Saturday, April 14, 2012

KEAM (Engineering) 2010 Questions (MCQ) on Electronics

"You may never know what results come of your actions, but if you do nothing, there will be no results."
–Mahatma Gandhi

Questions on electronics will be generally interesting to most of you. Today we will discuss questions in this section which appeared in Kerala engineering entrance (KEAM - Engineering) 2010 question paper. Here are the questions with their solution:
(1) A full wave rectifier with an a.c. input is shown:

The output  voltage across RL is represented as

The rectified output voltage will be a direct voltage but there will be very large amount of ripples. The capacitor C acts as a filter to remove the ripples; but there will still be a small amount of  ripples in the output. Therefore the correct option is (e).
(2) In the given circuit the current through the battery is
(a) 0.5 A
(b) 1 A
(c) 1.5 A
(d) 2 A
(e) 2.5 A
Since the diode D1 is reverse biased, no current will flow through the D1 branch. Diodes D2 and D3 are forward biased and hence the battery drives currents through the 20 Ω resistor and the series combination of the two 5 Ω resistors.
The current driven through the 20 Ω resistor is 10 V/20 Ω = 0.5 A.
The current driven through the 10 Ω resistor is 10 V/10 Ω = 1 A.
Therefore, total current through the battery is 0.5 A + 1 A = 1.5 A
(3) The collector supply voltage is 6 V and the voltage drop across a resistor of 600 Ω in the collector circuit is 0.6 V, in a transistor connected in common emitter mode. If the current gain is 20, the base current is
(a) 0.25 mA
(b) 0.05 mA
(c) 0.12 mA
(d) 0.02 mA
(e) 0.07 mA
We have ICRC = 0.6 V where IC is the collector current and RC is the resistance in the collector circuit.
Therefore,  IC×600 Ω = 0.6 V from which IC = 0.6/600 A = 10–3 A = 1 mA.
Since the current gain β is given by
            β = IC/IB where IB is the base current, we have
            IB = IC/β = 1 mA/20 = 0.05 mA.
(4) A pure semiconductor has equal electron and hole concentration of 1016 m–3. Doping by indium increases nh to 5×1022 m–3. Then the value of ne in the doped semiconductor is
(a) 106 m–3
(b) 1022 m–3
(c) 2×106 m–3
(d) 1019 m–3
(e) 2×109 m–3
According to the law of mass action we have
            ni2 = nenh where ni is the electron concentration as well as the hole concentration in the intrinsic (pure) semiconductor, ne is the electron concentration in the doped semiconductor and nh is the hole concentration in the doped semiconductor.
Therefore ne = ni2/nh = (1016)2/(5×1022) = 2×109 m–3

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