6.1 Ohm's Law, Power Law & DC Circuit Analysis

6.1 Ohm's Law, Power Law & DC Circuit Analysis

Quick Answer: Fire alarm field calculations rely on two core electrical relationships: Ohm's Law ($V = I \times R$) and Watt's Power Law ($P = V \times I = I^2 R = V^2 / R$). Commercial systems operate primarily on 24VDC nominal secondary power. In series circuits (such as supervisory wiring loops), current is constant throughout and resistances sum directly ($R_{\text{total}} = R_1 + R_2 + \dots$). In parallel circuits (such as notification appliance circuits and detector alarm contacts), voltage is constant across branches and branch currents sum ($I_{\text{total}} = I_1 + I_2 + \dots$). Conventional Initiating Device Circuits (IDCs) combine both into a series-parallel topology: a series End-of-Line Resistor (EOLR) establishes a 3–6 mA supervisory current, while closing a parallel normally-open detector contact creates a low-resistance shunt that surges current to 40–90 mA, driving the Fire Alarm Control Unit (FACU) into alarm.

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Foundational Electrical Units in Fire Alarm Systems

Mastering commercial fire alarm installation and passing the Oklahoma Commercial Fire Alarm Technician examination (CareerTech Assessment 4011) requires fluent comprehension of electrical units and their physical interactions. Life safety signaling infrastructure consists of low-voltage DC control circuits, power-limited notification circuits, and high-voltage AC utility branch circuits.

Every electrical circuit involves four interrelated quantities:

1. Electromotive Force / Voltage ($V$ or $E$): The electrical potential difference that drives electrons through a conductor, measured in Volts (V). Commercial fire alarm control units operate primarily on 24VDC nominal filtered or full-wave rectified direct current derived from dual secondary power supplies (internal transformers and sealed secondary batteries). Line-voltage AC power supplies operate at 120VAC nominal, 60 Hz single-phase.
2. Current ($I$): The rate of electrical charge flow past a point in a circuit, measured in Amperes (A) or milliamperes (mA). One Ampere represents one Coulomb of charge moving per second ($1\text{ A} = 1,000\text{ mA}$). Fire alarm supervisory standby currents are generally measured in milliamperes (e.g., 35 mA for an LCD annunciator), whereas notification loads and active alarm currents are measured in Amperes (e.g., 1.75 A for a strobe appliance branch).
3. Resistance ($R$): The opposition to electron flow within a material or device, measured in Ohms ($\Omega$). In fire alarm circuits, resistance originates from wire conductor composition, copper gauge thickness, circuit length, internal component coils, relay contacts, and supervisory End-of-Line Resistors (EOLRs).
4. Power ($P$): The rate at which electrical energy is transformed into another energy form (such as heat, light, or acoustic sound waves), measured in Watts (W) or milliwatts (mW).

┌─────────────────────────────────────────────────────────────────────────────┐
│                     CORE ELECTRICAL PARAMETERS IN FIRE ALARM                 │
├──────────────┬────────┬──────────────┬──────────────────────────────────────┤
│ Quantity     │ Symbol │ Unit         │ Typical Fire Alarm Application       │
├──────────────┼────────┼──────────────┼──────────────────────────────────────┤
│ Voltage      │ V or E │ Volts (V)    │ 24VDC system bus; 120VAC primary     │
│ Current      │ I      │ Amperes (A)  │ Standby: 10–50 mA; Alarm: 1.0–3.0 A  │
│ Resistance   │ R      │ Ohms (Ω)     │ Cable loop: 2–20 Ω; EOLR: 2.2–10 kΩ  │
│ Power        │ P      │ Watts (W)    │ Audio amps: 25–300 W; EOLR: 0.5 W    │
└──────────────┴────────┴──────────────┴──────────────────────────────────────┘

Metric Conversions: Amperes vs. Milliamperes

The most pervasive source of mathematical errors on the Oklahoma licensing exam is failing to convert between milliamperes (mA) and Amperes (A) before applying circuit formulas. All standard electrical formulas require current to be expressed in Amperes:

$\text{Current in Amperes (A)} = \frac{\text{Current in Milliamperes (mA)}}{1,000}$ $\text{Current in Milliamperes (mA)} = \text{Current in Amperes (A)} \times 1,000$

[!WARNING] Exam Trap: Multiplying milliamperes directly into Ohm's Law or battery capacity formulas will produce answers off by a factor of 1,000. For instance, calculating battery standby Amp-Hours with 350 mA as $350 \times 24 = 8,400\text{ Ah}$ will result in choosing a catastrophic distractor answer on the exam. The calculation must always begin by converting: $350\text{ mA} = 0.350\text{ A}$, yielding $0.350 \times 24 = 8.4\text{ Ah}$.

