Kerala Engineering Entrance Questions (MCQ) on Communication Systems

Questions on communication systems are generally simple and interesting and so you can easily score marks by attempting them. The following three multiple choice questions were included in Kerala Engineering Entrance 2005 question paper and will be able to answer them in less than three minutes:

(1) If a radio receiver is tuned to 855 kHz radio wave, the frequency of local oscillator in kHz is

(a) 1510

(b) 455

(c) 1310

(d) 1500

(e) 855

Since the tuned frequency is 855 kHz, the receiver is an AM (amplitude modulation) receiver. Modern AM receivers are super heterodyne receivers employing a higher frequency local oscillator. On mixing the local oscillator output with the incoming amplitude modulated carrier, an amplitude modulated wave at intermediate frequency (IF) of 455 kHz (by convention) is produced. Since the intermediate frequency is 455 kHz, it follows that the frequency of local oscillator is 855 kHz + 455 kHz = 1310 kHz.

(2) If n₁ and n₂ are the refractive indices of the core and the cladding respectively of an optical fibre,

(a) n₁ = n₂

(b) n₁ < n₂

(c) n₂ < n₁

(d) n₂ = 2n₁

(e) n₂ = √(2n₁)

Since the optical fibre confines the light signal within the fibre by total internal reflection, the refractive index of the cladding should be less than that of the core. Therefore, n₂ < n₁ [Option (c)]

(3) A TV tower has a height of 100 m. What is the maximum distance up to which TV transmission can be received? (Radius of the earth, R = 8×10⁶ m)

(a) 34.77 km

(b) 32.7 km

(c) 40 km

(d) 40.7 km

(e) 42.75 km

We have maximum distance, d ≈ √(2Rh) where h is the height of the antenna.

Substituting given values, d ≈ √(2×8×10⁶ ×100) = 40×10³ m = 40 km.

[The mean radius of the earth is nearly 6400 km. The value is given as 8000 km in the problem to make your calculation simple.

You should remember that the height of the transmitting antenna (or receiving antenna) is the height with respect to the ground level. If an antenna is mounted on a mast of height h₁ and the mast is erected on a hill or building of height h₂, the height of the antenna will be h = h₁ + h₂]

Multiple Choice Questions (MCQ) on Alternating Currents

Most of you might have noted that electric power stations generate alternating current rather than direct current even though the majority of electrical and electronic appliances require direct current. At the power generation stage, alternating current is preferred since the current can be controlled, without power loss, using reactive elements (inductors and capacitors). Transmission of electric power over long distances without appreciable losses is possible by using transformers. The idea, as most of you should know, is to transmit electrical energy at low current and high voltage so that the Joule heating (which is proportional to the square of the current) in the transmission lines is minimized.

You will find some earlier posts on alternating currents on this site. You can access all the posts by trying a search for ‘alternating current’ using the search box provided on this page. Or, you may click on the label ‘alternating current’ below this post.

Let us discuss a few more questions on alternating currents:

(1) A short circuited coil is placed in a time varying magnetic field. Electric power is dissipated due to the current induced in the coil. If the number of turns were quadrupled and the wire radius halved, the electric power dissipated would be

(a) halved

(b) same

(c) doubled

(d) quadrupled

The above question appeared in IIT 2002 screening test paper.

Electric power dissipation is given by P = V²/R. where V is the voltage induced in the coil and R is the resistance of the coil. When the number of turns is quadrupled, the induced voltage V is quadrupled so that V² becomes 16 times. But the resistance of the coil also becomes 16 times since resistance R = ρL /A where ρ is the resistivity (specific resistance), L is the length and A is the area of cross section of the wire. The length becomes 4 times when the number of turns is quadrupled and the area of cross section becomes one-fourth when the radius is halved. The resistance therefore becomes 16 times.
The power dissipated is therefore unchanged [Option (b)].

(2) An electric heater consumes 1000 watts power when connected across a 100 volt D.C. supply. If this heater is to be used with 200V, 50 Hz A.C.supply, the value of the inductance to be connected in series with it is

(a) 5.5 H

(b) 0.55 H

(c) 0.055 H

(d)1.1 H

(e)11 H

The current drawn by the heater is 1000 W/100V = 10 A . When the heater is used with A.C. supply, it will draw 10 A itself. (Note that the current and the voltage values are R.M.S. values when you deal with electric power). If ‘L’ is the inductance required, the expression for the current I is

I = V/√(R² + L²ω²)] where V, R and ω are respectively the alternating voltage, the resistance of the heater and the angular frequency of the A.C.

Substituting, 10 = 200/√[10² + L² (100π)²], since ω = 2πf where f’ is the frequency of the A.C.

Squaring and rearranging, L² = 300/100π)² from which L = √3/10π = 0.055 H.
(3) An alternating emf V = 6 cos1000t is applied across a series LR circuit of 3 mH inductance and 4 Ω resistance. The amplitude of the current is

(a) 0.6 A

(b) 1.2 A

(c) 1.4 A

(d) 1.8 A

Amplitude (maximum value or peak value) of current (I_max) is given by

I_max = V_max/√(R²+L² ω²) = 6/√[4²+(3×10^-3)²×1000²] = 1.2 A

[Note that the values of V_max and ω are obtained from the expression for the emf V which is in the form, V = V_max cos ωt].

