Monday, September 30, 2013

The variation of drain current with gate-to-source voltage of a MOSFET is shown in fig.The MOSFET is (GATE-EE-2003)

The variation of drain current with gate-to-source voltage of a MOSFET is shown in fig.The MOSFET is (GATE-EE-2003)

A. an n-channel depletion mode device
B. an n-channel enhancement mode device
C. a p-channel depletion mode device
D. a p-channel enhancement mode device



ANSWER:   C

It is the transfer characteristics of a depletion mode device and it  must be p-channel, since the Vgs is positive in p-channel depletion region.

Saturday, September 28, 2013

The steady state error due to step input for a type 1 system is ........ (GATE-EE-1995)

The steady state error due to step input for a type 1 system is ........  (GATE-EE-1995)


Answer :  0

All the resistances in the figure are 1 ohm each.The value of current 'I' is

All the resistances in the figure are 1 ohm each.The value of current 'I' is  ( GATE-EE-1992)

A. 1/15A
B. 2/15A
C. 4/15A
D. 8/15A




Answer:

Given R = 1, v = 1
After reduction , we get  Req = 1.875

I = 1 / 1.875 = 0.533 = 8/15 A

Friday, September 27, 2013

The uncontrolled electronic switch used in power electronic converter is

The uncontrolled electronic switch used in power electronic converter is (GATE-EE-1998)

A. thyristor
B. bipolar junction transistor
C. diode
D. MOSFET



Explanatory answer:

Thyristor - semi controlled device
BJT and Mosfet -fully controlled device

Diode - uncontrolled device, since its on/off depends on external circuit.we cannot control

Thursday, September 26, 2013

The concept of an electrically short, medium and long transmission line is primarily based on the (GATE-EE-2006)

The concept of an electrically short, medium and long transmission line is primarily based on the   (GATE-EE-2006)

A. nominal voltage of the line
B. physical length of line
C. wave length of line
D. power transmitted over the line

Ans:

Physical length of line

Wednesday, September 25, 2013

A low-pass filter with a cut-off frequency of 30HZ is cascaded with a high-pass filter with a cut-off frequency of 20HZ. The resultant system of filters will function as

A low-pass filter with a cut-off frequency of 30HZ is cascaded with a high-pass filter with a cut-off frequency of 20HZ. The resultant system of filters will function as   (GATE-EE-2011)

A. an all-pass filter
B. an all-stop filter
C. a band stop (band reject) filter
D. a band-pass filter 

Explanatory answer :

Low-pass filter with a cut-off frequency of 30HZ:

High-pass filter with a cut-off frequency of 20HZ:

The cascade combination filter is

A band-pass filter







An inductor is connected in series with SCR to protect it against

An inductor is connected in series with SCR to protect it against

A. damage due high dv/dt
B. damage due to high di/dt
C. damage due to large forward current
D. none of these

Explanatory answer :

An inductor is connected in series with SCR to protect it against damage due to high di/dt

At the instant of turn on, the current flowing through SCR is not distributed uniformly and the rate of rise of current (di/dt) is very fast.This produces localized hot spots and damages the device,hence SCR should be protected against this high di/dt. This can be achieved by connecting an inductor in series with the SCR, because inductance(L) opposes the high di/dt variations. 


Tuesday, September 24, 2013

The transfer function of a system is given as

The transfer function of a system is given as   
This system is                                      (GATE-EE-2008)

A. an overdamped system
B. an underdamped system
C. a critically damped system
D. an unstable system


Explanatory answer:

Systems are classified according to their damping ratio( Î¶ ) vales,

If Î¶ < 1, underdamped system

if  Î¶ = 1, critically damped system         

if  Î¶ > 1, overdamped system .

Comparing   s^2+ 2ζwn +wn^2   with   s^2+ 20s +100

wn^2=100, wn = 10


2ζwn = 20,   Î¶ = 20/20= 1

 Î¶=1, hence this system is a critically damped system. 

