Practical source with internal resistance: An 18 V Thevenin source has internal resistance RS = 70 Ω and is connected to a load RL = 33 Ω. Compute the load power PL delivered to RL.

Introduction / Context:
This problem tests application of Thevenin's model and the voltage-divider concept to determine power delivered to a load when a real (non-ideal) voltage source with internal resistance supplies a resistive load. Computing load voltage first and then using P = V^2 / R is a standard, reliable approach in basic circuit analysis.

Given Data / Assumptions:

• Thevenin source voltage, VS = 18 V.
• Internal (series) resistance, RS = 70 Ω.
• Load resistance, RL = 33 Ω.
• Direct current conditions and purely resistive elements.

Concept / Approach:

The load sees a voltage set by a simple divider: VL = VS * RL / (RS + RL). Once VL is known, compute load power using PL = VL^2 / RL. Alternatively, compute current I = VS / (RS + RL), then PL = I^2 * RL. Both methods should agree.

Step-by-Step Solution:

Verification / Alternative check:

Alternative: PL = I^2 * RL ≈ (0.174757^2) * 33 ≈ 0.03053 * 33 ≈ 1.01 W. This matches the previous result, validating the computation.

Why Other Options Are Wrong:

175 mW and 18 mW undervalue power by ignoring the divider effect correctly. 0 W would require VL = 0 or an open circuit, which is not the case. 3.5 W far exceeds what an 18 V source can deliver into 33 Ω with 70 Ω in series.

Common Pitfalls:

Using VS^2 / RL directly (ignores RS), or adding powers instead of using the divider and load formula. Also, rounding too early can skew the final choice.

Final Answer:

1 W