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Ohm's Law: Formulas, Derivations, and The Ohm's Wheel

Codified by Georg Ohm in 1827, Ohm's Law states that the current flowing through a conductor between two points is directly proportional to the voltage across the two points and inversely proportional to the resistance between them.

                         ┌───────────────┐
                         │   VOLTAGE     │
                         │     (V)       │
                         │    V = I×R    │
                         └───────┬───────┘
                                ╱ ╲
                               ╱   ╲
                              ╱     ╲
                    ┌────────┴───────┴────────┐
                    │ CURRENT     RESISTANCE  │
                    │   (I)           (R)     │
                    │ I = V/R       R = V/I   │
                    └─────────────────────────┘

The three mathematical derivations of Ohm's Law are:

1. Solving for Voltage ($V$): $V = I \times R$ Application: Determining the voltage drop across a notification circuit's wire resistance under maximum alarm current.
2. Solving for Current ($I$): $I = \frac{V}{R}$ Application: Calculating the supervisory current flowing through a Class B circuit terminated with an End-of-Line Resistor.
3. Solving for Resistance ($R$): $R = \frac{V}{I}$ Application: Determining the internal DC resistance of a solenoid, relay coil, or unknown wire loop using multimeter voltage and current measurements.

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Watt's Power Law and Component Wattage Ratings

Watt's Law defines the relationship between electrical power ($P$), electromotive force ($V$), and current ($I$). In direct current circuits, power is the direct product of voltage and current:

$P = V \times I$

By substituting Ohm's Law ($V = I \times R$ and $I = V / R$) into Watt's Law, two additional power equations are derived:

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

┌─────────────────────────────────────────────────────────────────────────────┐
│                     WATT'S LAW & OHM'S LAW FORMULA MATRIX                   │
├──────────────────┬──────────────────┬──────────────────┬────────────────────┤
│ To Calculate:    │ Using V and I    │ Using I and R    │ Using V and R      │
├──────────────────┼──────────────────┼──────────────────┼────────────────────┤
│ Voltage (V)      │ V = P / I        │ V = I × R        │ V = √(P × R)       │
│ Current (I)      │ I = P / V        │ I = √(P / R)     │ I = V / R          │
│ Resistance (R)   │ R = V² / P       │ R = P / I²       │ R = V / I          │
│ Power (P)        │ P = V × I        │ P = I² × R       │ P = V² / R         │
└──────────────────┴──────────────────┴──────────────────┴────────────────────┘

Practical Application: EOLR Wattage Dissipation and Safety Derating

When a resistor is placed in an electrical circuit, it converts electrical energy directly into thermal energy (heat). Resistors are rated by their ohmic resistance and their maximum continuous power dissipation capability (e.g., $1/8\text{ W}$ [0.125W], $1/4\text{ W}$ [0.25W], $1/2\text{ W}$ [0.50W], $1\text{ W}$, $2\text{ W}$).

If a resistor dissipates power close to its maximum rating, its internal temperature rises significantly. This thermal stress causes the resistance value to drift over time, which destabilizes the FACU's supervisory threshold, triggers intermittent "Trouble" signals, and can cause the resistor body to crack or char.

[!IMPORTANT] The 2× Power Derating Rule: In professional fire alarm engineering and field practice, a resistor must have a physical power rating at least two times (2×) greater than the maximum continuous power it will dissipate under normal operating conditions: $P_{\text{rated}} \ge 2 \times P_{\text{actual}}$

Worked Example: Sizing an EOLR on a 24VDC Circuit

Problem: A conventional zone requires a $2.2\text{ k}\Omega$ ($2,200,\Omega$) End-of-Line Resistor connected across a 24VDC supervisory circuit. What is the actual power dissipated by the resistor, and what minimum wattage rating must the installer select?

