Reverse-biased diode in a simple series circuit A DC source drives a series combination of a resistor and an ideal diode. If the diode is reverse biased (no conduction), what voltage does the resistor drop relative to the source voltage in steady state?

Introduction / Context:
Series diode-resistor circuits are common in clamping, steering, and protection networks. Knowing which element bears the source voltage under forward versus reverse bias prevents misinterpretation of measured node voltages.

Given Data / Assumptions:

• Ideal diode model for clarity.
• Series DC source, resistor R, and diode D.
• Diode is reverse biased so current ≈ 0 A in steady state.

Concept / Approach:
In steady state with an ideal reverse-biased diode, no current flows. The resistor’s voltage drop is V_R = I * R = 0, so it drops essentially 0 V. The entire source voltage appears across the open element—the diode—because it is the break in the path. This is consistent with KVL: the sum of element drops must equal the source; if the resistor drop is ~0, the diode must account for nearly the full source voltage.

Step-by-Step Solution:

Verification / Alternative check:
Measure with a voltmeter: the resistor ends read nearly the same potential; across the diode you read nearly the full source voltage (polarized in reverse). Any small leakage current in real diodes can produce a negligible resistor drop.

Why Other Options Are Wrong:

Common Pitfalls:
Forgetting that series elements share the same current; if the current is zero, every resistive element drops zero volts.

Final Answer:
Approximately 0 V (the diode drops nearly the full source voltage)