Multiple Choice Questions (including KEAM 2009) on Transistor Amplifiers

Questions from electronics are generally simple and interesting at the level expected from you. Here are a few questions:

(1) A transistor amplifier circuit is operated with an emitter current of 2 mA. The collector current is 1.98 mA. The common emitter current gain (β_dc) of the transistor used in the circuit is

(a) 45

(b) 50

(c) 100

(d) 125

(e) 200

Current gain (β_dc) in the common emitter configuration is given by

β_dc = I_C /I_B = 2 mA/(2 – 1.98) mA = 100

(2) A common emitter low frequency amplifier has an effective input resistance of 1 kΩ. The collector load resistance is 5 kΩ. On applying a signal, the base current changes by 10 μA and the collector current changes by 1 mA. What is the power gain of this amplifier?

(a) 100

(b) 500

(c) 1000

(d) 10000

(e) 50000

The power gain or power amplification (A_P) is the product of the current gain β_ac and the voltage gain A_v:

A_P = β_ac× A_v

Here β_ac = ∆I_C /∆I_B = (1×10^–3)/(10×10^–6) = 100 and

A_v = – β_ac×R_L/R_i where R_L is the load resistance and R_i is the input resistance. The negative sign just indicates that in the common emitter configuration the output signal is phase shifted by 180º.

[We use the load resistance R_L instead of the output resistance R_o (of the amplifier stage) in the above expression since the load resistance is small compared to the transistor output resistance. (The output resistance of the amplifier stage is the parallel combined value of R_L and the transistor output resistance)].

Ignoring the negative sign, A_v = 100×5/1 = 500

Therefore, power gain A_P = β_ac× A_v = 100×500 = 50000.

(3) For a common emitter amplifier, the audio signal voltage across the collector resistance 2 kΩ is 2V. If the current amplification factor of the transistor is 200 and the base resistance is 1.5 kΩ, the input signal voltage and base current are

(a) 0.1 V and 1 μA

(b) 0.15 V and 10 μA

(c) 0.015 V and 1 A

(d) 0.0015 V and 1 mA

(e) 0.0075 V and 5 μA

This MCQ appeared in KEAM (Engineering) 2009 question paper.

The collector signal current i_c is given by

i_c = v_c/R_c = 2 V/ 2 kΩ = 1 mA.

Since the current amplification (β_ac) of the transistor is 200, the base signal current i_b is given by

i_b = i_c /β_ac = 1 mA /200 = 0.005 mA = 5 μA

The input signal is the voltage drop (i_bR_b) produced across the base resistance by the flow of the base current.

Therefore, the input signal voltage = i_bR_b = 5 μA×1.5 kΩ = 7.5×10^–3 V = 0.0075 V.

(4) In an NPN transistor 108 electrons enter the emitter in 10^–8 s. If 1% electrons are lost in the base, the fraction of current that enters the collector and current amplification are respectively

(a) 0.8 and 49

(b) 0.9 and 90

(c) 0.7 and 50

(d) 0.99 and 99

(e) 0.88 and 88

This MCQ appeared in KEAM (Medical) 2009 question paper.

The number of electrons (108) entering the base and the time (10^–8 s) given in this question are not required to solve the problem. They can serve as distraction while attempting this simple question!

Since 1% electrons are lost in the base, 99% reach the collector so that the fraction of current that enters the collector is 99/100 = 0.99.

The current amplification is the ratio of the collector current to the base current and is equal to 99/1 = 99.

To access all questions from electronics on this blog you may click on the label ‘electronics’ below this post.

Kinematics- KEAM (Engineering) 2009 Questions on One dimensional Motion

Today we will discuss some multiple choice questions on one dimensional motion which appeared in Kerala Engineering Entrance 2009 question paper:

(1) A ball is thrown up vertically with speed u and at the same instant another ball B is released from a height h. At time t, the speed of A relative to B is

(a) u

(b) 2u

(c) u – gt

(d) √(u² – gt)

(e) gt

If you know the concept of relative velocity correctly, this question will be quite simple for you.

