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NEET PHYSICSSEMICONDUCTOR ELECTRONICSMedium

Question

Two ideal diodes are connected to a battery as shown in the circuit. The current supplied by the battery is:

A

0.75 A0.75 \text{ A}

B

zero

C

0.25 A0.25 \text{ A}

D

0.5 A0.5 \text{ A}

Step-by-Step Solution

Since the circuit diagram is missing, we can deduce its structure from the standard AIPMT 2012 question. The circuit consists of a 5 V5 \text{ V} battery connected in parallel to two branches. The first branch has an ideal diode D1D_1 in series with a 10 Ω10 \text{ } \Omega resistor, and the second branch has an ideal diode D2D_2 in series with a 20 Ω20 \text{ } \Omega resistor. Diode D1D_1 is forward-biased and conducts, offering zero resistance. Diode D2D_2 is reverse-biased and acts as an open circuit, so no current flows through the 20 Ω20 \text{ } \Omega resistor. The total current supplied by the battery flows only through the 10 Ω10 \text{ } \Omega resistor. According to Ohm's law, I=VR=5 V10 Ω=0.5 AI = \frac{V}{R} = \frac{5 \text{ V}}{10 \text{ } \Omega} = 0.5 \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 ELECTRONICSdiodesconnectedbatterycircuitcurrent

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