Monday, September 23, 2013

An Enhancement tye n-channel MOSFET is represented by the symbol

An Enhancement tye n-channel MOSFET is represented by the symbol   (GATE-EE-1999)





Explanatory answer:

 The dashed line between drain and source only used in E-mosfet symbol and that is the basic symbol difference between  D-mosfet and E-mosfet .

From the given options only option A and B has dashed line, but in n- channel arrow mark will be pointed in,so the correct answer will be option A.

Sunday, September 22, 2013

The colour code of 1K ohm resistor is

The colour code of 1K ohm resistor is   (GATE-EE-1999)

A. black,brown,red
B. red, brown,brown
C. brown,black,red
D. black,black,red

Explanatory answer: 

for
Brown :  1
Black   :  0
red       :  2

therefore ,    10*10^2  =  1000K

Ans:  brown,black,red



Saturday, September 21, 2013

The winding of a 4-pole alternator having 36 slots and a coil span of 1 to 8 is short-pitched by........ degrees.

The winding of a 4-pole alternator having 36 slots and a coil span of 1 to 8 is short-pitched by........ degrees.

A.140
B. 80
C. 20
D. 40

Explanatory answer:

Pole pitch:
                Centre to centre distance between two adjacent poles
                Pole picth= 180 deg electrical
                Pole pitch= slots/Pole= n

Coil span :
               The distance between two coil sides of acoil
               Expressed in electrical deg or by no. of slots
Given:
          P=4, S=36

Pole pitch,n= 36/4= 9 slots

i.e. 9 slots contributes one pole pitch(9 slots=180 deg)

therefore 1 slots = 180/9 = 20 deg

i.e 1 slots contributes 20 deg

given coil span (1 to 8 slots) i.e.8-1=7 slots

so the winding is short pitched by  = 2 slots ( pole pitch- actual coil span=9slots-7slots)

short pitch in deg = (deg for 1 slot * no of slots short pitched)

                          =  20 * 2
                          = 40 deg

In a 6 pole dc machine,90 mechanical degrees corresponds to ......... electrical degrees.

In a 6 pole dc machine,90 mechanical degrees corresponds to ......... electrical degrees.

A. 30
B. 180
C. 45
D. 270


Explanatory answer:

The relationship between mechanical and electrical degrees is

1 degree mechanical = (P/2) electrical degrees

i.e. for 1 deg of mechanical revolution of conductor inside a magnetic field produces (P/2) degree electrical output.

Given, P=6

90 deg mechanical = 90 (P/2) electrical

                             = 90*(6/2)
                             = 270 degree electrical

i.e. for 90 deg mechanical revolution ,it produces 270 deg electrical output.

Friday, September 20, 2013

How much RMS current does a 300W, 200V bulb take from the 200V, 50HZ power lines ?

How much RMS current does a 300W, 200V bulb take from the 200V, 50HZ power lines ?


A. 0.5A
B. 1.5A
C. 2A
D. 3A

Explanatory answer:

P=V^2/R

Therefore R= V^2/ P

R= 200^2 / 300

R= 133.33 ohm (resistance of lamp)

Rms current 
taken by lamp = rms value of supply voltage /resistance of  lamp

                                         =  200  /  133.33
                                      I   =  1.5A

Thursday, September 19, 2013

A 50 HZ alternator will run at the greatest possible speed if it is wound for ...... poles

A 50 HZ alternator will run at the greatest possible speed if it is wound for ...... poles

A. 8
B. 6
C. 4
D. 2

Explanatory answer:

Ns=120F/P

Alternators always runs at synchronous speed and which is inversely proportional to no.of poles.

so lesser the poles higher the speed and the minimum possible no. of poles is 2

for P=2,  Ns=3000 rpm
for P=4, Ns=1500 rpm
for P=6, Ns=1000 rpm
for P=8, Ns=750 rpm

The correct ans is P=2

Tuesday, September 17, 2013

To maximize any quantity we should differentiate that with respect to variable one and equate to zero,why?

To maximize any quantity we should differentiate that with respect to variable one and equate to zero,why?

To obtain the maximum point for a function, we should differentiate it with respect to variable one and equate it to zero because once a quantity is at its maximum/minimum value there wont be any further change so the change (differentiation) is zero.

