"I object to violence because when it appears to do good, the good is only temporary; the evil it does is permanent"

– Mahatma Gandhi

Here are three questions on electronics. These were included in the NEET 2016 (July) question paper:

(1) For CE transistor amplifier, the audio signal voltage across the collector resistance of 2 kΩ is 4V. If the current amplification factor of the transistor is 100 and the base resistance is 1 kΩ, then the input signal voltage is

(1) 30 mV

(2) 15 mV

(3) 10 mV

(4) 20 mV

The output signal current

*i*_{c}through the collector resistance R_{c}is the ratio of output signal voltage*v*_{c}to the collector resistance R_{c}:*i*

_{c}=

*v*

_{c}/R

_{c}= 4V/2 kΩ = 2 mA

The ratio of the collector current to the current amplification factor

*β*_{ac}gives the signal current*i*_{b}in the base circuit:*i*

_{b }=

*i*

_{c}/

*β*

_{ac}= 2 mA/100 = 0.02 mA

[Note that

*β*_{ac}is the amplification factor for small*signal currents*]The input signal voltage

*v*_{i}is the voltage drop across the base resistance R_{b}due to the base current:*v*

_{i}=

*i*

_{b}R

_{b}= 0.02 mA×1 kΩ = 0.02 volt =

**20 mV**

[You can obtain the answer in no time if you remember the expression for the voltage gain

*A*_{v}of a common emitter amplifier given by*A*

_{v}=

*v*

_{o}/

*v*

_{i}=

*β*

_{ac}

*R*

_{o}/

*r*

_{i}= 100×(2 kΩ/1kΩ) = 200

Note that

*R*_{o }is the output resistance and*r*_{i }is the input resistance.This gives

(2) The given circuit has two ideal diodes connected as shown in the figure below. *v*_{i }=*v*_{o}/200 = 4V/200 = 0.02 V = 20 mV]The current flowing through the resistance R

_{1}will be(1) 1.43 A

(2) 3.13 A

(3) 2.5 A

(4) 10.0 A

Since D

_{1}is reverse biased, no current will flow through it. We can therefore forget about the branch containing D_{1}and R_{2}. The battery sends a current through the series combination of R_{1}and R_{3}since the diode D_{2}is forward biased. Current through R

_{1}= 10 V/(2Ω + 2Ω) = 2.5 A(3) What is the output Y in the following circuit, when all the three inputs A,B,C are first 0 and then 1?

(1) 1,0

(2) 1,1

(3) 0,1

(4) 0.0

When all the three inputs are 0 the NAND gate following the AND gate has both inputs at 0 level. The output Y of the NAND is therefore 1. When all the three inputs are 1, the NAND gate has both inputs at 1 and hence the output is 0. The correct option is (1).