The relative velocity of A with respect to B is given by

V_AB = V_A – V_B where V_A and V_B are respectively the velocities of A and B (with respect to the common frame of reference chosen).

The velocity of A and B at time t are respectively u – gt and – gt, taking upward quantities positive and downward quantities negative.

The speed of A relative to B is u – gt – (– gt) = u [Option (a)].

(2) A body is falling freely under gravity. The distances covered by the body in the first, second and third minutes of its motion are in the ratio

(a) 1 : 4 : 9

(b) 1 : 2 : 3

(c) 1 : 3 : 5

(d) 1 : 5 : 6

(e) 1 : 5 : 13

Since you require the ratio of distances, it is enough to consider the times in seconds (instead of minutes). Strictly, it must be mentioned that the body starts from rest. In the case of such a freely falling body, the distance covered is directly proportional to t² since s = ut + ½ gt² with usual notations where u = 0

The distances covered by the body in one second, two seconds and three seconds are in the ratio 1 : 4 : 9.

Therefore, the distances covered by the body in the first, second and third seconds are in the ratio 1 : (4 – 1) : (9 – 4) which is 1 : 3 : 5

(3) A bullet fired into a fixed wooden block loses half of its velocity after penetrating 40 cm. It comes to rest after penetrating a further distance of

(a) ²²/₃ cm

(b) ⁴⁰/₃ cm

(c) ²⁰/₃ cm

(d) ²²/₅ cm

(e) ²⁶/₅ cm

We have v² = u² + 2as where u is the initial velocity, v is the final velocity after suffering a displacement s and a is the acceleration.

Therefore we have

u²/4 = u² + 2a × 0.4

Or, – 3u²/4 = 2a × 0.4 …………..(i)

If the bullet comes to rest after penetrating a further distance of s₁ we have

0 = u²/4 + 2a × s₁

Or, – u²/4 = 2a × s₁ …………….(ii)

Dividing Eq(i) by Eq(ii) we get

3 = 0.4/s₁ from which s₁ = 0.4/3 m = ⁴⁰/₃ cm.

[The above question can be worked out using the work energy principle as well].

You will find similar useful multiple choice questions on kinematics (with solution) here.

IIT-JEE 2009 Linked Comprehension Type Multiple Choice Questions involving de Broglie Waves

The following three questions [(i), (ii) and (iii)] involving de Broglie’s matter waves were included under Linked Comprehension Type Multiple Choice Questions (single answer type) in the IIT-JEE 2009 question paper (Paper I):

Paragraph for Questions (i), (ii) and (iii)

When a particle is restricted to move along x–axis between x = 0 and x = a, where a is of nanometer dimension, its energy can take only certain specific values. The allowed energies of the particle moving in such a restricted region, correspond to the formation of standing waves with nodes at its ends x = 0 and x = a. The wavelength of this standing wave is related to the linear momentum p of the particle according to the de–Broglie relation. The energy of the particle of mass m is related to its linear momentum as

E = p²/2m

Thus, the energy of the particle can be denoted by a quantum number n taking values 1, 2, 3, … (n = 1, called the ground state) corresponding to the number of loops in the standing wave.

Use the model described above to answer the following three questions for a particle moving in the line x = 0 to x = a. Take h = 6.6 × 10^–34 J s and e = 1.6 × 10^–19 C.

Question i

The allowed energy for the particle for a particular value of n is proportional to

(A) a^–2

(B) a^–3/2

C) a^–1

(D) a^2.

If there are n loops in the standing wave, we have

a = n λ/2 from which λ = 2a/n

[Note that the distance between consecutive nodes is λ/2]

From de Broglie relation, momentum p = h/λ = nh/2a on substituting for λ.

Now, energy E = p²/2m = n²h²/8a²m

Therefore, E α a^–2 [Option (A)].

