The Problem: Loaded DC Voltage Divider with Source Resistance
Exam Problem Statement:
A 24V DC power source has an internal resistance (Rs) of 2Ω. It feeds a voltage divider network consisting of R1 = 10Ω and R2 = 40Ω. A load resistor (RL) of 120Ω is connected in parallel with R2. Calculate the exact DC voltage across the load resistor (VL).
Step-by-Step Solution: Applying KVL and Series-Parallel Reduction
Which theorem/method applies and why? We will use Kirchhoff’s Voltage Law (KVL) combined with series-parallel resistance reduction. This method is ideal because the circuit is a single-source, purely resistive DC network. By reducing the parallel load and divider leg into a single equivalent resistance, we can find the total circuit current and subsequently the voltage drop across the load.
⚠️ The Trap in This Problem
The most common mistake in these DC voltage examples is calculating the unloaded voltage divider ratio (24V × 40Ω / 50Ω = 19.2V) and ignoring both the load resistor and the source's internal resistance. Adding a load always pulls the voltage down, and real-world sources always drop voltage under current draw.
Step 1: Find the equivalent resistance of the parallel branch (R2 || RL).
Req = (R2 × RL) / (R2 + RL)
Req = (40 × 120) / (40 + 120)
Req = 4800 / 160
Req = 30Ω
Step 2: Calculate the total series resistance of the circuit (Rtotal).
Rtotal = Rs + R1 + Req
Rtotal = 2Ω + 10Ω + 30Ω
Rtotal = 42Ω
Step 3: Determine the total current supplied by the source (Itotal).
Itotal = Vsource / Rtotal
Itotal = 24V / 42Ω
Itotal ≈ 0.5714 A
Step 4: Calculate the voltage drop across the parallel equivalent resistance (VL).
Since Req represents the parallel combination of R2 and RL, the voltage across Req is identical to the voltage across the load.
VL = Itotal × Req
VL = 0.5714 A × 30Ω
VL = 17.14 V
Sanity Check and Independent Verification via Thevenin
Answer Sanity Check: Let's check the order of magnitude and logical bounds. If the circuit were unloaded (RL removed) but we accounted for internal resistance, the voltage across R2 would be 24V × [40 / (2 + 10 + 40)] = 18.46V. Because adding a parallel load always reduces the equivalent resistance and draws more current (causing a larger drop across Rs and R1), our final answer must be less than 18.46V. Our result of 17.14V fits this physical constraint perfectly. The units are in Volts, and the order of magnitude (101) matches the 24V source.
How to verify the answer independently: We can use Thevenin’s Theorem. First, we remove RL and find the open-circuit Thevenin voltage (Vth) across the R2 terminals. As calculated in our sanity check, Vth = 18.4615V. Next, we zero out the 24V source (short circuit) and look back into the terminals to find the Thevenin resistance (Rth). Rs and R1 are in series (12Ω), and that combination is in parallel with R2 (40Ω). Rth = (12 × 40) / (12 + 40) = 9.2307Ω.
Finally, we reconnect RL to the Thevenin equivalent circuit (a 18.4615V source in series with 9.2307Ω and 120Ω):
VL = Vth × [RL / (Rth + RL)]
VL = 18.4615 × [120 / (9.2307 + 120)]
VL = 18.4615 × 0.92857 = 17.14 V. The independent verification matches exactly.
Real-World DC Voltage Examples: Ideal vs. Actual Sources
In textbook problems, DC sources are often treated as ideal (0Ω internal resistance). On the workbench, every power source has internal impedance. Understanding these real-world DC voltage examples is critical when sizing wires or predicting brownouts in embedded systems like ESP32 or Arduino projects. For deeper reading on how real components behave, refer to the All About Circuits DC textbook chapter on voltage dividers.
| DC Source Type | Nominal Voltage | Typical Internal Resistance | Practical Application |
|---|---|---|---|
| AA Alkaline Cell | 1.5V | 0.15Ω - 0.30Ω | Low-drain remotes, clocks |
| 18650 Li-ion (e.g., Samsung 30Q) | 3.6V - 4.2V | 0.02Ω - 0.05Ω | High-drain flashlights, DIY battery packs |
| Bench PSU (e.g., Rigol DP832) | 0V - 30V | < 0.01Ω (regulated) | Bench prototyping, testing |
| 12V Lead-Acid Automotive | 12.6V (resting) | 0.01Ω - 0.02Ω | Starter motors, car audio, solar banks |
FAQ: Practical DC Voltage Examples and Exam Questions
What are common real-world DC voltage examples in bench electronics?
The most standard DC rails you will encounter are 3.3V (modern logic, ESP32, STM32), 5V (legacy logic, Arduino Uno, USB power), 12V (PC fans, LED strips, op-amp dual supplies), and 24V (industrial PLCs, stepper motor drivers, HVAC controls). When designing circuits, always assume a 5% tolerance on these rails unless you are using a precision voltage reference IC like the LM4040.
Why does my measured DC voltage drop when I connect a motor or load?
This is the exact phenomenon demonstrated in our problem walkthrough. When a high-current load like a DC motor starts, it draws a massive inrush current. According to Ohm's Law (V = I × R), this high current multiplied by the internal resistance of your battery or power supply creates a large internal voltage drop. The remaining voltage available to the external circuit sags. To mitigate this, use thicker wires, lower-ESR batteries, or place local decoupling capacitors near the motor driver.
How do I calculate DC voltage drop across long wire runs?
Wire resistance acts exactly like the source resistance (Rs) in our example. To calculate it, use the formula: Vdrop = 2 × I × Rwire_per_foot × Length. The multiplier of 2 accounts for both the positive and negative (return) conductors. For example, if you are pulling 10A over 50 feet of 12 AWG copper wire (which has a resistance of roughly 0.001588 Ω/ft), the voltage drop is 2 × 10 × 0.001588 × 50 = 1.588V. For comprehensive field measurement techniques, the Fluke guide on voltage drop is an excellent practical resource.