For example, in dc motor to obtain the condition for maximum power developed by armature,the power equation is differentiated with respect to variable quantity Ia and equated to zero.

(i.e) power developed in armature, Pm = VI-Ia^2Ra


To get that condition for maximum power developed, we should do, dPm / dIa =0 ( Ia is varying quantity(because as Ia varies Pm varies).

dPm / dIa =0  (under this condition, there wont be any further change of Pm with respect to Ia(i.e the change is zero)

Monday, September 16, 2013

A sine wave voltage source is connected across a series combination of a resistor and an inductor.The frequency is set so that inductive reactance is equal to resistance.If the frequency is increased then

A sine wave voltage source is connected across a series combination of a resistor and an inductor.The frequency is set so that inductive reactance is equal to resistance.If the frequency is increased then

A.  Vr>Vl
B.  Vr<Vl
C.  Vl=Vr
D.  none of the above

Explanatory answer:

XL=2*pi*f*L

as f increases XL increases

so voltage drop across inductor, IXL increases but voltage drop across resistance remains same because resistance not depends on frequency.

The correct answer will be: Vr< Vl

Sunday, September 15, 2013

A three- phase, three-stack variable reluctance stepper motor has 20 poles on each rotor and stator stack. The step angle of this motor is

A three- phase, three-stack variable reluctance stepper motor has 20 poles on each rotor and stator stack. The step angle of this motor is   (GATE-EE-2007)

A. 3
B. 6
C. 9
D. 18


EXPLANATORY ANSWER:

Step angle= 360/ m*Nr

where
Nr=no.of rotor poles

m=no.of phases

therefore   step angle=360/3*20

step angle=6

Saturday, September 14, 2013

The slip of an induction motor normally does not depend on

The slip of an induction motor normally does not depend on  (GATE-EE-2012)



A. Rotor speed

B. Synchronous speed

C. Shaft torque

D. Core loss component


EXPLANATORY  ANSWER :


Slip= Ns-Nr / Ns

So slip depends on both rotor speed(Nr) and synchronous speed (Ns)

Torque varies as rotor speed (Nr) varies

Hence the correct answer will be  Core loss component

Friday, September 13, 2013

A 8 pole, DC generator has a simplex wave wound armature containing 32 coils of 6 turns each. It flux/pole is 0.06 wb.The machine is running at 250 rpm, calculate the induced armature voltage.

A 8 pole, DC generator has a simplex wave wound armature containing 32 coils of 6 turns each. It flux/pole is 0.06 wb.The machine is running at 250 rpm, calculate the induced armature voltage. (GATE-EE-2004)

Given:

P=8, DC gen,Wave(A=2),Coils C=32, turns per coil = 6, flux ø= 0.06 wb, speed N=250 rpm
EMF E=?

Solution:

E=øZNP/60A , Turns T= coils * turns per coil =32*6= 192

Z=2T= 2*192 = 384 conductors

E = 0.06*384*250*8 / 60*2

Ans : E=384 volts


Thursday, September 12, 2013

The vector j5E is same as vector

The vector  j5E is same as vector

A.   jE
B . j E
C.  JE
D.  JE


Explanatory answer :


We know j2=-1


and     j5=j2*j2*j


j5= (-1) *(-1)* (j)= j

therefore  jE can be written as  j E

A 4 pole dynamo with wave wound armature has 51 slots containing 20 conductors in each slots.The induced emf is 357 volt and the speed is 8500 rpm . calculate the flux per pole. (GATE-EE-1996)

 A 4 pole dynamo with wave wound armature has 51 slots containing 20 conductors in each slots.The induced emf is 357 volt and the speed is 8500 rpm . calculate the flux per pole.  (GATE-EE-1996)



Explanatory answer :

 Given:



 P=4 , Wave(A=2) , slots S= 51 , cond/slot = 20 , E= 357 volt
 N = 8500 rpm , Ø = ?

