Thevenin source with internal resistance: A 120 V source with internal resistance RS = 12 Ω feeds a load RL = 470 Ω. What is the load voltage across RL?

Introduction / Context:
Real voltage sources can be modeled as an ideal source in series with an internal resistance (Thevenin model). The load voltage is then found using a simple voltage-divider relationship between the internal resistance and the load resistance.

Given Data / Assumptions:

• Ideal source voltage V_S = 120 V.
• Internal resistance R_S = 12 Ω (series).
• Load resistance R_L = 470 Ω.
• Steady-state DC conditions.

Concept / Approach:

For a series divider, the load voltage is V_L = V_S * (R_L / (R_S + R_L)). Because R_L ≫ R_S here, the drop across R_S is small, so V_L will be close to the source voltage but slightly lower.

Step-by-Step Solution:

Verification / Alternative check:

Calculate current: I ≈ 120 / 482 ≈ 0.249 A. Drop on R_S: I * R_S ≈ 0.249 * 12 ≈ 2.99 V. Then V_L = 120 − 2.99 ≈ 117.01 V, matching the divider result.

Why Other Options Are Wrong:

120 V ignores internal resistance. 12 V is the small internal drop, not the load voltage. 0 V is only with open circuit at the load (not the case). 112 V is too low for these values.

Common Pitfalls:

Forgetting the internal resistance drop, or misapplying the divider formula by swapping R_S and R_L.

Final Answer:

117 V