## 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.

## Saturday, October 27, 2012

### Multiple Choice Questions on Metre Bridge

“If I have seen a little further it is by standing on the shoulders of Giants.”
– Sir Isaac Newton

The metre bridge is basically a Wheatstone bridge, the condition of balance of which is conveniently used in measuring unknown resistances. Today we shall discuss a few questions (MCQ) involving metre bridge:
(1) Resistances R1 and R2 are connected respectively in the left gap and right gap of a metre bridge. If the balance point is located at 55 cm, the ratio R2/R1 is
(a) 4/5
(b) 5/4
(c) 9/11
(d) 11/9
(e) 6/7
The distance of the balance point from the left end of the wire in the metre bridge is 55 cm. Therefore the distance of the balance point from the right end of the wire is 45 cm. Thus we have
R1/R2 = 55/45
Or, R2/R1 = 45/55 = 9/11
(2) A 15 Ω resistance is connected in the left gap and an unknown resistance less than 15 Ω is connected in the right gap of a metre bridge. When the resistances are interchanged, the balance point is found to shift by 20 cm. The unknown resistance is
(a) 5 Ω
(b) 6 Ω
(c) 8 Ω
(d) 10 Ω
(e) 12 Ω

Initially the balance point will be at J1 as indicated in the figure. On interchanging the resistances the balance point will shift to J2 so that the length of the bridge wire between J1 and J2 is 20 cm. The balance points J1 and J2 must be equidistant from the mid point (50 cm mark) of the wire so that J1 is at 60 cm and J2 is at 40 cm.
Therefore we have
R/X =  60/40 = 6/4
Or, 15/X = 6/4
This gives X = 10 Ω
(3) Resistances 4 Ω and 6 Ω are connected across the left gap and right gap respectively of a metre bridge. When a 2 Ω resistance is connected in series with the 4 Ω resistance in the leftt gap, the shift in the balance point is
(a) 10 cm
(b) 15 cm
(c) 20 cm
(d) 25 cm
(e) 30 cm
Initially when the left gap and right gap contain 4 Ω and 6 Ω respectively, the condition of balance is
4/6 = L/(100 L) where L is the balancing length
This gives L = 40 cm.
When a 2 Ω resistance is connected in series with the 4 Ω resistance in the leftt gap, the balancing length becomes 50 cm (since the gaps contain equal resistances).
Therefore the shift in the balance point is (50 cm – 40 cm) = 10 cm.

## Tuesday, October 09, 2012

### Questions Involving Charging of Capacitors

“Science distinguishes a Man of Honour from one of those Athletic Brutes whom undeservedly we call Heroes.”
– John Dryden

Today we will discuss a few multiple choice questions involving charging of capacitors. These questions will serve as practice questions for various entrance examinations including the National Eligibility Cum Entrance Test (NEET) for admission to MBBS and BDS Courses and the Joint Entrance Examination (JEE) for Admission to Undergraduate Engineering Programmes in IITs, NITs and other Technical Institutions. Here are the questions:
(1) An initially uncharged capacitor of capacitance 100 μF is charged at a steady rate of 20 μC/s. What is the time (in seconds) required for the potential difference across the capacitor to build up to 12 V?
(a) 240 s
(b) 120 s
(c) 60 s
(d) 30 s
The final charge Q on the capacitor is given by
Q = CV = 100 μF×12 V = 1200 μC
Since the charging rate is 20 μC/s, the time t required for the potential difference across the capacitor to build up to 12 V is given by
t = 1200 μC/20 μC/s = 60 s

(2) A battery of emf V volt is connected to a combination of four identical capacitors (each of capacitance C farad) arranged as shown in the adjoining figure. What is the amount of energy  supplied by the battery to the capacitor combination?
(a) CV2/4
(b) CV2/2
(c) CV2/3
(d) 2CV2/3
The effective capacitance connected across the battery is C + (C/3) = 4C/3
The energy E of the capacitor combination on getting charged to V volt is given by
E = ½ × (4C/3) V2 = 2CV2/3
(3) A parallel plate capacitor with air as dielectric has capacitance C. It is charged fully using a battery of emf V. The battery is then disconnected and the separation between the plates of the capacitor is doubled. What is the final energy stored in the capacitor?
(a) ½ CV2
(b) CV2
(c) 3CV2/2
(d) 2 CV2
(e) CV2/4
The charge Q on the capacitor is unchanged since the battery is disconnected. The charge Q is given by
Q = CV ------------- (i)
When the separation between the plates is doubled, the capacitance is halved (C/2).
[Note that the capacitance of a parallel plate air capacitor is given by C = ε0A/d where A is the plate area and d is the separation between the plates.
The energy E stored in the capacitor is given by
E = Q2/2C1 where C1 = C/2
Substituting for Q from Eq. (i), we have
E = C2V2/(2C/2) = CV2, as given in option (b).
[Note that the increased energy of the capacitor is the result of the external work done for increasing the separation between the plates, overcoming the electrostatic attractive force between the plates]
(4) In the above question, suppose the battery remains connected to the capacitor while doubling the separation. What will be the final energy stored in the capacitor>
(a) ½ CV2
(b) CV2
(c) 3CV2/2
(d) 2 CV2
(e) CV2/4
In this case charges can flow through the wires connecting the battery to the capacitor. The capacitor will hold a smaller amount of charge (let us say, Q1) since the capacitance is decreased from C to C/2. We have\
Q1 = (C/2) V
The final energy (E1) stored in the capacitor is given by
E1 = Q12/2C1 = [(C/2)V]2/(2C/2) = CV2/4

## Wednesday, September 26, 2012

### Joint Entrance Examination (JEE) 2013 for Admission to Undergraduate Engineering Programmes in IITs, NITs and other Technical Institutions

Central Board of Secondary Education has notified that admission to Undergraduate Engineering Programmes in IITs, NITs and other Centrally Funded Technical Institutions will be based on a Joint Entrance Examination (JEE) with effect from 2013.
JEE will be conducted in two parts – JEE (Main) and JEE (Advanced).
JEE (Main) will replace AIEEE and JEE (Advanced) will replace IIT-JEE, with the important criterion that only the top 150,000 candidates performing in JEE (Main) will qualify for appearing in JEE (Advanced). Admission to IITs, IT-BHU and ISM Dhanbad will be based only on category-wise All India Rank in JEE (Advanced), subject to the additional condition that such candidates are in the top 20 percentile of successful candiates in Class XII/equivalent examination conducted by their Boards.
For admission to Undergraduate Engineering Programmes in NITs and other Centrally Funded Technical Institutions the rank list will be prepared based on 40% weightage to school board marks in Class XII and 60% weightage to JEE (Main) marks.
JEE (Main) 2013 Paper I (for admission to B.E./B.Tech courses) which replaces Paper I of earlier AIEEE, will be of 3 hours, consisting of objective type questions from Physics, Chemistry and Mathematics.
Paper II for admission to B.Arch/B.Planning courses also will be of 3 hours and will consist of Mathematics, Aptitude Test and Drawing Test as per past practice of AIEEE.
The JEE (Main) (Paper I of earlier AIEEE) for B.E./B.Tech will be held in two modes, viz., offline and online (CBT). The offline examination for JEE (Main) will be conducted on 7th April, 2013 and the online examinations will be conducted thereafter in April, 2013.
For details you may visit