ESE2012E&T(1)

Engineering Services exam is considered as an equivalent to civil services exam in the engineering field.It offers most prestigious positions in Indian government service which can go up to the rank of Secretary to Govt. of India.This exam is easier compared to the GATE exam, but challenging because it requires wide and in depth theoretical knowledge in engineering subjects.Some of the important question asked in Engineering Services exam 2012 exam is solved in detail here.Feel free to report any mistakes or suggestion.

 

1) The maximum power that a 12 V DC source with an internal resistance of 2Ω can supply to a resistive load is

 (a) 72 W  (b) 48 W  (c) 24 W  (d) 18 W

 

Solution

Given 12V supply in series with 2ohm internal resistance

 

For maximum power to be transferred from a source to a load resistance, the value of value of load resistance should be equal to the internal resistance of the source.

 

 

 

current through the circuit = 12/(2+2) = 3 A

power transferred = I²R = 9 x 2 = 18 W

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2) The total resistance faced by the voltage source having zero internal resistance in the circuit is

 

(a) 10 Ω   (b) 5 Ω   (c) 2.5 Ω   (d) 1.5 Ω

 

Solution

We have three meshes, and let current flowing through them be I_{1 ,} I_{2 ,} I₃ 

Total resistance faced by source = voltage/current drawn from source

Mesh1:                      5 - 5(I₁ -  I₂) = 0

                                I₁ -  I₂  = 1   ---------(1)

Mesh2:                     5(I₂ –  I₁) + 10(I₂ –  I₃) = 0

                                 

                              -5 I₁ +15 I₂ = 10 I₃

                                  

                                  -5 I₁ +15 I₂ = 10 x 0.5        (because I₃ = 0.5A)

                   

                               -I₁ +3 I₂ =1  ---------(2)

   (1) + (2) =>          2 I₂ = 2

                             

                                 I₂ = 1 A

                              

                                 I₁ = 2A

Total resistance faced by source = voltage/current drawn from source

                                                                = 5/2 = 2.5Ω

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 3) In the circuit, Thevenin’s voltage and resistance across the terminals XY will be

 

(a) 20 V and 100 Ω 

(b) 40 V and 93.33 Ω

(c) 60 V and 93.33 Ω

(d) 100 V and 100 Ω

 

 Solution

Thevenin’s voltage is the open circuit voltage at XY. This is same as voltage appearing on node (1) since no current flows through 80Ω.

Applying KCL at node (1)

(V-60)/20 – 2 = 0

(V-60)/20 = 2

V/20 = 5

=> V_th = 100V

Thevenin’s resistance is found out by closed circuiting all voltage sources and open circuiting all current sources

R_th = 80+20 = 100Ω

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4) In the circuit, the value of the resistance Rs required for maximum power transfer from the 10 V source to the 10Ω  load is given by

 

(a) 5 Ω   (b) 10 Ω   (c) 0 Ω   (d) 30 Ω 

Solution

For maximum power transfer from a circuit to a load resistance, the load resisitance should be equal to the Thevenin’s resistance of the circuit which drives the load.

Here 10Ω is the load and the rest of  circuit is the driving circuit

For finding Thevenin’s equivalent resistance of driving circuit:

1    (1) Remove 10Ω resistor(open circuit)

       (2)Short circuit the voltage source

       (3)Find the equivalent resistance of the resultant circuit

R_th = 30 || 30 || (30 + R_s)

      

     = 15 || (30 + R_s)

R_th = 10Ω   for maximum power transfer

15 x (30 + R_s)/(15 + 30 + R_s) = 10

If we solve this R_s = 0

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5) The output impedance of a BJT under common-collector configuration is
 

    (a) low   (b) high   (c) medium   (d) very high 

 

Solution

Common collector or emitter follower have high input impedance(since it is multiplied by a factor of 1+β – resistance reflection rule) and low output impedance(since it is divided by a factor of 1+β – reverse resistance reflection rule). Because of this property they are used as buffer amplifiers (gain is also unity).

