When studying for an electrical exam or designing an industrial control panel, a textbook example of series circuit problems often ignores real-world wire resistance. They assume ideal conductors, which works on paper but causes mysterious field failures on the jobsite. In this walkthrough, we bridge that gap by analyzing a long-run 24V DC solenoid circuit, applying rigorous algebra, and making a concrete wire-sizing decision.
The Problem Statement: 24V Long-Run Solenoid Circuit
Exam Practice Problem
Given:
- Source: 24V DC Power Supply (ideal, zero internal resistance).
- Load: 24V DC Solenoid Valve. Coil resistance ($R_{load}$) = $80\Omega$. Minimum pull-in voltage = 19V.
- Wiring: 500 feet one-way distance from the PLC output to the valve, using 24 AWG solid copper wire.
Find:
- The total circuit current ($I$).
- The actual voltage delivered to the solenoid ($V_{load}$).
- Determine if the valve will reliably actuate.
Method Selection and Identifying the "Trap"
Which theorem applies and why? Because this is a single-loop DC network with components connected end-to-end, we use Ohm’s Law combined with Kirchhoff’s Voltage Law (KVL). KVL states that the directed sum of the potential differences around any closed loop is zero ($\Sigma V = 0$). In plain terms: the source voltage must equal the sum of the voltage drops across every resistive element in the loop.
The Trap: There are two classic traps in this specific KVL application. First, students frequently use the "one-way" distance (500 ft) to calculate wire resistance, forgetting that current must return to the power supply, doubling the effective wire length. Second, they assume any voltage close to 24V will run a 24V solenoid. Inductive loads require a high initial current to overcome the spring and close the air gap (pull-in); if voltage sags below the manufacturer's threshold (19V here), the valve will chatter, overheat, and burn out the coil.
Step-by-Step Algebraic Solution
Let's solve this with every algebraic step shown. We will use standard copper resistance values at 20°C (68°F) per NEC Chapter 9, Table 8.
- Calculate Total Wire Resistance ($R_{wire}$):
The loop length is $500 \text{ ft (out)} + 500 \text{ ft (return)} = 1000 \text{ ft}$.
24 AWG copper resistance = $25.67 \Omega / 1000 \text{ ft}$.
$$R_{wire} = \left( \frac{1000 \text{ ft}}{1000 \text{ ft}} \right) \times 25.67 \Omega = 25.67 \Omega$$ - Calculate Total Series Resistance ($R_{total}$):
In a series circuit, resistances simply add.
$$R_{total} = R_{load} + R_{wire}$$
$$R_{total} = 80 \Omega + 25.67 \Omega = 105.67 \Omega$$ - Calculate Circuit Current ($I$):
Using Ohm's Law ($I = V / R$):
$$I = \frac{24 \text{ V}}{105.67 \Omega} = 0.22712 \text{ A} \text{ (or } 227.1 \text{ mA)}$$ - Calculate Load Voltage ($V_{load}$):
Using the voltage drop across the load ($V = I \times R$):
$$V_{load} = 0.22712 \text{ A} \times 80 \Omega = 18.17 \text{ V}$$
Sanity Check and Independent Verification
Before finalizing an exam answer or a bill of materials, always run a sanity check to ensure your math aligns with physical reality.
Order of Magnitude & Units Check:
If the wire had zero resistance, the current would be $24\text{V} / 80\Omega = 300\text{ mA}$. Our calculated 227.1 mA is lower, which logically tracks because we added series resistance. The units resolve correctly (Volts / Ohms = Amperes; Amperes $\times$ Ohms = Volts).
KVL Loop Verification:
Let's calculate the voltage dropped by the wire: $V_{wire} = 0.22712 \text{ A} \times 25.67 \Omega = 5.83 \text{ V}$.
Does $V_{load} + V_{wire} = V_{source}$?
$18.17 \text{ V} + 5.83 \text{ V} = 24.00 \text{ V}$. The loop balances perfectly.
Jobsite Verification (How to test independently):
If you are troubleshooting this in the field, grab a digital multimeter like a Fluke 87V. Measure the DC voltage directly at the power supply terminals while the circuit is energized (it should read ~24.0V). Then, move your probes to the solenoid's connection terminals. If the meter reads ~18.1V and the valve is humming without pulling in, you have empirically verified the math. The fix is not a new solenoid; the fix is addressing the voltage drop.
Decision Tree: Wire Gauge Selection
We know 24 AWG fails. How do we choose the correct replacement? Use this decision path to terminate on a concrete part selection.
| Condition / Test | Outcome | Action |
|---|---|---|
| Is $V_{load} \ge 19\text{V}$? | No (18.17V). 24 AWG is too small. | Upsize wire to 20 AWG and re-test. |
| Recalculate for 20 AWG ($10.15 \Omega$/kft): $R_{wire} = 10.15\Omega$. $R_{tot} = 90.15\Omega$. $I = 266\text{mA}$. $V_{load} = 21.3\text{V}$. |
Yes (21.3V > 19V). Passes basic threshold. | Check thermal derating margin. |
| Does 21.3V provide enough margin for high ambient temps (copper resistance increases ~0.4% per °C)? | Marginal. In a 50°C panel, resistance spikes, dropping voltage closer to the 19V cliff. | Upsize to 18 AWG for robust industrial margin. |
| Recalculate for 18 AWG ($6.385 \Omega$/kft): $R_{wire} = 6.385\Omega$. $R_{tot} = 86.385\Omega$. $I = 277\text{mA}$. $V_{load} = 22.2\text{V}$. |
Yes. 22.2V provides a safe 3.2V buffer above the 19V pull-in threshold. | Terminate Decision. |
Concrete Default Pick: Specify 18 AWG THHN Copper for this run. It guarantees reliable solenoid actuation, minimizes $I^2R$ heating in the cable, and provides adequate thermal headroom for industrial environments.
FAQ: Common Series Circuit Exam Mistakes
Q: Do I need to account for the solenoid's inductance in this DC calculation?
A: Only for transient analysis (the exact millisecond the switch closes). For steady-state DC analysis—which dictates whether the valve stays pulled in—the inductor acts as a short circuit, leaving only the coil's DC wire resistance ($80\Omega$) to limit current.
Q: What happens if I just increase the power supply voltage to 28V to force more current through the 24 AWG wire?
A: You will likely fry the PLC output card or the solenoid coil once the valve finally closes. When the solenoid air gap closes, its inductance spikes and the current requirement drops to a lower "hold-in" state. Overvoltage will cause the coil to overheat and fail prematurely. Always fix voltage drop by reducing series resistance (thicker wire), not by overvolting the source.
Q: Did we forget any protective components in this series example?
A: Yes. In a real-world schematic, you must place a flyback diode (e.g., 1N4007) in parallel with the solenoid coil (reverse-biased). When the series circuit is broken, the collapsing magnetic field generates a massive reverse voltage spike that will destroy your PLC's output transistor if not clamped by the diode.
