QJEEM24Ph961d: JEE Main 2024 Physics - Class 12 JEEM24Ph961d Solution: The truth table for this given circuit is :

Question

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The truth table for this given circuit is :

JEE Main 2024 (Online) 29th January Evening Shift Physics - Semiconductor Question 24 English

A https://ik.imagekit.io/novatra/photos/ad24eb88322d4688af3302d54cc1dcc8cedba6d24eb7c9d87bdbad905dc0b703.webp
B https://ik.imagekit.io/novatra/photos/ad8ecab0c3e23f1eac750c5961f0433bc71986e5499d1d6c926c5e86ec6337e9.webp
C https://ik.imagekit.io/novatra/photos/47900d36249b67bdc9d505e7e96d02ea9bd0b2a7e6117316fe86c5547c818b50.webp
D https://ik.imagekit.io/novatra/photos/16f70ca233ce7701af6c3f8e93cb1abcbe3883128ed7e9b691fe412e71c712ea.webp

Accepted Answer

JEE Main 2024 (Online) 29th January Evening Shift Physics - Semiconductor Question 24 English Explanation

$\begin{aligned} Y & = A \cdot B + \overset{―}{A} \cdot B \\ = (A + \overset{―}{A}) \cdot B \\ Y & = 1 \cdot B \\ Y & = B \end{aligned}$

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