Thursday, July 28, 2016

NEET Questions on Electronics

    – 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 ic through the collector resistance Rc is the ratio of output signal voltage vc to the collector resistance Rc:
            ic = vc/Rc = 4V/2 kΩ = 2 mA
The ratio of the collector current to the current amplification factor βac gives the signal current ib in the base circuit:
             ib = ic/βac = 2 mA/100 = 0.02 mA
[Note that βac is the amplification factor for small signal currents]
The input signal voltage vi is the voltage drop across the base resistance Rb due to the base current:
             vi = ibRb = 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 Av of a common emitter amplifier given by
            Av = vo/vi = βacRo/ri  = 100×(2 kΩ/1kΩ) = 200
Note that Ro is the output resistance and ri is the input resistance.
This gives vi = vo/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).

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