorean Relationship

Apparent Power2=Real Power2+Reactive Power2\text{Apparent Power}^2 = \text{Real Power}^2 + \text{Reactive Power}^2 S=P2+Q2S = \sqrt{P^2 + Q^2}

Power MetricSymbolUnitPhysical Manifestation in HVAC
Real Power$P$Watts (W) / kWHeat from electric strip elements, mechanical shaft torque of motors, utility meter billing energy.
Apparent Power$S$Volt-Amps (VA) / kVARating of step-down control transformers (40VA, 75VA), total ampacity demand on wires and breakers.
Reactive Power$Q$VAR / kVARMagnetic flux sustaining stator rotation in induction motors (inductive VARs) vs electrostatic field in run capacitors (capacitive VARs).

Power Factor (PF) and Its Practical Importance

Power Factor is the ratio of Real Power (Watts) doing actual work to the total Apparent Power (VA) supplied by the electrical source: Power Factor (PF)=Real Power (Watts)Apparent Power (VA)=cos(θ)\text{Power Factor (PF)} = \frac{\text{Real Power (Watts)}}{\text{Apparent Power (VA)}} = \cos(\theta) Percentage PF=(WattsVA)×100%\text{Percentage PF} = \left(\frac{\text{Watts}}{\text{VA}}\right) \times 100\%

Power Factor Correction in HVAC

  • Inductive Loads (Lagging PF): Single-phase and three-phase induction motors draw magnetizing current that causes the current waveform to lag behind the voltage waveform, resulting in a poor power factor (typically 0.70 to 0.85). This low PF forces the motor to draw higher total line current (Amps) for the same delivered mechanical horsepower.
  • Capacitive Correction (Leading PF): Motor permanent split capacitor (PSC) run capacitors supply leading reactive power (capacitive VARs) that directly counteracts and cancels the lagging inductive VARs of the motor windings. This brings total voltage and current closer into phase, raising the operating power factor to 0.90–0.98, which reduces total line current, lowers conductor $I^2 R$ heat losses, and prevents premature nuisance tripping of circuit breakers.
Test Your Knowledge

A technician measures a single-phase heat pump compressor drawing 15.0 Amperes at 240 VAC. A wattmeter connected to the circuit reads a real power consumption of 3,060 Watts. What is the apparent power and power factor of this operating compressor?

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Test Your Knowledge

Two electric heater elements with resistances of 12 Ohms and 24 Ohms are connected in parallel across a 240 VAC supply. What is the total equivalent circuit resistance and the total current drawn from the line?

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Test Your Knowledge

A technician troubleshooting a non-functioning air conditioning condensing unit finds the 24 VAC compressor contactor coil is de-energized. While the thermostat is calling for cooling (Y energized), a voltmeter placed across the high-pressure switch reads 24.0 VAC, while across the low-pressure switch it reads 0.0 VAC. What is the cause of the system failure?

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Test Your Knowledge

Why are permanent split capacitor (PSC) motor run capacitors connected across the start and run windings of HVAC fan and compressor motors?

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