Here is a very simple question which you should answer carefully:

(4) The voltage V applied across an A.C. circuit and the current I flowing in it are given by

V = 12 cos ωt volt and I = 20 sin ωt milliampere respectively.

The power dissipated in the circuit is

(a) 120 watt

(b) 120 milliwatt

(c) 240 watt

(d) 249 milliwatt

(e) zero

In alternating current circuits the power is given by P = V_rms I_rms cosΦ where Φ is the phase difference between the applied voltage and the resulting current. Since the voltage is a cosine function and the current is a sine function, the phase difference Φ is π/2. [Note that the voltage can be written as V = 12 sin(ωt + π/2) volt]. Therefore cosΦ (which is called the power factor) is zero. The correct option is (e).

All India Engineering/Architecture Entrance Examination 2011 (AIEEE 2011) Postponed

The Central Board of Secondary Education has rescheduled the All India Engineering/Architecture Entrance Examination (online/pen & paper) 2011 (AIEEE 2011) from 24-4-2011 to 1-5-2011 in consideration of the Easter festival. The sale of Information Bulletins from all selling centres will start from 22.12.2010 instead of 15.12.2010. Rest of the things/schedule will remain unchanged.

Visit the web sites www.cbse.nic.in and www.aieee.nic.in for details.

Electronics - Multiple Choice Questions on Transistor Amplifiers

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“How strange is the lot of us mortals! Each of us is here for a brief sojourn; for what purpose we know not, though sometimes sense it. But we know from daily life that we exist for other people first of all for whose smiles and well-being our own happiness depends.”

– Albert Einstein

Questions on transistor amplifiers at the class 12 (plus two or higher secondary) level are generally simple and interesting even though some among you may have unclear ideas. Today we will discuss a few questions on common emitter transistor amplifiers:

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(1) The adjoining figure shows a common emitter transistor amplifier which uses a silicon transistor. If the quiescent emitter current is 1 mA what is the base biasing voltage?

(a) 4.7 V

(b) 3.7 V

(c) 2.7 V

(d) 1.7 V

(e) 0 V

Because of the emitter current the voltage drop across the 1 KΩ resistor connected to the emitter is 1 V.

[1 mA×1 KΩ = (1/1000) A×1000 Ω = 1 V].

The voltage drop across the base-emitter junction of the silicon transistor is 0.7 V. Therefore, the base voltage under no signal (quiescent) condition is 1 V + 0.7 V = 1.7 V.

(2) In the amplifier circuit shown in Question No.1 what is the function of the capacitor C₁ connected across the 1 KΩ emitter resistor?

(a) To produce positive feed back.

(b) To produce negative feed back.

(c) To pass the excess signal to the ground.

(d) To act as filter capacitor for the transistor supply voltage.

(e) To bypass the signal current so that it will not flow through the emitter resistor.

The capacitor C₁ provides an easy path (bypass) for the signal component of the emitter current. If C₁ is absent the signal component of the emitter current will produce signal voltage drop across the emitter resistor, thereby reducing the signal output at the collector.

The correct option is (e).

(3) If the common emitter current gain β_dc of the transistor used in the amplifier circuit shown in Question No.1 is 200, the quiescent base current of the transistor is very nearly equal to

(a) 1 mA

(b) 1 μA

(c) 2 μA

(d) 4 μA

(e) 5 μA

In the common emitter mode, the current amplification factor (current gain) under no signal condition (β_dc) is given by

β_dc = I_C/I_B where I_C is the collector current and I_B is the base current (both under no signal conditions).

Since the collector current is almost equal to the emitter current I_E (because of large value of β_dc), we have

β_dc ≈ I_E/I_B

Therefore I_B ≈ I_E/β_dc = 1 mA/200 = 0.005 mA = 5 μA.

(4) If the common emitter current gain β_dc of the transistor used in the amplifier circuit shown in Question No.1 is 200, what is the voltage drop across the base biasing resistor R under quiescent conditions?

(a) 12 V

(b) 11 V

(c) 10.3 V

(d) 5.4 V

(e) 4.7 V

The quiescent base current is 5 μA as shown in answering question no.3 above. The base biasing voltage is 1.7 V as shown in answering question no.1. The power supply voltage is 12 V. Therefore, the voltage drop across the base biasing resistor R under quiescent conditions is 12 V – 1.7 V = 10.3 V.

(5) The base biasing resistor in the circuit shown in Question No.1 is

(a) 1 KΩ

(b) 4.7 KΩ

(c) 1.03 MΩ

(d) 1.87 MΩ

(e) 2.06 MΩ

The quiescent base current of the transistor is very nearly equal to 5 μA as shown in answering Question No.3. The voltage drop across the base biasing resistor R under quiescent conditions is 10.3 V as shown in answering Question No.4. Therefore, the base biasing resistor is given by

R = (10.3)V/(5 μA) = 2.06 MΩ

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You can access all questions (with solution) on electronics posted on this site by clicking on the label ‘electronics’ below this post or by trying a search for ‘electronics’ using the search box provided on this page

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