6.3 Power, Energy, and Watt’s Law

Why power and energy show up on an installer exam

Ohm’s law tells you volts, amps, and ohms. Module 1.III also expects power (how hard a load works, in watts) and energy (power over time, in watt-hours). You will use these numbers to read nameplates, estimate heat in a resistor or in copper, and to preview battery calculations. Full secondary-power arithmetic — 24-hour standby, 5-minute alarm, 20% safety factor, and occupancy exceptions — is Chapter 9. This section teaches the laws those later hours sit on.

Power also explains a design choice you already used: 24 V fire circuits versus 12 V security circuits. For the same wattage of horns, doubling voltage halves current (I = P / V) and therefore halves I × R voltage drop in the same copper. Fire NACs run 24 VDC for a reason.

Watt’s law — three forms, one relationship

Power P is the rate of energy conversion. The unit is the watt (W). One watt equals one joule per second.

P = V × I

P = I² × R

P = V² / R

These three are the same law once Ohm’s law is substituted (because V = I × R). Pick the form that matches the values you already know. P = V × R is not a form of Watt’s law — that product is not power.

Worked example I — P = V × I (one appliance and a NAC)

Horn-strobe: 24 V, 75 mA.

1. 75 mA = 0.075 A.
2. P = 24 × 0.075 = 1.80 W.

Six such appliances in parallel: 6 × 1.80 = 10.8 W at the loads.

Check with total current from Section 6.2: P = 24 × 0.450 = 10.8 W. Same number.

If those 10.8 W of appliances were fed at 12 V instead of 24 V, current would be I = P / V = 10.8 / 12 = 0.90 A — twice the 0.45 A you had at 24 V. Wire drop and copper heating would both get worse. That is the 24 V advantage in one line of Watt’s law.

Worked example J — P = I²R (heat in the wire)

From Section 6.1: I = 0.450 A, R_wire = 2.556 Ω.

P_wire = (0.450)² × 2.556 = 0.2025 × 2.556 = 0.52 W.

That half-watt is heat along 400 ft of 18 AWG — not a fire by itself, but it is exactly the 1.15 V drop expressed as power:

P = V_drop × I = 1.15 × 0.450 = 0.52 W.

Copper that is too small gets warm and starves the last strobe of voltage. Doubling current quadruples I²R heating. A NAC short is violent compared with 5 mA of EOL supervision for that reason: current may jump from milliamps toward the panel’s short-circuit limit, and heating in the fault and in the copper spikes with the square of I.

Worked example K — P = V² / R (EOL resistor wattage)

EOL 4.7 kΩ on 24 V:

P = (24)² / 4,700 = 576 / 4,700 = 0.123 W.

A typical listed EOL is a ½ W (0.5 W) resistor — about four times this dissipation, which is why substituting a ¼ W part is a poor field habit. The resistor still only sees about 24 V in alarm on a NAC as well; the appliances take the real power. The danger is installing the wrong resistance, not the EOL sharing horn wattage.

Wrong part: a technician grabs 470 Ω (yellow-violet-brown) instead of 4.7 kΩ (yellow-violet-red).

I = 24 / 470 = 0.0511 A = 51.1 mA (looks like a perpetual alarm or a trouble, depending on the panel).

P = (24)² / 470 = 576 / 470 = 1.23 W — more than twice a ½ W resistor’s rating. The part can overheat. Same Ohm’s law, now with Watt’s law telling you the resistor is in trouble.

On a 12 V security EOL of 2.2 kΩ: P = (12)² / 2,200 = 144 / 2,200 = 0.065 W. Still a ½ W part in most kits, with even more thermal margin at 12 V.

Energy, watt-hours, and a battery preview

Energy = power × time. If P is in watts and t is in hours, energy is in watt-hours (Wh).

Wh = V × I × t (hours)

Ah (ampere-hours) = I × t (hours) — the language of battery capacity.

