Physics Electronics MDCAT 2000
PMDC Verified Question 486 of 494
RD1D2+6V0V

In the circuit shown above, diodes \( D_1 \) and \( D_2 \) have an ideal forward resistance of \( 50\ \Omega \) and infinite reverse resistance. If a \( +6.0\text{ V} \) DC potential is applied at the left terminal and \( R = 100\ \Omega \), what is the total current flowing through \( R \)?
A
0.040 A
B
0.020 A
C
0.000 A
D
0.060 A
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option A: 0.040 A
Concept:

When a positive potential is applied to the common input, analyze the biasing state of each parallel diode branch.

Formula:

$$I_{\text{total}} = \frac{V}{R_{\text{branch}} + R_L}$$

Solution:

  • \( D_1 \) has its anode connected to \( +6\text{ V} \), so it is forward-biased with \( R_{D1} = 50\ \Omega \).


  • \( D_2 \) has its cathode connected to \( +6\text{ V} \), so it is reverse-biased and acts as an open circuit (\( R_{D2} = \infty \)).


  • Current flows exclusively through the branch with \( D_1 \).


  • Total resistance: \( R_{\text{total}} = R_{D1} + R = 50\ \Omega + 100\ \Omega = 150\ \Omega \).


  • Circuit current: \( I = \frac{6.0\text{ V}}{150\ \Omega} = 0.040\text{ A} = 40\text{ mA} \).


Why other options are incorrect:

  • Option B: 0.020 A incorrectly treats both diodes as conducting in series (\( 6/300 \)).
  • Option C: 0.000 A assumes both diodes are reverse-biased.
  • Option D: 0.060 A neglects the diode forward resistance (\( 6/100 \)).

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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