Question ii

If the mass of the particle is m = 1.0×10^–30 kg and a = 6.6 nm, the energy of the particle in its ground state is closest to

(A) 0.8 meV

(B) 8 meV

(C) 80 meV

(D) 800 meV

In the ground state, n = 1 so that energy E = h²/8a²m

Therefore, E = (6.6 × 10^–34)²/[8×(6.6×10^–9)²×10^–30]

=10^–20/8 joule = 10^–20/(8×1.6×10^–19) electron volt

= 10^–1/(8×1.6) eV = 0.0078 eV, nearly = 7.8 meV (milli electron volt)

The correct option (nearest to the answer) is (B).

[Note that the symbol for million electron volt is MeV]

Question iii

The speed of the particle, that can take discrete values, is proportional to

(A) n^–3/2

(B) n^–1

(C) n^1/2

(D) n

The energy of the particle is given by E = p²/2m = n²h²/8a²m

Therefore, ½ mv² = n²h²/8a²m

From this the speed, v α n [Option (D)]

You will find IIT-JEE 2009 Linked Comprehension Type Multiple Choice Questions on nuclear physics (with solution) here

IIT-JEE 2009 and Kerala Engineering Entrance 2009 Questions involving Friction

Today we will discuss two questions (MCQ) involving friction. The first one appeared in IIT-JEE 2009 question paper while the second one appeared in KEAM (Engineering) 2009 question paper. Here are the questions with their solution:

(1) A block of base 10 cm × 10 cm and height 15 cm is kept on an inclined plane. The coefficient of friction between them is √3 . The inclination θ of this inclined plane from the horizontal plane is gradually increased form 0º. Then

(a) at θ = 30º, the block will start sliding down the plane

(b) the block will remain at rest on the plane up to certain θ and then it will topple

(c) at θ = 60º, the block will start sliding down the plane and continue to do so at higher angles

(d) at θ = 60º, the block will start sliding down the plane and further increasing θ, it will topple at certain θ.

The block will not slide down if the coefficient of friction μ is greater than tan θ. In other words, sliding begins only when θ becomes the angle of friction λ.

Since μ = tan λ we have √3 = tan λ from which λ = 60º.

So the block can start sliding down when θ = 60º.

But the block can topple when the line of action of the weight Mg of the block passes through the edge A of the block (fig.).

This can happen when tan θ = (AB/2) /(BC/2) = AB/BC = 10/15. Therefore, toppling occurs at an angle less than 45º (and greater than 30º).

The correct option therefore is (b).

(2) A mass of 1 kg is just able to slide down the slope of an inclined rough surface when the angle of inclination is 60º. The minimum force necessary to pull the mass up the inclined plane is

(a) 14.14 N

(b)17.32 N

(c) 10 N

(d) 16.66 N

(e) 0.866 N

The angle of friction is given as 60º and hence the coefficient of friction μ is given by

μ = tan 60º = √3

Normally you may assume that the force is applied parallel to the inclined plane. (If the force can be applied without this restriction, the case becomes the most general one and the minimum force necessary to pull the mass up the inclined plane will be less than the value you will get if the force is applied parallel to the plane).

The weight Mg of the mass M acts vertically downwards. The components of this weight along the plane and normal to the plane are Mg sinθ and Mg cosθ respectively. The normal force exerted by the plane on the mass is Mg cos θ and the frictional force is μMg cos θ.

When a force is applied to pull the mass up the plane, the frictional force will oppose the motion and will therefore act down the plane. The total force opposing the tendency of motion of the mass will be Mg sinθ + μMg cos θ.

This is the the minimum force F necessary to pull the mass up the inclined plane.

Substituting proper values, F = (1×10×sin 60º)+ (√3 ×1×10×cos 60º) = 5√3 +5√3 = 17.32 newton [Option (b)].

