# MCQs on Digital Electronics

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01․ Which of the following operations is commutative but not associative?
AND
NOR
OR
X-OR

Let A, B and C are inputs, AND gate: 1. AB = BA, therefore it is following cumulative law. 2.f ABC = (AB)C = (AC)B = (BC)A, therefore it is following associative law. OR gate: 1. A + B = B + A, therefore it is following cumulative law. 2. A + B + C = (A + B) + C = (A + C) + B = (B + c) +A, therefore it is following associative law. In similar way X-Or gate will follow the both laws. NAND, NOR and X-NOR will follow only commutative law but not follow the associative law.

02․ A 32 × 10 ROM contains a decoder of size
5 × 32
32 × 32
32 × 10
10 × 32

Decoder is a combinational circuit which has many inputs and many outputs. Decoder is used to convert binary to other codes. Such as, binary to octal, binary to hexa decimal and BCD to decimal. The size of the decode is N×2N. Therefore, for given problem size of the decoder is 5 × 32.

03․ An 8 × 4 ROM contains a decoder of size
3 × 8
8 × 4
8 × 8
3 × 4

Decoder is a combinational circuit which have many inputs and many outputs. Decoder is used to convert binary to other codes. Such as , binary to octal, binary to hexa decimal and BCD to decimal. The size of the decode is N×2N. Therefore, for given problem size of the decoder is 3 × 8.

04․ The address bus with a ROM of size 1024 × 8 bits is
8 bits
10 bits
12 bits
16 bits

The size of the ROM is 1024 × 8 = 210× 8 Here 10 indicate the address bus. Therefore, the address bus with a ROM of size 1024 × 8 bits is 10 bits.

05․ The decimal equivalent of the highest possible address for an 8-bit address bus is
8
128
255
256

The decimal equivalent of the highest possible address for an 8-bit address bus is = 28-1 = 255

06․ The data bus width of a ROM of size 2048 × 8 bits is
8
10
12
16

The size of the ROM is 2048 × 8 = 211× 8 Here 11 indicate the address bus and 8 indicates the data bus width.

07․ A ROM has a 16-bit address bus. The number of locations in this memory is
16
32
1024
65536

The number of locations of a n-bit address bus = 2n The number of locations of a 16-bit address bus = 216 = 65536

08․ It is desired to have a 64 × 8 ROM. The ROMs available are of 16 × 4 size. The number of ROMs required will be
8
6
4
2

The number of ROMs required = (64 × 8)/( 16 × 4) = (128× 4)/( 16 ×4) = 8

09․ Four ROM chips of 16 × 4 size have their address buses connected together. This system will be of size
64 × 4
16 × 16
32 × 8
256 × 1

Four ROM chips of 16 × 4 size have their address buses connected together. This system will be of size 64 × 4.

10․ In a read-only memory information can be stored
at the time of fabrication
by the user only once during its life time
by the user a number of times
in any of the above ways depending upon the type of memory

1. Permament ROM’s a) MROM b) PROM
2. Erasable ROM’s a) EPROM b) EEPROM
Therefore, depending upon the type of memory can be stored or erased.

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