Textbook circuits rarely survive contact with the workbench. In an academic setting, wires are perfect conductors and batteries have infinite current capacity. In reality, every wire has resistance, and every power source sags under load. If you are looking for a realistic example of Ohm's law that bridges the gap between exam theory and actual troubleshooting, you need to account for parasitic resistance.
Below, we will dissect a classic mixed-circuit problem. We will identify the correct methodology, execute every algebraic step without skipping the messy fractions, and verify the results using independent circuit theorems.
The Problem Statement: 12V DC Distribution with Parasitic Resistance
📝 Exam Problem Statement
A 12.0V DC lead-acid battery with an internal resistance of 0.2Ω powers a remote sensor station. The connection is made using a long-run 20 AWG copper cable that introduces a total loop resistance of 0.8Ω.
At the remote end, the circuit splits into two parallel loads: a heating element (Load A, 4.0Ω) and a DC cooling fan motor modeled at stall (Load B, 12.0Ω).
Calculate:
- The total current drawn from the battery.
- The actual voltage delivered to the parallel load bank.
- The individual current through Load A and Load B.
Before solving, let's map the schematic values to their real-world physical equivalents. Understanding what these components actually look like on a bench prevents you from accepting mathematically correct but physically impossible answers.
| Component | Schematic Label | Nominal Value | Real-World Bench Equivalent |
|---|---|---|---|
| Power Source | $V_{source}$ / $R_{int}$ | 12.0V / 0.2Ω | 12V 7Ah SLA Battery (Yuasa NP7-12) |
| Feeder Cable | $R_{cable}$ | 0.8Ω | ~75 ft loop of 20 AWG stranded copper |
| Load A (Heater) | $R_{A}$ | 4.0Ω | 36W 12V polyimide flexible heating pad |
| Load B (Motor) | $R_{B}$ | 12.0Ω | 12V DC brushless fan (stall/starting state) |
Step-by-Step Solution: Method, Algebra, and Execution
Which method applies and why? We must use Series-Parallel Equivalent Reduction combined with Ohm's Law ($V = IR$). You cannot apply Ohm's law to the total circuit until you have reduced the entire network to a single equivalent resistance ($R_{total}$). Furthermore, because the loads are in parallel with each other, but in series with the cable and battery internal resistance, we must collapse the parallel bank first, then add the series resistances.
Step 1: Reduce the Parallel Load Bank
Load A (4.0Ω) and Load B (12.0Ω) are in parallel. We use the product-over-sum formula for two resistors:
$$R_{parallel} = \frac{R_A \times R_B}{R_A + R_B}$$
$$R_{parallel} = \frac{4.0 \times 12.0}{4.0 + 12.0}$$
$$R_{parallel} = \frac{48.0}{16.0} = 3.0\Omega$$
Step 2: Calculate Total Circuit Resistance
Now, the circuit is a simple series loop consisting of the battery's internal resistance, the cable resistance, and our newly calculated $R_{parallel}$.
$$R_{total} = R_{int} + R_{cable} + R_{parallel}$$
$$R_{total} = 0.2\Omega + 0.8\Omega + 3.0\Omega = 4.0\Omega$$
Step 3: Calculate Total Current (Ohm's Law)
With the total resistance and the source voltage known, we apply Ohm's law to find the total current leaving the battery.
$$I_{total} = \frac{V_{source}}{R_{total}}$$
$$I_{total} = \frac{12.0V}{4.0\Omega} = 3.0A$$
Step 4: Calculate Actual Load Voltage
This is where we find the voltage actually delivered to the remote sensor station. We apply Ohm's law specifically to the parallel equivalent resistance.
$$V_{load} = I_{total} \times R_{parallel}$$
$$V_{load} = 3.0A \times 3.0\Omega = 9.0V$$
Step 5: Calculate Individual Branch Currents
Now that we know the parallel bank sees exactly 9.0V (not 12.0V), we apply Ohm's law to each individual branch.
