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

ideal diode and 1 kOhm resistor between A (+4 V) and B (-6 V)

1. 0 A

2. \(10^{-2}\) A

3. \(10^{-1}\) A

4. \(10^{-3}\) A

Doubtly Asked 30 Oct 2026 NEET UG 2016 Phase 1 Practice test neet Save WhatsApp Report

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Answer: 2. \(10^{-2}\) A

Key idea: check whether the ideal diode is forward or reverse biased.

A is at +4 V (on the p-side) and B at -6 V, so the diode is forward biased and acts like a plain wire.
Voltage across the resistor \(=4-(-6)=10\) V.
\(I=\dfrac{10}{1000}=10^{-2}\) A.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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