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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Four inputs A, B, C, D are fed to a NOR gate. The output of NOR gate is fed to an inverter. The output of inverter is |
| Answer» Output = = A + B + C + D. | |
| 2. |
For the logic circuit of the given figure, the minimized expression is |
| Answer» + A + C = A + B + C + A + C = A + C + B = | |
| 3. |
A 4 bit synchronous counter uses flip flops with a delay time of 15 ns each. The time required for change of state is |
| Answer» In a synchronous counter clock input is applied to all flip flops simultaneously. Hence total delay time is 15 ns. | |
| 4. |
110.11 x 110 = __________ |
| Answer» 110.11 =6.75 and 110 = 6 and 6.75 x 6 = 40.5 in decimal. | |
| 5. |
Y = A + B is the same as |
| Answer» Verify by preparing truth table. | |
| 6. |
The inputs to a NAND gate are as shown in the given figure. The waveform of output is |
| Answer» NAND gate gives high output when both inputs are Low. | |
| 7. |
In the decimal number 27, the digital 2 represents |
| Answer» 2 x 10 = 20. | |
| 8. |
In the circuit of the given figure, V = |
| Answer» Transistor is off and circuit current is zero. | |
| 9. |
A parallel in-parallel out shift register can be used to introduce delay in digital circuits. |
| Answer» No time delay in this shift register since all bits are loaded and read simultaneously. | |
| 10. |
E7F6 = __________ . |
| Answer» E7 F6 in hexadecimal = 14 x 163 + 7 x 162 + 15 x 16 + 6 = 59382 in decimal. | |
| 11. |
A 14 pin AND gate IC has __________ AND gates. |
| Answer» Each AND gates requires 3 pins, 2 inputs and one output, 1 supply pin and 1 ground pin. Hence 4 AND gates. | |
| 12. |
For the design of a sequential circuit having 9 states, minimum number of memory elements required is |
| Answer» 23 = 8 and 24 = 16. Hence minimum of 4 memory elements. | |
| 13. |
ICs are |
| Answer» Op-amp is an analog IC. | |
| 14. |
The hexadecimal number 'A0' has the decimal value |
| Answer» 'Ao' = 10 x 161 + 0 = 160. | |
| 15. |
A 4 bit ripple counter is in 0000 state. The clock pulses are applied and then removed. The counter reads 0011. The number of clock pulses which have occurred are |
| Answer» After 16 pulses the counter resets. 0011 is 3. Hence clock pulses can be 3 or 3 + 16 or 3 + 16 + 16. | |
| 16. |
If an analog voltage is expressed in binary using 4 bits, each successive binary count would represent |
| Answer» Since 1111= decimal 15, each successive binary count represent 1/15 of total voltage. | |
| 17. |
In a 4 input AND gate, the total number of High outputs for 16 input states are |
| Answer» In only one combination are all inputs high. | |
| 18. |
A digital system is required to amplify a binary encoded audio signal. The user should be able to control the gain of the amplifier from a minimum to a maximum in 100 increments. The minimum number of bits required to encode, in straight binary is, |
| Answer» Let 6 bit required = 26 = 64, which indicate 64 increment. If 7 bit = 27 = 128, for 100 increment, 7 bit required. | |
| 19. |
A 4 bit transistor register has output voltage of high-low-high-low. The binary number stored and its decimal equivalent are |
| Answer» 1010 in binary and 8 + 2 = 10 in decimal. | |
| 20. |
Hexadecimal number F is called to octal number |
| Answer» F = 15 in decimal notation = 17 in octal notation (8 + 7 = 15). | |
| 21. |
A half adder can be used only for adding |
| Answer» | |
| 22. |
Symmetrical square wave of time period 100 μs can be obtained from square wave of time period 10 μs by using |
| Answer» Frequency has to be divided by 10. | |
| 23. |
The initial state of a Mod-16 counter is 0110. After 37 clock pulses the state of counter will be |
| Answer» After 37 clock pulses the state will be the same as after (37 - 32), i.e., 5 clock pulses. 6 + 5 = 11. Hence 1011. | |
| 24. |
The circuit in the figure is has two CMOS-NOR gates. This circuit functions as a |
| Answer» It is simple monostable multivibrator. | |
| 25. |