$$I_A = \frac{V_{load}}{R_A} = \frac{9.0V}{4.0\Omega} = 2.25A$$
$$I_B = \frac{V_{load}}{R_B} = \frac{9.0V}{12.0\Omega} = 0.75A$$
The Exam Trap and Independent Verification
⚠️ The Trap: Assuming Ideal Voltage Delivery
The most common mistake on this type of exam problem is ignoring $R_{int}$ and $R_{cable}$. Students will look at the parallel bank, see a 12V source, and calculate $I_A = 12V / 4\Omega = 3.0A$. This is physically wrong. The parasitic resistances act as a voltage divider, stealing 3.0V from the system before it ever reaches the loads. In real-world 12V automotive or solar systems, a 3V drop (25% loss) is massive and will cause microcontrollers to brownout or motors to fail to start.
Answer Sanity Check
Before moving on, always perform a sanity check on your order of magnitude and units.
- Units: Volts divided by Ohms yields Amperes. Volts divided by Amperes yields Ohms. All units balance.
- Order of Magnitude: The total resistance must be greater than the largest series component (0.8Ω) but less than the sum of all components if they were in series (17Ω). Our 4.0Ω fits perfectly. The total current (3.0A) is reasonable for a 12V SLA battery, which can typically supply 10A+ continuously without severe voltage sag.
How to Verify the Answer Independently
Never trust a single calculation path. We can verify our results using Kirchhoff’s Voltage Law (KVL) and Kirchhoff's Current Law (KCL), as outlined in standard circuit theory texts like All About Circuits.
Verification 1: KVL (Voltage Drops must equal Source)
Let's calculate the voltage dropped across the parasitic resistances:
- $V_{drop(int)} = 3.0A \times 0.2\Omega = 0.6V$
- $V_{drop(cable)} = 3.0A \times 0.8\Omega = 2.4V$
Sum of all drops: $0.6V + 2.4V + 9.0V (load) = 12.0V$. This matches our $V_{source}$ perfectly.
Verification 2: KCL (Branch Currents must equal Total Current)
Sum of branch currents: $I_A + I_B = 2.25A + 0.75A = 3.0A$. This matches our calculated $I_{total}$ perfectly.
Frequently Asked Questions (FAQ)
How do I measure this parasitic voltage drop on the bench?
Do not rely on your power supply's digital display; it only measures voltage at its internal terminals. To measure the true voltage drop of the cable, take your multimeter (like a Fluke 87V) and measure the DC voltage directly across the battery terminals while the circuit is active. Then, measure the DC voltage directly at the load terminals. The difference between these two readings is the exact voltage lost to $R_{int}$ and $R_{cable}$. For highly accurate cable resistance measurements without the influence of contact resistance, use a 4-wire Kelvin measurement technique.
Does wire resistance change as the circuit runs?
Yes. The 0.8Ω value for our 20 AWG wire is calculated at standard room temperature (20°C). Copper has a positive temperature coefficient of approximately 0.00393 per °C. As the 3.0A current heats the wire (dissipating $I^2R = 7.2W$ of heat into the cable jacket), the copper's resistance will rise. According to Georgia State University's HyperPhysics, if the wire temperature rises to 60°C, the resistance will increase by roughly 15%, pushing $R_{cable}$ to ~0.92Ω and further dropping your load voltage. This is why thermal derating is critical in enclosed conduit runs.
What if the loads are not purely resistive?
In this example, we modeled the DC fan motor as a 12Ω resistor. In reality, a motor is an inductive load that generates Back-EMF as it spins. The 12Ω value represents the stall current (the exact moment power is applied before the rotor moves). Once the fan spins up, its effective resistance increases dramatically, and the total current draw will drop well below 3.0A. When sizing wires and fuses for motors, always calculate using the stall or locked-rotor current to ensure your protection devices do not trip during startup.
