A forward source of 9 V is applied to a silicon diode in series with a 1 kΩ load resistor. The measured voltage across the load resistor is 0 V. What does this observation most strongly indicate about the diode or circuit condition?

Introduction / Context:In a simple series circuit (source → diode → resistor → return), the load resistor drop equals I * R. If the voltmeter across the resistor reads 0 V under an applied forward voltage, then the series current must be zero. The diagnostic task is to identify which fault is consistent with zero current and zero resistor drop.

Given Data / Assumptions:

• Supply voltage nominally +9 V forward-biasing the diode in series with a 1 kΩ resistor.
• Measured V_R across the resistor = 0 V.
• Ideal meter with negligible burden; wiring otherwise correct.

Concept / Approach:Ohm's law requires a nonzero voltage drop across the resistor if current flows. Therefore, V_R = 0 implies I = 0 A. In a series path, I = 0 indicates an open circuit somewhere in the chain. If the diode were shorted, the current would be approximately 9 V / 1 kΩ = 9 mA and the resistor drop would be about 9 V, not 0 V. Hence, a shorted diode is inconsistent with the observation. An open diode, however, completely stops current, producing zero resistor drop and leaving nearly all the source voltage across the open device.

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