## Invitation to Computer Science 8th Edition

Published by Cengage Learning

# Chapter 4 - 4.4 - Building Computer Circuits - Practice Problems - Page 209: 1

#### Answer

Each 1 -CE circuit contains two NOT gates, two AND gates, and one OR gate. This will require $(2 \times 1)+(2 \times 3)+(1 \times 3)=2+6+3=11$ transistors. The 32 -bit compare-for-equality circuit of Figure 4.28 has the following components: 32 1-CE circuits $=32 \times 11=352$ transistors 31 AND gates $=31 \times 3=93$ transistors Total $=352+93=445$ transistors

#### Work Step by Step

Each 1 -CE circuit contains two NOT gates, two AND gates, and one OR gate. This will require $(2 \times 1)+(2 \times 3)+(1 \times 3)=2+6+3=11$ transistors. The 32 -bit compare-for-equality circuit of Figure 4.28 has the following components: 32 1-CE circuits $=32 \times 11=352$ transistors 31 AND gates $=31 \times 3=93$ transistors Total $=352+93=445$ transistors

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