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NEET PHYSICSSEMICONDUCTOR ELECTRONICSMedium

Question

Consider the junction diode as ideal. The value of current flowing through AB is:

A

102 A10^{-2} \text{ A}

B

101 A10^{-1} \text{ A}

C

103 A10^{-3} \text{ A}

D

0 A0 \text{ A}

Step-by-Step Solution

Although the specific circuit diagram is missing from the question text, this relies on a standard NEET 2016 problem. In the corresponding circuit, point A is at a potential of +4 V+4 \text{ V} and is connected to the p-side of the diode, while point B is at 6 V-6 \text{ V} and connected to the n-side through a 1 kΩ1 \text{ k}\Omega (1000 Ω1000 \text{ } \Omega) resistor. Since the p-side is at a higher potential than the n-side (VA>VBV_A > V_B), the diode is forward-biased. An ideal diode offers zero resistance in forward bias. The potential difference across the resistor is V=VAVB=4 V(6 V)=10 VV = V_A - V_B = 4 \text{ V} - (-6 \text{ V}) = 10 \text{ V}. According to Ohm's law, the current is I=VR=10 V1000 Ω=0.01 A=102 AI = \frac{V}{R} = \frac{10 \text{ V}}{1000 \text{ } \Omega} = 0.01 \text{ A} = 10^{-2} \text{ A}.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from SEMICONDUCTOR ELECTRONICS. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSSEMICONDUCTOR ELECTRONICSconsiderjunctioncurrentflowingthrough

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