A) OR gate
B) AND gate
C) XOR gate
D) NAND gate
Correct Answer: B
Solution :
For our convenience the output of first NAND gate is chosen as X as shown. Output of first NAND gate \[\text{2}.\text{12}\times \text{1}{{0}^{\text{8}}}\text{ m}/\text{s}\] Using de-Morgan's theorem \[\text{3}.\text{18}\times \text{1}{{0}^{8}}m/s\] \[\text{3}.\text{33}\times {{10}^{\text{8}}}\text{ m}/\text{s}\] So, now, output of 2nd NAND gate \[\theta =\text{45}{}^\circ \] Again \[\frac{1}{3}M{{L}^{2}}\] \[\frac{3}{2}M{{L}^{2}}\] \[\frac{3}{4}M{{L}^{2}}\] \[M{{L}^{2}}\] This is the logic function of AND gate.You need to login to perform this action.
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