1. Identify Known Values: $V = 24\text{ VDC}$, $R = 2,200,\Omega$.
2. Calculate Actual Power Dissipation ($P$): $P = \frac{V^2}{R} = \frac{24^2}{2,200} = \frac{576}{2,200} = 0.2618\text{ W} \approx 0.262\text{ W}$
3. Apply 2× Safety Derating Factor: $P_{\text{minimum rating}} = 0.2618\text{ W} \times 2 = 0.5236\text{ W}$
4. Select Commercial Resistor Rating:
   • A $1/4\text{ W}$ ($0.25\text{ W}$) resistor will overheat and fail because $0.262\text{ W} > 0.25\text{ W}$.
   • A $1/2\text{ W}$ ($0.50\text{ W}$) resistor is marginally undersized under the strict 2× rule ($0.50\text{ W} < 0.524\text{ W}$).
   • A $1\text{ W}$ resistor provides robust thermal stability and complies with professional standards.

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DC Series Circuit Analysis

A series circuit provides only a single electrical pathway for current flow. In fire alarm systems, supervisory loops, tamper switch circuits, and notification appliance wiring loops operate as series circuits during supervisory monitoring.

        ┌───────[ + 24VDC Source ]───────┐
        │                                │
        ▼                                ▼
     ( + OUT )                       ( - RETURN )
        │                                │
        ├───/\[ R_wire1 = 12 Ω ]/\───────┤
        │                                │
        ├───/\[ R_device1 = 8 Ω ]/\──────┤
        │                                │
        └───/\[ R_EOLR = 4700 Ω ]/\──────┘
                 Single Current Loop

The Three Fundamental Rules of DC Series Circuits

1. Current is Constant at All Points: Because there is only one continuous path for electrons to travel, the current entering any component must equal the current leaving that component: $I_{\text{total}} = I_1 = I_2 = I_3 = \dots = I_n$
2. Total Resistance is the Sum of All Resistances: The total resistance opposing current flow is the algebraic sum of every individual resistance in the loop: $R_{\text{total}} = R_1 + R_2 + R_3 + \dots + R_n$
3. Voltage Drops Sum to the Source Voltage (Kirchhoff's Voltage Law): Kirchhoff's Voltage Law (KVL) dictates that the sum of all individual voltage drops across components in a closed loop must equal the total applied source voltage: $V_{\text{source}} = V_1 + V_2 + V_3 + \dots + V_n$ $V_k = I_{\text{total}} \times R_k$

Step-by-Step Worked Example: Class B Supervisory Circuit

Scenario: A 24VDC Class B initiating circuit has $1,500\text{ feet}$ of 18 AWG solid copper wire. The total round-trip wire resistance ($R_{\text{wire}}$) measures $24,\Omega$. The circuit is terminated with a $4.7\text{ k}\Omega$ ($4,700,\Omega$) End-of-Line Resistor.

Calculate:

1. The total circuit resistance ($R_{\text{total}}$).
2. The normal quiescent supervisory current ($I_{\text{supervisory}}$).
3. The voltage drop across the wire conductors ($V_{\text{wire}}$).
4. The voltage measured directly across the EOLR ($V_{\text{EOLR}}$).

Step 1: Calculate Total Circuit Resistance ($R_{\text{total}}$): $R_{\text{total}} = R_{\text{wire}} + R_{\text{EOLR}} = 24\,\Omega + 4,700\,\Omega = 4,724\,\Omega$

Step 2: Calculate Supervisory Current ($I$): $I = \frac{V_{\text{source}}}{R_{\text{total}}} = \frac{24\text{ V}}{4,724\,\Omega} = 0.0050804\text{ A} = 5.08\text{ mA}$

Step 3: Calculate Voltage Drop Across the Wire Conductor ($V_{\text{wire}}$): $V_{\text{wire}} = I \times R_{\text{wire}} = 0.0050804\text{ A} \times 24\,\Omega = 0.1219\text{ V} \approx 0.12\text{ V}$