Note that the mass can be pulled up the plane using a smaller force if the force is applied at an angle α with respect to the plane. The force required will be minimum when α = λ where λ is the angle of friction. The minimum force F_min is given by

F_min = Mg sin(θ + λ)

You can find all posts related to friction on this site by clicking on the label ‘friction’ below this post.

Additional questions related to friction can be found here.

Kerala Engineering Entrance 2009 Questions (MCQ) on Alternating Current Circuits

Imagination is more important than knowledge
– Albert Einstein

Alternating current circuits were discussed on this site in the post dated 26th October 2006. You will find the post here.

Today we will discuss the questions from alternating currents which appeared in Kerala Engineering Entrance 2009 question paper. Here are the questions:

(1) In an LCR series ac circuit the voltage across L, C and R is 10 V each. If the inductor is short circuited, the voltage across the capacitor would become

(a) 10 V

(b) 20/√2 V

(c) 20√2 V

(d) 10/√2 V

(e) 20 V

Since the voltages across L and C are of the same value, the circuit is at resonance and Lω = 1/Cω where ω is the angular frequency. Under resonance the voltage across the resistor is equal to the supply voltage. So the supply voltage is 10 V.

When the inductor is short circuited, the LCR circuit reduces to an RC circuit and the supply voltage is the vector sum of the voltages across R and C so that we have

10 = √(V_C² + V_R²)

The voltage drops V_C and V_R across the capacitor and resistor respectively are equal in value since Lω = 1/Cω = R as judged from the equal voltage drops across L, C and R in the series LCR circuit. Therefore, from the above equation V_C = V_R = 10/√2 V

(2) A transformer of efficiency 90% draws an input power of 4 kW. An electrical appliance connected across the secondary draws a current of 6 A. The impedance of the device is

(a) 60 Ω

(b) 50 Ω

(c) 80 Ω

(d) 100 Ω

(e) 120 Ω

Efficiency, η = P_o/P_i with usual notations. Therefore, the output power is given by

P_o = η P_i = 0.9 ×4000 W = 3600 W.

In the problem nothing is mentioned about the power factor cos φ. At your level the load is usually assumed to be effectively resistive so that the power factor may be assumed to be unity. Therefore we have

P_o = I_o²R where R is the load resistance (or, the load impedence in this case).

This gives R = P_o /I_o² = 3600 /6² = 100 Ω.

(3) The impedance of an R-C circuit is Z₁ for a frequency f and Z₂ for frequency 2f. Then Z₁/Z₂ is

(a) between 1 and 2

(b) 2

(c) between ½ and 1

(d) ½

(e) 4

The impedance of an R-C circuit is √(R² + 1/C²ω²).

Since the angular frequency ω = 2πf we have

Z₁/Z₂ = [√(R² + 1/C²ω²)] /[√(R² + 1/4C²ω²)]

For large value of ω the value of Z₁/Z₂ tends to 1.

For small value of ω the value of Z₁/Z₂ tends to 2.

Therefore Z₁/Z₂ lies between 1 and 2.

(4) When a circular coil of radius 1 m and 100 turns is rotated in a horizontal uniform magnetic field, the peak value of emf induced is 100 V. The coil is unwound and then rewound into a circular coil of radius 2 m. If it is rotated now, with the same speed, under similar conditions, the new peak value of emf developed is

(a) 50 V

(b) 25 V

(c) 100 V

(d) 150 V

(e) 200 V

When a coil having N turns and area A rotates in a uniform magnetic field B with angular velocity ω, the peak value of the emf (V_max) induced in the coil is given by

(V_max) = NABω

The area of the coil in the second case is 4A since the radius is doubled. But the number of turns will be N/2 in the second case since the length of the wire required for a turn is doubled.

If V₁ and V₂ are the peak values of the emf in the two cases, we have

V₁/V₂ = 100 /V₂ = NABω / [(N/2)×(4A)Bω]

Or, 100 /V₂ = ½ from which V₂ = 200 volt.