6.2 Series and Parallel Circuits

Why series and parallel decide alarm behavior

Ohm’s law describes one resistor. Field circuits are networks. A conventional initiating circuit is a series wiring path that must stay continuous out to the end-of-line (EOL) resistor, with normally-open devices in parallel across that pair. A notification appliance circuit (NAC) is a parallel load of horns, strobes, and speakers on a shared 24 V pair, with an EOL at the last appliance for supervision. If you mix up current is the same with voltage is the same, you will mis-size wire, leave the EOL on the panel terminals, and misread trouble versus alarm.

The DOS exam covers security and fire. A burglary loop of normally-closed contacts is often a true series string — one open door or one cut wire opens the whole loop. A conventional fire IDC is usually the opposite geometry: devices across the pair, EOL at the end. Do not copy a burglary series habit onto a fire zone.

Series rules — current the same; voltages add

In a series circuit there is one path. The same current I flows through every element. Voltages across elements add up to the source (Kirchhoff’s voltage law). Total resistance is the arithmetic sum:

R_T = R1 + R2 + R3 + …

I = V_T / R_T (this I is the same everywhere)

V_n = I × R_n and V_T = V1 + V2 + …

The largest resistor drops the largest voltage. If any element opens, current in the whole loop becomes zero.

Worked example E — EOL in series with copper

Take the 24 V source, the 4.7 kΩ typical listed EOL, and 3.2 Ω of copper (18 AWG at the Section 6.1 rate, about 500 ft round-trip).

1. R_T = 4,700 + 3.2 = 4,703.2 Ω.
2. I = 24 / 4,703.2 = 0.005103 A = 5.10 mA. That current is the same through the wire and through the EOL.
3. V_wire = 0.005103 × 3.2 = 0.016 V (16 mV — negligible on a meter).
4. V_EOL = 0.005103 × 4,700 = 23.98 V.
5. Check KVL: 0.016 + 23.98 = 24.00 V.

Teaching point: during supervision, the EOL dominates. Almost all of the 24 V appears across the 4.7 kΩ resistor. Wire resistance barely matters until the circuit goes into a low-resistance alarm state or you are calculating NAC appliance voltage.

Voltage-divider form of the same math: V_EOL = V_T × (R_EOL / R_T) = 24 × (4,700 / 4,703.2) ≈ 23.98 V.

Three resistors in series (pure arithmetic)

24 V, resistors 100 Ω, 200 Ω, and 300 Ω.

R_T = 100 + 200 + 300 = 600 Ω.

I = 24 / 600 = 0.040 A = 40 mA (same in each).

V_100 = 0.040 × 100 = 4 V; V_200 = 8 V; V_300 = 12 V; 4 + 8 + 12 = 24 V.

If the 200 Ω resistor opens, I = 0 everywhere and the full 24 V appears across the open — the qualitative open preview from Section 6.1, now with KVL underneath it.

Conventional IDC — series path, parallel contacts

Do not memorize a cartoon that says the smoke detectors are in series. On a typical conventional Class B IDC:

1. Two conductors leave the IDC terminals.
2. Each initiating device (manual station, 4-wire dry contact, waterflow, 2-wire smoke) lands across the pair — electrically in parallel with the other devices and with the EOL.
3. The EOL resistor lands across the pair at the last device, so supervision current is forced to travel the entire installed length.
4. In the normal condition (contacts open), the intended load is the EOL plus the tiny 2-wire smoke standby currents.
5. A break anywhere in the series wiring path stops EOL current → trouble.
6. A closed contact (or a 2-wire smoke in alarm) places a low resistance across the pair → alarm.

The EOL value is a typical listed value printed on the panel door. 4.7 kΩ is a widely used example in this guide; 2.2 kΩ is also common. Installing the wrong value, or leaving the factory EOL on the terminals instead of moving it to the last device, is a classic failed inspection.

If you install the EOL at the panel, the panel still sees a resistor and may show normal even when the field wiring is cut. That is why the resistor belongs at the end of the line.

Resistor color code, because the wrong part looks similar: 4.7 kΩ is yellow-violet-red (4, 7, ×100). 2.2 kΩ is red-red-red. A 470 Ω part is yellow-violet-brown — ten times too small. Section 6.3 shows that 470 Ω on 24 V dissipates more than a typical ½ W EOL can survive.