Step 4: Calculate Voltage Dropped Across the EOLR ($V_{\text{EOLR}}$): $V_{\text{EOLR}} = I \times R_{\text{EOLR}} = 0.0050804\text{ A} \times 4,700\,\Omega = 23.878\text{ V} \approx 23.88\text{ V}$

Verification via Kirchhoff's Voltage Law: $V_{\text{source}} = V_{\text{wire}} + V_{\text{EOLR}} = 0.12\text{ V} + 23.88\text{ V} = 24.00\text{ V}$ (Proof holds)

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DC Parallel Circuit Analysis

A parallel circuit provides two or more independent branches across a common voltage source. In fire alarm systems, notification appliances (horns, strobes) and initiating detector alarm contacts connect in parallel across circuit conductors.

         ┌───────────────[ + 24VDC Supply ]───────────────┐
         │                       │                        │
         ▼                       ▼                        ▼
     Branch 1                Branch 2                 Branch 3
     [ Strobe 1 ]            [ Strobe 2 ]             [ Strobe 3 ]
     I1 = 120 mA             I2 = 150 mA              I3 = 180 mA
     R1 = 200 Ω              R2 = 160 Ω               R3 = 133.3 Ω
         ▲                       ▲                        ▲
         │                       │                        │
         └───────────────[ - Common Ground ]──────────────┘
               Itotal = I1 + I2 + I3 = 450 mA (0.45 A)

The Three Fundamental Rules of DC Parallel Circuits

1. Voltage is Identical Across All Branches: Every parallel branch connects directly across the common supply rails. Therefore, each branch experiences identical electromotive force: $V_{\text{total}} = V_1 = V_2 = V_3 = \dots = V_n$
2. Total Current is the Sum of Branch Currents (Kirchhoff's Current Law): Kirchhoff's Current Law (KCL) dictates that the electrical current entering any junction must equal the total current leaving that junction. Total circuit current is the direct sum of currents drawn by each branch: $I_{\text{total}} = I_1 + I_2 + I_3 + \dots + I_n$
3. Equivalent Resistance is Always Less than the Smallest Branch Resistance: Adding parallel branches creates additional paths for electron flow, decreasing overall circuit resistance: $\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots + \frac{1}{R_n}$ $R_{\text{eq}} = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2} + \dots + \frac{1}{R_n}}$

[!NOTE] Two-Resistor Shortcut: For exactly two parallel branches, use the product-over-sum formula: $R_{\text{eq}} = \frac{R_1 \times R_2}{R_1 + R_2}$

Step-by-Step Worked Example: Parallel Notification Branch

Scenario: An auxiliary NAC power supply supplies 24VDC to three notification appliances connected in parallel:

• Appliance 1 (Corridor Strobe): draws $120\text{ mA}$ ($0.120\text{ A}$)
• Appliance 2 (Office Strobe): draws $150\text{ mA}$ ($0.150\text{ A}$)
• Appliance 3 (Lobby Horn/Strobe): draws $180\text{ mA}$ ($0.180\text{ A}$)

Calculate:

1. The total current demand on the power supply ($I_{\text{total}}$).
2. The internal equivalent DC resistance of each individual appliance ($R_1, R_2, R_3$).
3. The total equivalent resistance of the entire parallel circuit ($R_{\text{eq}}$).

Step 1: Calculate Total Current ($I_{\text{total}}$): $I_{\text{total}} = 0.120\text{ A} + 0.150\text{ A} + 0.180\text{ A} = 0.450\text{ A} = 450\text{ mA}$

Step 2: Calculate Individual Device Equivalent Resistances: $R_1 = \frac{V}{I_1} = \frac{24\text{ V}}{0.120\text{ A}} = 200\,\Omega$ $R_2 = \frac{V}{I_2} = \frac{24\text{ V}}{0.150\text{ A}} = 160\,\Omega$ $R_3 = \frac{V}{I_3} = \frac{24\text{ V}}{0.180\text{ A}} = 133.33\,\Omega$

Step 3: Calculate Equivalent Circuit Resistance ($R_{\text{eq}}$): Using Ohm's Law with total circuit values: $R_{\text{eq}} = \frac{V_{\text{total}}}{I_{\text{total}}} = \frac{24\text{ V}}{0.450\text{ A}} = 53.33\,\Omega$

