Semiconductors

JEE Main level, single correct, medium

161 of 209

Question 161 of 209

The logic performed by the circuit shown in figure is equivalent to

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Answer: (A) AND

All the gates used in the given configuration are NORNOR gates. Thus, the output can be found as follows-
Y=(a+b)=(ab)=ab\begin{matrix}Y & = & \left(a+b\right) \\ & = & \left(ab\right) \\ & = & ab\end{matrix}
The output of the given configuration can also be verified by the following truth table: aa bb OutputfromfirstgateOutputfromfirstgate OutputfromsecondgateOutputfromsecondgate YY 00 00 11 11 00 00 11 11 00 00 11 00 00 11 00 11 11 00 00 11 The calculation and the truth table clearly shows that the configuration represents ANDAND gate.

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