1
MCQ (Single Correct Answer)

NEET 2016 Phase 1

Consider the junction diode as ideal. The value of current flowing through AB is

A
10$$-$$1 A
B
10$$-$$3 A
C
0 A
D
10$$-$$2 A

Explanation

Here, the p-n junction diode is forward biased, hence it offers zero resistance.

So IAB = $${{4 - \left( { - 6} \right)} \over {1 \times {{10}^3}}}$$ = 10-2 A
2
MCQ (Single Correct Answer)

NEET 2016 Phase 1

A npn transistor is connected in common emitter configuration in a given amplifier. A load resistance of 800 $$\Omega $$ is connected in the collector circuit and the voltage frop across it is 0.8 V. If the current amplification factor is 0.96 and the input resistance of the circuit is 192 $$\Omega $$, the voltage gain and the power gain of the amplifier will respectively be
A
4, 4
B
4, 3.69
C
4, 3.84
D
3.69, 3.84

Explanation

Voltage gain = Current gain × Resistance gain

= 0.96 $$ \times $$ $${{800} \over {192}}$$ = 4

Power gain = [Current gain] × [Voltage gain]

= 0.96 × 4 = 3.84
3
MCQ (Single Correct Answer)

NEET 2016 Phase 1

To get output 1 for the following circuit, the correct choice for the input is

A
A = 1, B = 1, C = 0
B
A = 1, B = 0, C = 1
C
A = 0, B = 1, C = 0
D
A = 1, B = 0, C = 0

Explanation

Output of the circuit, Y = (A + B)·C

Y = 1 if C = 1 and A = 0, B = 1

or A = 1, B = 0 or A = B = 1
4
MCQ (Single Correct Answer)

NEET 2016 Phase 2

What is the output Y in the following circuit, when all the three inputs A, B, C are first 0 and then 1 ?

A
0, 1
B
0, 0
C
1, 0
D
1, 1

Explanation

Applying De Morgan’s law:

Output Y = [(A ⋅ B) ⋅ C]' = A' + B' + C'

When A, B, C are 0 $$ \Rightarrow $$ Y = 1

When A, B, C are 1 $$ \Rightarrow $$ Y = 0

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