Verification via Reciprocal Formula: $\frac{1}{R_{\text{eq}}} = \frac{1}{200} + \frac{1}{160} + \frac{1}{133.33} = 0.0050 + 0.00625 + 0.0075 = 0.01875$ $R_{\text{eq}} = \frac{1}{0.01875} = 53.33\,\Omega$ (Exact match; $53.33,\Omega$ is lower than the smallest branch resistance of $133.33,\Omega$)

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Series-Parallel Dynamics: Class B Initiating Device Circuits (IDCs)

Commercial conventional fire alarm systems utilize a hybrid series-parallel circuit topology on Initiating Device Circuits (IDCs). Understanding how the Fire Alarm Control Unit (FACU) monitors circuit states through voltage and current transitions is a foundational competency tested on the Oklahoma exam.

┌─────────────────────────────────────────────────────────────────────────────┐
│                     CLASS B IDC SERIES-PARALLEL CIRCUIT                     │
├─────────────────────────────────────────────────────────────────────────────┤
│                                                                             │
│   FACU (+) ────┬───────────────────┬───────────────────┐                    │
│   (24VDC)      │                   │                   │                    │
│              ┌─┴─┐               ┌─┴─┐               ┌─┴─┐                  │
│              │   │ Pull Station  │   │ Smoke Det.    │   │ End-of-Line      │
│              │   │ (N.O. Contact)│   │ (N.O. Contact)│ R │ Resistor (EOLR)  │
│              │   │               │   │ [Alarm Res.]  │   │ (e.g., 4.7 kΩ)   │
│              └─┬─┘               └─┬─┘               └─┬─┘                  │
│   FACU (-) ────┴───────────────────┴───────────────────┘                    │
│                                                                             │
│   • SUPERVISORY: All contacts open; trickle current flows through EOLR.     │
│   • ALARM: Contact closes; parallel low-resistance shunts circuit.          │
│   • TROUBLE: Wire breaks; series loop opens; current drops to zero.         │
└─────────────────────────────────────────────────────────────────────────────┘

The Three Electrical Operating States of an IDC

1. Normal Standby (Quiescent) State

Under normal conditions, all manual pull stations, heat detectors, and smoke detector alarm contacts remain normally-open (N.O.). Current leaving the positive terminal cannot cross any initiating device. It travels through the entire length of the outgoing conductor, flows across the series End-of-Line Resistor (EOLR) at the furthest point, and returns to the FACU via the negative conductor.

• With a $4.7\text{ k}\Omega$ EOLR and 24VDC source, a steady supervisory current of approximately $5.08\text{ mA}$ flows continuously.
• The FACU microprocessor senses this stable supervisory current and confirms circuit continuity.

2. Alarm State (Parallel Shunt)

When an occupant activates a manual pull station or a smoke detector detects particulate matter, internal solid-state switches or mechanical contacts close across the two conductors. This places a low-resistance path in parallel with the high-resistance EOLR.

• Current follows the path of least resistance, bypassing the downstream EOLR.
• Conventional smoke detectors incorporate an internal current-limiting alarm resistor (typically $300,\Omega$ to $470,\Omega$) to prevent an absolute dead short that would damage the FACU power supply.
• Total circuit resistance immediately plunges from $\approx 4,720,\Omega$ down to $\approx 400,\Omega$.
• Applying Ohm's Law: $I = 24\text{ V} / 400,\Omega = 0.060\text{ A} = \mathbf{60\text{ mA}}$.
• The control panel detects this sudden tenfold current surge, latches into Alarm, lights the zone indicator, activates notification appliances, and dials the supervising station.

3. Trouble State (Open Series Loop)

If a wire is severed, a terminal screw loosens, or a detector head is removed from its base without a supervisory continuity spring, the series loop breaks.

• Circuit resistance becomes infinite ($R = \infty$).
• Current flow immediately drops to $0.00\text{ mA}$.
• Under NFPA 72 Section 10.19.1, the FACU must detect this open circuit and initiate an audible and visual trouble signal within 200 seconds.

