## Friday, November 23, 2012

### Multiple Choice Questions on Transistors including EAMCET Engineering 2004 and 2005 Questions

“The pursuit of truth and beauty is a sphere of activity in which we are permitted to remain children all our lives.”
– Albert Einstein

A very important semiconductor device you need to study is the transistor, the invention of which brought about a revolution in electronics. Questions involving transistors have been discussed earlier on this site. You can access them by trying a search for ‘transistor’ using the search box provided on this page or by clicking on the label ‘transistor’ below this post. Today we shall discuss a few more questions on transistors
(1) In an amplifier circuit using a transistor the collector current changes by1.224 mA when the emitter current changes by 1.228 mA. What is the common emitter small signal current gain of the transistor used in the circuit?
(a) 0.997 (very nearly)
(b) 1.003 (very nearly)
(c) 122 (very nearly)
(d) 307
(e) 306
The small signal common emitter current gain is the ratio of the change in collector current to the change in base current. Since the emitter current is the sum of the collector current and the base current, the change in base current in the present case is 1.228 mA 1.224 mA which is equal to 0.004 mA.
Therefore common emitter small signal current gain (βac) = 1.224/0.004 = 306.
(2) In an n-p-n transistor in CE configuration
(i) the emitter is more heavily doped than the collector
(ii) emitter and collector can be interchanged
(iii) the base region is very thin but is heavily doped
(iv) the conventional current flows from base to emitter
(a) (i) and (ii) are correct
(b) (i) and (iii) are correct
(c) (i) and (iv) are correct
(d) (ii) and (iii) are correct
The above question appeared in EAMCET Engineering 2004 question paper.
Option (c) is the answer. Note that the configuration deoes not matter.
(3) An n-p-n transistor power amplifier in CE configuration gives
(a) voltage amplification only
(b) current amplification only
(c) both current and voltage amplifications
(d) only power gain of unity
This question appeared in EAMCET engineering 2005 question paper. In common emitter configuration a transistor (n-p-n as well as p-n-p) gives both current and voltage amplifications so that there will be appreciable power gain. Therefore the correct option is (c).

(4) The adjoining figure shows an n-p-n transistor operated in the common emitter configuration in which the potential divider resistors R1 and R2 provide the base voltage required for forward biasing the base emitter junction. A student wrongly selected too small a value for R1 so that the transistor got saturated. (When a transistor is saturated, its emitter to collector voltage is nearly zero). The collector load resistor RC is 5 KΩ and the emitter resistor RE (used for bias stabilization) is 1 KΩ. If βdc of the transistor is 200 what is the minimum base current required for saturating the transistor?
Even though this question may appear to be difficult at the first glance, it is really simple. You have to first calculate the current I which will produce a voltage drop of 12 volts across 6 KΩ.
[Since the emitter to collector voltage is nearly zero when the transistor is saturated, the entire supply voltage of 12 volts must be dropped across the resistors RC and RE].
Therefore we have
I × 6 KΩ = 12 V so that I = 2mA
The emitter current and collector current are nearly the same (because of large current gain βdc) so that we can take the collector current to be 2 mA. The base current required for producing a collector current (IC) of 2 mA is the saturating base current in this circuit. Therefore the minimum base current IB required for saturating the transistor is given by
IB = IC/βdc = 2 mA/200 = 0.01 mA = 10 μA.

## Monday, November 12, 2012

### Questions from Kinematics (Including Karnataka CET 2008 Question)

“God used beautiful mathematics in creating the world.”
– P.A.M. Dirac

Today we shall discuss a few questions (MCQ) in the section, ‘kinematics in one dimension’. The questions I give you are meant for testing your knowledge, comprehension and the ability for applying what you have learned in this section.

(1) The velocity-time graphs of a car and a motor bike traveling along a straight road are shown in the adjoining figure. At time t = 0 they have the same position co-ordinate.  Pick out the correct statement from the following:
(a) At time t = 0 the car and the motor bike are at rest.
(b) At time t = 0 the car is at rest but the motor bike is moving.
(c) The distances traveled by the car and the motor bike in time t1 are equal.
(d) The distance traveled by the bike in time t1 is twice the distance traveled by the car in the same time.
(e) The distance traveled by the bike in time t1 is half the distance traveled by the car in the same time.
You can easily conclude that options (a) and (b) are wrong.
To check the remaining options we use the equation of linear motion, s = ut + ½ at2 where s is the displacement in time t, u is the initial velocity and a is the uniform acceleration.
The distance traveled by the car in time t1 is vct1 where vc (let us say) is the constant velocity of the car.
[The velocity of the car is constant since the velocity-time graph of the car is parallel to the time axis].
The motion of the motor bike is uniformly accelerated and the acceleration a is given by
a =  vc/t1
[Note that the velocity of the motor bike changes from 0 to vc in time t1.
The distance s traveled by the motor bike in time t1 is given by
s = ut + ½ at2 = 0 + (½)×(vc/t1) t12
Or, s = vct1/2
Therefore the distance traveled by the bike in time t1 is half the distance traveled by the car in the same time [Option (e)].
(2) A student standing at the edge of a cliff throws a stone of mass m vertically upwards with speed v. It strikes the ground at the foot of the cliff with speed v1. The student then throws another stone of mass m/4 vertically downwards with speed v. It strikes the ground at the foot of the cliff with speed v2. If air resistance is negligible, v1 and v2 are related as
(a) v1 = v2
(b) v1 = v2/2
(c) v1 = 2v2
(d) v1 = v2/4
(e) v1 = 4v2
While returning, the sphere of mass m has downward speed v when it passes the edge of the cliff. For calculating the speed with which it strikes the ground at the foot of the cliff, the situation is similar to that of the stone of mass m/4 thrown downwards. Obviously both stones will strike the ground with the same speed [Option (a)].
(3) A body is projected vertically upwards. The times corresponding to height h while ascending and descending are t1 and t2 are respectively. Then the velocity of projection is (g is acceleration due to gravity)
(1) g√(t1t2)
(2) gt1t2/(t1+t2)
(3) g√(t1t2)/2
(4) g(t1+t2)/2
The above question appeared in Karnataka CET 2008 question paper.
We have the following two equations to give h:
h = ut1 –  (½) g t12…………….(i)
h = ut2 –  (½) g t22…………….(ii)
(We have taken the displacement h and the initial velocity u (both upwards) as positive and that’s why the acceleration due to gravity g is negative).
(ii) – (i) gives u(t2 t1) = (½) g(t22 t12)
Therefore u = (½) g(t22 t12)/(t2 t1) = g(t1+t2)/2

You can find a few more multiple choice practice questions (with solution) in this section here.