A 6 bit DAC uses binary weighted resistors. If MSB resistor is 20 k ohm, the value of LSB resistor is |
| Answer» Resistances are R, 2R, 4R, 8R, 16R and 32R. LSB resistance = 32R = 32 x 20 = 640 K ohm. | |
| 26. |
A 6 bit dual slope A/D converter uses a reference of -6v and a 1 MHz clock. It uses a fixed count of 40 (101000). Then, what will be input, if the output register shows 100111 at the end of conversion. |
| Answer» I/P voltage and reference voltage arc of opposite polarity. | |
| 27. |
In the TTL circuit in the figure, S to S are select lines and X to X are input lines. S and X are LSBs. The output Y is |
| Answer» The MUX is made up of TTL circuit. For TTL circuit open terminal is taken high, since S2 select line is connected to OR gate whose one terminal connected to C and the other is open (high) so OR gate output is S2 = 1 + C = 1. S2 = 1 S1(B) S0(A) Y 1 0 0 0 1 0 1 1 1 1 0 1 1 1 1 0 Y = S0⊕S1 => A⊕B. | |
| 28. |
A 12 bit ADC is employed to convert an analog voltage of zero to 10 volts. The resolution of the ADC is |
| Answer» Full scale measurement range = 0 to 10 volts. ADC resolution is 12 bits: 2^12 = 4096 quantization levels. ADC voltage resolution is: (10-0)/4096 = 0.00244 volts = 2.44 mV. | |
| 29. |
A device which converts BCD to seven segment is called |
| Answer» Decoder converts binary/BCD to alphanumeric. | |
| 30. |
In 2's complement representation the number 11100101 represents the decimal number |
| Answer» A = 11100101. Therefore A = 00011010 and A' = A + 1 = 00011011 = 16 + 8 + 2 + 1 = 27. Therefore A = -27. | |
| 31. |
A decade counter skips |
| Answer» A decade counter counts from 0 to 9. It has 4 flip-flops. The states skipped are 10 to 15 or 1010 to 1111. | |
| 32. |
BCD input 1000 is fed to a 7 segment display through a BCD to 7 segment decoder/driver. The segments which will lit up are |
| Answer» 1000 equals decimal 8 Therefore all segments will lit up. | |
| 33. |
Octal number 12 is equal to decimal number |
| Answer» 12 in octal = 8 + 2, i.e., 10 in decimal. | |
| 34. |
Hexadecimal number E is equal to binary number |
| Answer» E = 14 in decimal or 1110 (8 + 4 + 2 + 0 = 14) is binary. | |
| 35. |
Decimal number 46 in excess 3 code = |
| Answer» Decimal 46 in excess 3 code = 46 + 33 = 79 in decimal = 0111 1001 in 4 bit binary. | |
| 36. |
A multilevel decoder is more costly as compared to single level decoder. |
| Answer» A multi level decoder may be cheaper than single level decoder. | |
| 37. |
EPROM can be |
| Answer» Both ultra violet and electric pulses are used for erasing. | |
| 38. |
A full subtractor has a total of |
| Answer» | |
| 39. |
A full adder adds |
| Answer» | |
| 40. |
For a Mod 64 parallel counter we need |
| Answer» 26 = 64 So we need 6 flip-flops and (6 - 2) = 4 AND gates. | |
| 41. |
If the inputs to a 2 input XOR gate are high then, the output is high. |
| Answer» Output is Low. | |
| 42. |
The number of select lines in a 16 : 1 multiplexer are |
| Answer» 24 = 16. | |
| 43. |
BCD number 1100111 = __________ |
| Answer» 0110 = 6 and 0111 = 7 Hence 67. | |
| 44. |
The K-map for a Boolean function is shown in figure. The number of essential prime implicants for this function is |
| Answer» Essential Prime implicates = 4. | |
| 45. |
When microprocessor processes both positive and negative numbers, the representation used is |
| Answer» 2's complement representation requires simple electronic circuitry. | |
| 46. |
TTL is used in electronic calculators. |
| Answer» Calculators use CMOS. | |
| 47. |
A 4 bit down counter can count from |
| Answer» It can count from 15 to 0 or 1111 to 0000. | |
| 48. |
For the circuit of the given figure, the output equation is |
| Answer» The inputs A and B are ANDed. Inputs B and C' so ANDed. The outputs of two AND gates are ORed. | |
| 49. |
Using 2's complement, the largest positive and negative number which can be stored with 8 bits are |
| Answer» Largest positive number is 0111 111 = + 127 and largest negative number is 1000 0000 = -128. | |
| 50. |
Three Mod-16 counters are connected in cascade. The maximum counting range is |
| Answer» 16 x 16 x 16 = 4096. | |