4. Ground Fault State

When physical damage causes a bare conductor to contact metallic conduit, a metal junction box, or structural building steel, a ground fault is created. Commercial fire alarm control units operate as "floating" or ground-referenced isolated DC power supplies. An internal ground-fault detection circuit continuously biases the positive and negative conductors relative to earth ground. When current leaks to earth, the panel detects the voltage imbalance and trips a Ground Fault Trouble.

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Step-by-Step Worked Field Calculation: Multimeter Diagnostics on an IDC

Field Scenario: An installer in Oklahoma City is commissioning an addressable monitor module supervising a manual pull station zone with a specified $10\text{ k}\Omega$ ($10,000,\Omega$) EOLR. The panel indicates an active "Open Circuit Trouble." The technician uses a Digital Multimeter (DMM) to diagnose the circuit.

┌─────────────────────────────────────────────────────────────────────────────┐
│                     DMM DIAGNOSTIC PROCEDURE FOR IDC TROUBLE                │
├─────────────────────────────────────────────────────────────────────────────┤
│  STEP 1: VOLTAGE MEASUREMENT AT FACU TERMINALS (CIRCUIT CONNECTED)          │
│  • Reading = 24.0 VDC (Full open-circuit voltage present).                  │
│  • Analysis: Panel supply is functional; zero current is causing no drop.   │
├─────────────────────────────────────────────────────────────────────────────┤
│  STEP 2: CURRENT MEASUREMENT IN SERIES                                      │
│  • DMM in DC mA mode placed in series with positive terminal.              │
│  • Reading = 0.00 mA. Normal supervisory should be:                         │
│    I = 24V / 10,000Ω = 2.4 mA.                                              │
│  • Analysis: Confirms complete physical break in the supervisory loop.      │
├─────────────────────────────────────────────────────────────────────────────┤
│  STEP 3: DE-ENERGIZED RESISTANCE MEASUREMENT                                │
│  • Disconnect loop conductors from FACU terminals (NEVER measure R live).   │
│  • DMM in Ohms mode across disconnected field wire pair.                    │
│  • Reading = "O.L." (Over Limit / Infinite Resistance).                     │
│  • Analysis: Confirms open conductor between panel and EOLR.                │
├─────────────────────────────────────────────────────────────────────────────┤
│  STEP 4: HALF-SPLIT TROUBLESHOOTING TECHNIQUE                               │
│  • Move to device at physical midpoint of circuit (Pull Station 3 of 6).    │
│  • Disconnect wire and measure resistance back toward panel and toward EOLR.│
│  • Toward EOLR: 10,012 Ω (EOLR and wire intact downstream).                 │
│  • Toward Panel with jumper at panel: O.L. (Break is between Panel and PS3).│
│  • Technician inspects Pull Station 2 and finds a loose terminal screw.     │
└─────────────────────────────────────────────────────────────────────────────┘

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Exam Watchouts & Common Traps

[!IMPORTANT] Critical Exam Watchouts:

• Never Measure Resistance on an Energized Circuit: Connecting an ohmmeter across energized 24VDC terminals will blow the meter's internal protective fuse or destroy the multimeter's input circuitry. The circuit must always be disconnected from power before taking resistance measurements.
• Current Measurements Must Be in Series: To measure current, the circuit must be opened and the meter placed in series so current flows through the meter. Placing an ammeter in parallel across 24VDC creates a direct short through the ammeter's low-impedance shunt, blowing the meter fuse or damaging the FACU.
• Conductor Loop Resistance vs. One-Way Resistance: Field wiring consists of two conductors (outgoing and return). If an exam question states that a run is 500 feet of cable with a conductor resistance of $3.07,\Omega$ per 1,000 feet, the loop resistance involves 1,000 feet of total conductor ($2 \times 500\text{ ft} = 1,000\text{ ft}$), yielding $3.07,\Omega$, NOT $1.54,\Omega$.
• Resistor Wattage Derating: Remember the 2× safety multiplier. A calculated dissipation of $0.15\text{ W}$ cannot use a $1/8\text{ W}$ ($0.125\text{ W}$) resistor and should not use a $1/4\text{ W}$ ($0.25\text{ W}$) resistor if strict 2× margin ($0.30\text{ W}$) is required; a $1/2\text{ W}$ ($0.50\text{ W}$) unit is the professional standard.