"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

(b) 1 A

(c) 1.5 A

(d) 2 A

(e) 2.5 A

Since the diode D

_{1}is reverse biased, no current will flow through the D_{1}branch. Diodes D_{2}and D_{3}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*_{C}*R*_{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 10

^{16}m^{–3}. Doping by indium increases*n*_{h}to 5×10^{22}m^{–3}. Then the value of*n*_{e}in the doped semiconductor is
(a) 10

^{6}m^{–3}
(b) 10

^{22}m^{–3}
(c) 2×10

^{6}m^{–3}
(d) 10

^{19}m^{–3}
(e) 2×10

^{9}m^{–3}
According to the law of mass action we have

*n*

_{i}

^{2}=

*n*

_{e}

*n*

_{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*_{i}^{2}/*n*_{h}= (10^{16})^{2}/(5×10^{22}) = 2×10^{9}m^{–3}
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