Accounting for source internal resistance: A 12 V source has internal resistance RS = 90 Ω. With a load RL = 20 Ω connected, what power PL is delivered to the load?

Introduction / Context:
Real voltage sources are modeled using Thevenin equivalents: an ideal source in series with an internal resistance. Load power is found by first computing the load voltage via the divider, then applying PL = VL^2 / RL, or by using current and PL = I^2 * RL.

Given Data / Assumptions:

• Source voltage VS = 12 V.
• Internal resistance RS = 90 Ω (series).
• Load resistance RL = 20 Ω.
• Purely resistive, steady-state conditions.

Concept / Approach:

Voltage divider: VL = VS * RL / (RS + RL). Then compute PL using PL = VL^2 / RL. Expect a relatively small VL because RS is much larger than RL, meaning a significant drop occurs inside the source.

Step-by-Step Solution:

Verification / Alternative check:

Using PL = I^2 * RL gives (0.10909^2) * 20 ≈ 0.0119 * 20 = 0.238 W, which agrees exactly with the voltage method.

Why Other Options Are Wrong:

23.8 W exceeds the source's capability with this resistance network. 2.38 W and 0.12 W result from arithmetic or divider mistakes. 2.38 mW is orders of magnitude too small.

Common Pitfalls:

Ignoring RS in the divider, or plugging VS directly into PL = V^2 / RL. Also, mixing milli and base units can lead to factor-of-1000 errors.

Final Answer:

238 mW