Wh = V × Ah for a constant-voltage estimate.

Alarm batteries are rated in ampere-hours at a stated discharge rate. Chapter 9 converts standby current and alarm current into Ah, then applies a safety factor. Here we only build the energy intuition. Do not treat the following numbers as a finished battery size.

Worked example L — 24-hour standby energy

A FACU draws 150 mA standby from a 24 V battery for 24 hours. 24-hour standby is a common NFPA 72 secondary-power concept for many fire alarm systems; confirm the required hours for the occupancy and system type in Chapter 9 and the adopted NFPA 72 edition (New York’s 2025 Uniform Code references NFPA 72—22). 150 mA is a teaching standby current, not a particular panel’s datasheet.

1. Ah_standby = 0.150 A × 24 h = 3.60 Ah.
2. Wh_standby = 24 V × 3.60 Ah = 86.4 Wh.
3. Or P = 24 × 0.150 = 3.6 W, then 3.6 W × 24 h = 86.4 Wh.

Worked example M — 5-minute alarm energy

Same system, alarm current 2.00 A at 24 V for 5 minutes (5/60 = 0.0833 h). Five minutes of alarm is the duration many fire alarm systems use for secondary-power alarm time; emergency voice/alarm communication systems often use a longer alarm period — again, Chapter 9.

1. Ah_alarm = 2.00 × (5/60) = 0.167 Ah.
2. Wh_alarm = 24 × 2.00 × (5/60) = 4.00 Wh.

Standby energy (86.4 Wh) dwarfs alarm energy (4 Wh) in this example because 24 hours is long and 5 minutes is short. A panel that is idle but hungry (high standby current from communicators, smoke-detector power, or a hungry keypad on a security panel) kills batteries faster than a loud 5-minute NAC.

Rough combined draw before a safety factor: 3.60 + 0.167 = 3.77 Ah. Chapter 9 will multiply by 1.20 (20% safety factor as industry/NFPA practice) and add all NAC, auxiliary, and communicator loads. 3.77 Ah is not a finished battery size.

Worked example N — do not double-count the EOL

The 4.7 kΩ EOL at 0.123 W for 24 hours: 0.123 × 24 = 2.95 Wh. That energy is already inside the panel’s standby current if supervision current comes from the battery. Do not add 2.95 Wh again in Chapter 9.

Kilowatt-hours (kWh) are utility energy: 1 kWh = 1,000 Wh. A 7 Ah, 24 V battery stores about 24 × 7 = 168 Wh = 0.168 kWh. This exam almost never discusses alarm batteries in kWh; stay in Wh and Ah.

Power, polarity, opens, and shorts

Power in a resistor is always positive (heat). On a battery, current out of the positive terminal is discharge (the battery supplies energy); current into the positive terminal is charge. Chapter 7.2 covers charging; here you only need the sign of energy flow.

Reversed polarity on a polarized appliance means the device dissipates about 0 W (it does not operate) even though voltage is present — a DMM can still show about 24 V. Measuring voltage without considering current is how technicians miss a reverse-polarity NAC.

Open versus short, restated in power language (still a preview of Chapter 7.3):

• Open: I ≈ 0, so P ≈ 0 in the load, even if voltage is present at the source.
• Short: V_load ≈ 0, so P_load ≈ 0 at the load, but power in the remaining resistance and in the source can be very large.

Exam traps

• Using P = V × R.
• Leaving milliamperes unconverted so 24 × 75 becomes 1,800 W instead of 1.8 W.
• Reporting 3.6 Ah as 3.6 Wh (missing the × 24 V).
• Treating the Chapter 9 20% factor as optional decoration rather than part of the later calc.
• Sizing a ½ W EOL, then installing 470 Ω at 24 V.

You now have voltage, current, resistance, series/parallel, power, and energy. Chapter 7 applies them to AC/DC supplies, batteries, and meters. Chapter 9 puts Ah on a job sheet.