The Problem Statement: A Practical Example of DC Power Delivery
Method Selection: Thevenin and Maximum Power Transfer
Which theorem applies here and why? We use Thevenin's Theorem combined with the Maximum Power Transfer Theorem. Any real-world DC voltage source (like a battery, solar panel, or linear power supply) can be modeled as an ideal Thevenin voltage source (Vth) in series with a Thevenin resistance (Rth). In this problem, Vth = 24V and Rth = 2.0 Ω.
The Maximum Power Transfer Theorem states that maximum power is delivered to the load when the load resistance exactly equals the Thevenin (internal) resistance of the source. While you can find this via calculus, recognizing the theorem immediately gives you the target condition: RL = Rth. According to All About Circuits, this principle is foundational for matching source and load impedances, though its application in high-power DC systems requires careful thermal management.
Step-by-Step Algebraic Solution
We will not skip the calculus that proves the theorem, as understanding the derivation prevents errors in non-standard circuits.
- Define the Power Equation: Power delivered to the load is PL = I² × RL.
- Define Current via Ohm's Law: The total circuit current is I = Vth / (Rth + RL).
- Substitute I into the Power Equation:
PL = [Vth / (Rth + RL)]² × RL
PL = (Vth² × RL) / (Rth + RL)² - Differentiate with respect to RL and set to zero: Using the quotient rule, where u = Vth² × RL and v = (Rth + RL)²:
dPL/dRL = Vth² × [ (Rth + RL)²(1) - RL × 2(Rth + RL) ] / (Rth + RL)⁴ = 0 - Solve for RL: Since Vth and the denominator cannot be zero, the numerator bracket must equal zero:
(Rth + RL) - 2RL = 0
Rth - RL = 0
RL = Rth = 2.0 Ω - Calculate Numerical Values:
Total Resistance = 2.0 Ω + 2.0 Ω = 4.0 Ω
Current (I) = 24V / 4.0 Ω = 6.0 A
Load Power (PL) = (6.0 A)² × 2.0 Ω = 72 W
Total Source Power (Ptotal) = 24V × 6.0 A = 144 W
Sanity Check and Independent Verification
Before finalizing component orders, we must perform a sanity check on the order of magnitude and units. 24V divided by roughly 4 ohms yields roughly 6 amps. 6 amps squared is 36; 36 times 2 ohms is 72 watts. The units resolve cleanly to Watts (A² × Ω = W). The numbers align with physical reality.
How to verify the answer independently on the bench: Do not rely solely on the math. If you build this, measure the open-circuit voltage (Voc) of the battery pack; it should read 24.0V. Next, connect your 2.0 Ω load and measure the voltage directly across the load terminals. According to HyperPhysics circuit principles, at maximum power transfer, the load voltage must drop to exactly half of Voc. If your multimeter reads 12.0V across the load, you have independently verified the 6A current (12V / 2Ω) and the 72W power (12V × 6A) without repeating the algebraic derivation.
The Trap: Efficiency vs. Maximum Power
In real-world power distribution, we never design for maximum power transfer. We design for maximum voltage transfer and high efficiency, meaning we want RL to be vastly larger than Rth (e.g., a 100Ω load on a 2Ω source yields >98% efficiency). Maximum power transfer is strictly reserved for low-power signal matching (like audio amplifiers to speakers), RF antenna tuning, or specific resistive heating applications where the source's internal heating is managed or irrelevant.
Decision Path: Selecting the Physical Resistor
We need a 2.0 Ω resistor capable of handling 72W continuously. Safety Callout: 72W of concentrated heat will instantly cause severe burns and can ignite nearby combustible materials. You cannot use standard 1/4W or 5W ceramic resistors. You must apply a derating factor. The industry standard for chassis-mounted power resistors is a 50% derating margin to keep the component surface temperature below 150°C in a 25°C ambient environment. Therefore, 72W × 2 = 144W minimum rating. We will specify a 150W resistor.
| Condition / Constraint | Component Topology | Reasoning |
|---|---|---|
| Space constrained, panel mount | Aluminum Housed Wirewound | Flat base allows direct bolting to a metal chassis or external heatsink for thermal transfer. |
| High surge current, open frame | Tubular Ceramic Wirewound | High thermal mass absorbs sudden joule heating; radiates heat omnidirectionally. |
| High precision, low inductance | Thick Film Planar | Non-inductive design prevents voltage spikes during fast DC switching transients. |
The Concrete Pick: For a standard bench or panel application, choose the Arcol THS1502R0J (or equivalent 150W, 2.0Ω, 5% tolerance aluminum-housed resistor). It features a robust metal housing designed to be bolted to a heatsink, easily surviving the 72W continuous load provided it is mounted to an appropriate thermal mass. Ensure you use thermal paste between the resistor base and the mounting surface.
Frequently Asked Questions
What happens if I use a 100W resistor instead of 150W?
The resistor will be operating at 72% of its rated capacity. While it might not fail immediately in a well-ventilated room, the surface temperature will likely exceed safe touch limits (>200°C), degrading the internal winding insulation over time and drastically shortening the component's lifespan. Always adhere to the 50% derating rule for enclosed or chassis-mounted power electronics.
Does this theorem apply to AC circuits?
Yes, but the math shifts from simple resistance to complex impedance. In AC circuits, maximum power transfer occurs when the load impedance is the complex conjugate of the source impedance (ZL = Zth*). This means the resistive parts must be equal, but the reactive parts (inductance/capacitance) must be equal and opposite to cancel each other out.
Can I wire multiple smaller resistors to get 2 ohms at 150W?
Absolutely. If you cannot source a single 150W 2Ω unit, you can wire four 8Ω 50W resistors in parallel. This yields exactly 2.0 Ω total resistance, and the 72W load is divided equally among the four resistors (18W each), keeping them well within their 50W ratings without requiring a massive external heatsink.
