Loaded voltage source with internal resistance: A 120 Ω load is connected to a source with VS = 12 V and source resistance RS = 8 Ω. What is the voltage across the load resistor?

Introduction / Context:Real voltage sources are modeled with an ideal source in series with an internal resistance (Thevenin form). When a load is attached, the output voltage divides between the source resistance and the load resistance according to the voltage-divider rule.

Given Data / Assumptions:

• Source voltage, VS = 12 V.
• Source (internal) resistance, RS = 8 Ω.
• Load resistance, RL = 120 Ω.
• All components are ideal and operated under steady-state DC conditions.

Concept / Approach:Use the voltage-divider relation for series elements: VL = VS * (RL / (RS + RL)). This follows directly from Ohm’s law in a simple series path where the same current flows through RS and RL.

Step-by-Step Solution:

Verification / Alternative check:Find current first: I = VS / RT = 12 / 128 = 0.09375 A. Then VL = I * RL = 0.09375 * 120 = 11.25 V, which matches the divider result.

Why Other Options Are Wrong:

• 12 V: Would be true only if RS = 0 Ω (ideal source with no internal drop).
• 1.13 V: Off by a factor of 10; likely from mistakenly using RS instead of RL in the divider.
• 0 V: Only if the source were shorted or the load were disconnected—neither applies.

Common Pitfalls:

• Forgetting the internal resistance in series with the load.
• Arithmetic errors when computing the fraction RL / (RS + RL).

Final Answer:11.25 V