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 R_L 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 I_CR_C = 0.6 V where I_C is the collector current and R_C is the resistance in the collector circuit.

Therefore,  I_C×600 Ω = 0.6 V from which I_C = 0.6/600 A = 10^–3 A = 1 mA.

Since the current gain β is given by

            β = I_C/I_B where I_B is the base current, we have

            I_B = I_C/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 n_h to 5×10²² m^–3. Then the value of n_e 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

            n_i² = n_en_h where n_i is the electron concentration as well as the hole concentration in the intrinsic (pure) semiconductor, n_e is the electron concentration in the doped semiconductor and n_h is the hole concentration in the doped semiconductor.

Therefore n_e = n_i²/n_h = (10¹⁶)²/(5×10²²) = 2×10⁹ m^–3