Class A versus Class B returns, including whether a particular circuit uses an EOL or a return pair, belong in Chapter 9. For this section, remember Class B: one path out, EOL at the last device.

Parallel rules — voltage the same; currents add

In a parallel circuit every branch sees the same voltage. Currents in the branches add (Kirchhoff’s current law). Total resistance is less than the smallest branch.

Two resistors: R_T = (R1 × R2) / (R1 + R2) (product over sum).

n equal resistors of value R: R_T = R / n.

General: 1/R_T = 1/R1 + 1/R2 + …

I_n = V / R_n and I_T = I1 + I2 + …

Opening one parallel branch stops only that branch. The others keep running. That is why five strobes still flash when the sixth is disconnected — and why a NAC still needs an EOL path to prove the wiring, not the appliances, is intact.

Worked example F — NAC appliances in parallel

24 V NAC. Six listed horn-strobes, each 75 mA at 24 V (each equivalent to 320 Ω from Section 6.1).

Because they share the pair, each still sees (ideally) 24 V, so each still draws 75 mA.

I_T = 6 × 0.075 = 0.450 A = 450 mA.

R_T = 320 / 6 = 53.3 Ω.

Check: I_T = 24 / 53.3 ≈ 0.450 A.

If you wrongly treated them as series, you would predict I = 24 / (6 × 320) = 12.5 mA — about 36 times too small — and both the battery and the NAC would be undersized. Notification appliances on a NAC are parallel loads.

Add a seventh appliance at 125 mA (equivalent 192 Ω):

I_T = 0.450 + 0.125 = 0.575 A.

Always add nameplate currents at the listed voltage, then apply voltage-drop (Ohm’s law on the wire) as a second step. Manufacturer NAC voltage-drop worksheets do the same arithmetic.

Worked example G — two unequal branches, product-over-sum check

24 V across 320 Ω (75 mA appliance) in parallel with 192 Ω (125 mA appliance).

I1 = 24 / 320 = 0.075 A; I2 = 24 / 192 = 0.125 A; I_T = 0.200 A.

R_T = 24 / 0.200 = 120 Ω.

Product over sum: (320 × 192) / (320 + 192) = 61,440 / 512 = 120 Ω. Same answer, two routes. Use current-addition on NACs; use product-over-sum when a problem hands you two resistors and no nameplate milliamperes.

Worked example H — two EOL resistors in parallel (fault)

Someone leaves a 4.7 kΩ on the panel terminals and installs another 4.7 kΩ at the last device. Those two resistors are in parallel across the IDC.

R_T = 4,700 / 2 = 2,350 Ω.

I = 24 / 2,350 = 10.2 mA — about twice the designed supervision current.

Some panels will show trouble (current too high). Some may appear normal but will not detect an open in the field if the panel-mounted resistor still completes a local path. Either way, it is wrong. One listed EOL, at the last device, value as specified on the door.

Combination circuits and alarm implications

Most field circuits are series-parallel: series copper plus parallel devices. Collapse the parallel loads into one equivalent resistance, add the series wire, apply Ohm’s law for the total current, then expand back to find branch currents and the voltage at the last device (Section 6.1 Example C is that method).

Opens and shorts, restated for these two circuits:

• An open in a series path is trouble on both IDC and NAC (supervision current cannot return).
• A short across a conventional IDC is typically an alarm (the panel sees a low-resistance initiating condition).
• A short across a NAC is a trouble (and often a short-circuit disconnect of that NAC), because the panel must not dump unprotected current into a bolted pair.

Chapter 7.3 puts a meter on those faults. Chapter 9 covers circuit classes and the return pair on Class A.

Security take-home: a string of normally-closed door contacts is a true series loop. One open contact opens the loop. Fire conventional initiating devices are usually normally open, paralleled, with EOL supervision. Same Ohm’s law; opposite rest condition.

Exam traps

• Adding parallel resistances instead of using the reciprocal (or current addition).
• Predicting NAC current as if appliances were in series.
• Leaving the EOL on the FACU terminals.
• Treating a burglary NC loop and a fire NO IDC as the same drawing.
• Forgetting that the wiring to the EOL is series even when the devices are parallel.

Section 6.3 turns these currents into watts and watt-hours so Chapter 9 can size batteries.