Physics Electronics PMDC Conceptual Practice
PMDC Verified Question 119 of 494
A silicon diode (\( V_D = 0.7\text{ V} \)) is connected in series with a \( 100\ \Omega \) resistor and a \( 10.0\text{ V} \) DC battery in forward bias. What is the current flowing in the circuit?
A
0.100 A
B
0.007 A
C
0.050 A
D
0.093 A
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option D: 0.093 A
Concept:

Applying Kirchhoff's Voltage Law (KVL) to a series diode-resistor circuit accounts for the diode's built-in forward voltage drop.

Formula:

$$V_B - V_D - I R = 0 \implies I = \frac{V_B - V_D}{R}$$

Solution:

  • Given: \( V_B = 10.0\text{ V} \), \( V_D = 0.7\text{ V} \), \( R = 100\ \Omega \).


  • Voltage across resistor: \( V_R = 10.0 - 0.7 = 9.3\text{ V} \).


  • Current: \( I = \frac{9.3\text{ V}}{100\ \Omega} = 0.093\text{ A} = 93\text{ mA} \).


Why other options are incorrect:

  • Option A: 0.100 A (100 mA) incorrectly neglects the 0.7 V forward drop across the diode (\( 10/100 \)).
  • Option B: 0.007 A incorrectly divides the diode drop by the resistance (\( 0.7/100 \)).
  • Option C: 0.050 A assumes an arbitrary 5.0 V drop across the resistor.

Quality & Fidelity Assurance: Every question on BeambePrep is rigorously curated against the official PMDC syllabus with zero filler, zero out-of-syllabus content, and zero typos. When an authentic past paper originally contained a historical mistake or ambiguity from the examining board (such as UHS or NUMS), BeambePrep faithfully reflects the original paper while detailing the nuance and scientific consensus in the autopsy above.

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