 

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6) Match List-I with List-II and select the correct answer using the code given below the
Lists:
          List – I                                        List – II

A. di / dt rating limits                    1. Snubber circuit

B. dv / dt rating                            2. Heat sink

C. i2t limit                                    3. Series reactor

D. Junction temperature limit         4. Fuse

Codes:
     A B C D                  A B C D
(a) 2 4 1 3              (b) 3 4 1 2
(c) 2 1 4 3              (d) 3 1 4 2

 

 

Solution

Snubber circuit - Snubbers are frequently used in electrical systems with an inductive load where the sudden interruption of current flow leads to a sharp rise in voltage across the current switching device, in accordance with Faraday's law. This transient can be a source of electromagnetic interference (EMI) in other circuits. Additionally, if the voltage generated across the device is beyond what the device is intended to tolerate, it may damage or destroy it. The snubber provides a short-term alternative current path around the current switching device so that the inductive element may be discharged more safely and quietly. Its rating is in terms of voltage change.

Series reactor - A reactor used in alternating-current power systems for protection against excessively large currents under short-circuit or transient conditions; it consists of coils of heavy insulated cable either cast in concrete columns or supported in rigid frames and mounted on insulators Such reactors are also used to limit the starting currents of synchronous electric motors and to compensate reactive power in order to improve the transmission capacity of power lines. Its limit is in terms of maximum rate of change of current.

answer is (d) 

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7)  An SCR can be turned off

(a) by passing a negative pulse to its gate
(b) by removing the gate supply
(c) by reverse biasing it
(d) by forcing the current through gate to become zero

 Solution

Three methods can be used to turn OFF a SCR

a) Natural Commutation

When the anode current is reduced be­low the level of the holding current, the SCR turns off. However, it must be noted that rated anode current is usually larger than 1,000 times the holding value

(b) Reverse-bias Turn-off

A reverse anode to cathode voltage (the cathode is positive with respect to the anode) will tend to interrupt the anode current. The voltage reverses every half cycle in an ac circuit, so that an SCR in the line would be reverse biased every negative cycle and would turn off. This is called phase commutation or ac line commutation. To create a reverse biased voltage across the SCR, which is in the line of a dc circuit, capacitors can be used. The method of discharging a capacitor in parallel with an SCR to turn-off the SCR is called forced commutation.

 (c) Gate Turn Off

In some specially designed SCRs the characteristics are such that a negative gate current increases the holding current so that it exceeds the load current and the device turns-off.

(a) and (c) are correct

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8)  A signal f(t) is described as

         f(t) = [1-|t|] when | t | <= 1

               = 0 when | t | > 1

This represents the unit

(a) sinc function           (b) area triangular function
(c) signum function      (d) parabolic function 

  Solution

The function f(t) can be plotted like the following

when t=0 , f(t) = 1

when t=0.1, f(t) = 0.9

when t= -0.1, f(t) = 0.9

when t= 0.2, f(t) = 0.8 

when t= -0.2, f(t) = 0.8 

.

.

when t=1,f(t) = 0 

when t= -1,f(t) = 0

 

 The graph represents an area triangular function

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9)  The I-V characteristics of a tunnel diode exhibit

(a) current-controlled negative resistance
(b) voltage-controlled negative resistance
(c) temperature-controlled positive resistance
(d) current-controlled positive resistance

  Solution

Tunnel diode exhibits voltage controlled negative resistance.Due to tunneling phenomenon tunnel diode conducts very early when compared to normal diodes.

As forward voltage is increased electrons from filled states in valance band of n-region will tunnel through junction to reach empty states in p-region conduction band.As forward bias increases, the misalignment between the above mentioned states increases, as a result current reduces, causing a negative resistance region.

For more detailed explanation on working of a tunnel diode click here

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10)  A gate to drain-connected enhancement mode MOSFET is an example of
 

(a) an active load                (b) a switching device
(c) a three-terminal device   (d) a three-terminal device 

 Solution

when gate is connected to drain, V_GS = V_DS

MOSFET will act like a load impedance with V_DS controlling  I_D

 

Answer is (a) 

For more detailed explanation on active load click here

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