Voltage division with source resistance: A load RL = 2 Ω is connected to a source VS = 110 V that has internal resistance RS = 24 Ω. What output voltage appears across the load?

Introduction / Context:
When a real source with internal resistance powers a low-ohmic load, a substantial portion of the applied voltage can drop inside the source. This question assesses your ability to apply the voltage-divider rule and reason about loading effects.

Given Data / Assumptions:

• Source voltage, VS = 110 V.
• Source resistance, RS = 24 Ω.
• Load resistance, RL = 2 Ω.
• DC steady-state analysis; resistances are ideal.

Concept / Approach:
For series RS and RL fed by VS, the load voltage is VL = VS * (RL / (RS + RL)). Because RL is small relative to RS, only a small fraction of the source voltage will appear across the load.

Step-by-Step Solution:

Verification / Alternative check:
Find circuit current: I = VS / RT = 110 / 26 ≈ 4.2308 A. Then compute VL = I * RL ≈ 4.2308 * 2 ≈ 8.4616 V; rounding to the nearest tenth gives 8.5 V, consistent with the divider calculation.

Why Other Options Are Wrong:

• 85 V or 110 V: These would require RS to be negligible compared to RL, which is not the case.
• 0 V: Only if the circuit were open or the load removed.

Common Pitfalls:

• Ignoring internal resistance and assuming the full source voltage appears across the load.
• Rounding too early; perform the division first, then round the final result.

Final Answer:
8.5 V