NEET Questions on Electronics

"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_bR_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_acR_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 v_i = v_o/200 = 4V/200 = 0.02 V = 20 mV]

 (2) The given circuit has two ideal diodes connected as shown in the figure below.

 

The current flowing through the resistance R₁ will be

(1) 1.43 A

(2) 3.13 A

(3) 2.5 A

(4) 10.0 A

Since D₁ is reverse biased, no current will flow through it. We can therefore forget about the branch containing D₁ and R₂. The battery sends a current through the series combination of R₁ and R₃ since the diode D₂ is forward biased.

            Current through R₁ = 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).