Loading effects: A smaller-value load resistor (lower resistance) will cause the measured output voltage to change more relative to no-load than a larger-value load. True or false?

Introduction / Context:
Any real source has a nonzero source (Thevenin) resistance. When a load is connected, the output voltage depends on the voltage divider formed by the source resistance and the load. Recognizing how load value affects output is key in power supplies, sensors, and measurement circuits.

Given Data / Assumptions:

• Source modeled as Vs in series with Rs.
• Load resistor RL connects across the output terminals.
• We compare the effect of small RL versus large RL on output voltage.

Concept / Approach:

The output voltage across RL is Vout = Vs * (RL / (Rs + RL)). For a fixed Rs, decreasing RL decreases Vout due to a larger share of the drop across RL's complement (Rs). This represents stronger loading (more current drawn), therefore greater deviation from the no-load (open-circuit) voltage Vs.

Step-by-Step Solution:

Verification / Alternative check:

Try numbers: Vs = 10 V, Rs = 100 Ω. With RL = 10 kΩ, Vout ≈ 9.90 V. With RL = 100 Ω, Vout = 5 V. The smaller load clearly produces a greater change from 10 V (no-load) to 5 V.

Why Other Options Are Wrong:

• Choosing “False” suggests larger loads disturb the source more, which contradicts the divider formula and practical observation.

Common Pitfalls:

Equating “smaller-value load” with “light loading.” In fact, a smaller resistance means heavier loading (more current) and bigger voltage sag from the source.

Final Answer:

True