Test: Transistors & Logic Gates (February 5) - NEET MCQ

# Test: Transistors & Logic Gates (February 5) - NEET MCQ

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## 10 Questions MCQ Test Daily Test for NEET Preparation - Test: Transistors & Logic Gates (February 5)

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Test: Transistors & Logic Gates (February 5) - Question 1

### The binary number 10101 is equivalent to decimal number

Detailed Solution for Test: Transistors & Logic Gates (February 5) - Question 1

10101 is the binary number to be converted to a decimal number.
To convert, multiply the numbers individually with 2n where n changes according to place value of the number.
(10101)2= (20× 1) + (0 × 21) + (1×22) + (0×23) + (1×24)
= 1 + 0 + 4 + 0 + 16
= 21

Test: Transistors & Logic Gates (February 5) - Question 2

### The NOR gate is OR gate followed by

Detailed Solution for Test: Transistors & Logic Gates (February 5) - Question 2

An OR gate followed by a NOT gate in a cascade is called a NOR gate. In other words, the gate which provides a high output signal only when there are low signals on the inputs such type of gate is known as NOR gate.

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Test: Transistors & Logic Gates (February 5) - Question 3

### For functioning of a transistor its emitter-base junction has to be:

Detailed Solution for Test: Transistors & Logic Gates (February 5) - Question 3

For a proper functioning of the transistor, the emitter-base region must be forward-biased and collector-base region must be reverse-biased. In semiconductor circuits, the source voltage is called the bias voltage. In order to function, bipolar transistors must have both junctions biased.

Test: Transistors & Logic Gates (February 5) - Question 4

A transistor is used in the common-emitter mode as an amplifier. Then

Detailed Solution for Test: Transistors & Logic Gates (February 5) - Question 4

When a transistor is used in the common emitter mode as an amplifier, the base-emitter junction acts as input junction and collector-emitter junction acts as output junction. The input junction is made forward biased by applying forward voltage and output junction is made reverse biased by applying reverse voltage. The input signal is connected in series with the voltage applied to forward bias the base-emitter junction.

Test: Transistors & Logic Gates (February 5) - Question 5

In an n-p-n transistor, the majority carriers in the emitter and collector are :

Detailed Solution for Test: Transistors & Logic Gates (February 5) - Question 5

As we can see from the diagram, the emitter in the npn transistor is of n-type. Therefore majority charge carriers in the emitter of an npn transistor are called electrons.

Test: Transistors & Logic Gates (February 5) - Question 6

In an n-p-n transistor, the emitter current is

Detailed Solution for Test: Transistors & Logic Gates (February 5) - Question 6

In the transistor circuit, the current flows from emitter to collector via base.
Ie​=Ib​+Ic​ where variables have their usual meanings.
Hence, emitter current is slightly more than the collector current.

Test: Transistors & Logic Gates (February 5) - Question 7

The binary number system has a base of​

Detailed Solution for Test: Transistors & Logic Gates (February 5) - Question 7

Binary number system, in mathematics, positional numeral system employing 2 as the base and so requiring only two different symbols for its digits, 0 and 1, instead of the usual 10 different symbols needed in the decimal system.

Test: Transistors & Logic Gates (February 5) - Question 8

Which of the following is the universal gate?

Detailed Solution for Test: Transistors & Logic Gates (February 5) - Question 8

A universal gate is a gate which can implement any Boolean function without need to use any other gate type. The NAND and NOR gates are universal gates. In practice, this is advantageous since NAND and NOR gates are economical and easier to fabricate and are the basic gates used in all IC digital logic families.

Test: Transistors & Logic Gates (February 5) - Question 9

If A = 1 and B = 0, what is the value of A.A+B?​

Detailed Solution for Test: Transistors & Logic Gates (February 5) - Question 9

A=1 and B=0
A.A+B
=A2+B [De-Morgan’s theorem)
=1+0
=1
=A

Test: Transistors & Logic Gates (February 5) - Question 10

Which among the following gates has output = A . B ?​

Detailed Solution for Test: Transistors & Logic Gates (February 5) - Question 10

The logic or Boolean expression given for a digital logic AND gate is that for Logical Multiplication which is denoted by a single dot or full stop symbol, ( . ) giving us the Boolean expression of:  A.B = Y.

Then we can define the operation of a digital 2-input logic AND gate as being:
“If both A and B are true, then Y is true”

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