1
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
2
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
3
MCQ (Single Correct Answer)

NEET 2016 Phase 2

The given circuit has two ideal diodes connected as shown in the figure. The current flowing through the resistance R1 will be

A
2.5 A
B
10.0 A
C
1.43 A
D
3.13 A

Explanation

Current will not flow through D1 as it is reversed biased.

Current will flow through resistor R1, diode D2 and resistor R3

Now current i = 10/(2 + 2) = 2.5 A
4
MCQ (Single Correct Answer)

NEET 2016 Phase 2

For CE transistor amplifier, the aufio signal voltage across the collector resistance of 2 k$$\Omega $$ is 4 V. If the current amplification factor of the transistor is 100 and the base resistance is 1 k$$\Omega $$, then the input signal voltage is
A
10 mV
B
20 mV
C
30 mV
D
15 mV

Explanation

Voltage gain, A = $$\beta {{{R_C}} \over {{R_B}}}$$ = $$100 \times {{2000} \over {1000}}$$ = 200

Also, A = $${{{V_0}} \over {{V_i}}}$$

$$ \Rightarrow $$ $${V_i} = {{{V_0}} \over A}$$ = $${4 \over {200}}$$ = 20 mA

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