Single-resistor drop equals source voltage: If a 175 V DC source is applied across a lone resistor R1 (no other series elements), the voltage across R1 equals 175 V under ideal conditions. Assess the correctness of this statement.

Introduction / Context:A common measurement scenario is applying a source across a single component. In such a case, the entire source voltage appears across that component because there is nowhere else for the drop to occur. This reinforces the basic node-voltage concept used in all series–parallel analyses.

Given Data / Assumptions:

• Ideal 175 V source.
• Only one element present: resistor R1 across the source terminals.
• Negligible lead/wiring drops; meter loading ignored.

Concept / Approach:With only a single element across the source, KVL dictates that the sum of drops around the loop equals the source voltage. Since R1 is the only drop, V_R1 must equal 175 V. Ohm’s law then determines the current (I = 175 / R1), but the resistor value does not change the magnitude of the voltage imposed by the source across it.

Step-by-Step Solution:

Verification / Alternative check:Bench test with a regulated supply and DMM shows the full source voltage across a single load, subject only to tolerance and negligible lead resistance effects.

Why Other Options Are Wrong:

• Incorrect: Violates KVL.
• True only if R1 = 175 Ω: Resistance value sets current, not applied voltage.
• Depends on current range or room temperature: Meter range and temperature do not alter the ideal circuit law.

Common Pitfalls:Imagining “voltage division” where no division exists; mixing up current settings on meters with voltage measurements; overlooking that all of the source voltage must appear across the sole component.

Final Answer:Correct