 Sol:


Total conductors = slots * cond/slot = 51 *20 = 1020 conductors

Ø = E * 60 * A / ZNP

Ø = 357 * 60 * 2 / 1020 * 8500 * 4

Ans: Ø = 1.23 mwb

Wednesday, September 11, 2013

Capacitive reactance is more when

Capacitive reactance is more when

A. capacitance is less and frequency of supply is less
B. capacitance is less and frequency of supply is more
C. capacitance is more and frequency of supply is less
D. capacitance is more and frequency of supply is more


Explanatory answer:

We know capacitive reactance reactance, Xc = 1 / 2ΠFC

Xc is inversely proportional to F and C.

Therefore Xc increases as frequency and capacitance decreases,hence the suitable answer is option A

Tuesday, September 10, 2013

The power consumed in a circuit element will be least when the phase difference between the current and voltage is

The power consumed in a circuit element will be least when the phase difference between the current and voltage is

A. 180
B. 90
C. 60
D. 0



Explanatory answer:

we know power in ac circuit,  P=VICOSø

when  Ã¸ = 90, COS(90)= 0

Therefore the power consumed by the circuit will becomes zero at  Ã¸ = 90  (least value).
  

ANNA UNIVERSITY UG/PG-2013 REVALUATION RESULTS

ANNA UNIVERSITY UG/PG-2013 REVALUATION RESULTS


anna university ug/pg revaluation results has been announced, check it here...




ANNA UNIVERSITY UG/PG REVALUATION RESULTS    (GRADE SYSTEMS)






Why silicon steel is preferred for magnetic core materials of electrical machines to reduce hysteresis losses ?



Explanatory answer :

by steinmetz formula, 
                              Hysteresis loss,    Wh=n(Bmax)1.6f v watts

Where

n= hysteresis co-efficient
f= frequency of reversal of magnetism
v= volumeof magnetic material

from the above above relation, we conclude that  hysteresis loss (Wh) is proportional to hysteresis co-efficient (n) hence by selecting a magnetic material with low hysteresis co-efficient value and high electrical resistivity , the hysteresis loss can be reduced.


Hysteresis co-efficient value for different magnetic materials :


 n= 502 J/m3   for good dynamic sheet steel

 n= 191 J/m3    for silicon steel
 n= 7040 J/ m3   for hard steel
 n= 750-3000 J/ m3   for  for cast steel
 n= 2700-4000 J/ m3   for  cast iron

           out of these listed hysteresis co-efficient value for different magnetic materials, silicon steel has low hysteresis co-efficient value hence silicon steel is mostly preferred for magnetic core material to reduce hysteresis losses.

The peak value of a sine wave is 200V.Its average value is

The peak value of a sine wave is 200V.Its average value is 

A. 127.4 V
B. 141.4 V
C. 282.8 V
D. 200 V

Ans: A


Explanatory answer :

For sinusoidal waveform,

peak factor= peak value/ rms value =1.414

therefore  rms value = 200/ 1.414 = 141.44

and form factor = rms value/ average value = 1.11

therefore, average value = 141.44/ 1.11 = 127.4 V

Monday, September 9, 2013

A sine wave has a frequency of 50 HZ. Its angular frequency is............radian/second.

A sine wave has a frequency of 50 HZ. Its angular frequency is............radian/second.

A. 100Ï€ 

B. 50Ï€  

c. 25Ï€  

D. 5Ï€   



Ans:   A

Explanatory answer :

The relationship between frequency and angular velocity is

                                    w=2Ï€F  rad/sec

where  w= angular velocity in rad/sec
            F= frequency in HZ

Therefore       w=2Ï€*50
               
                      w=100Ï€  rad/sec

Sunday, September 8, 2013

An ammeter has a current range of 0 - 5 A, and its internal resistance is 0.2Ω . In order to change the range to 0 - 25 A, we need to add a resistance of

An ammeter has a current range of 0 - 5 A, and its internal resistance is 0.2Ω . In
order to change the range to 0 - 25 A, we need to add a resistance of  
                 (GATE-EE-2010)

(A) 0.8Ω in series with the meter

(B) 1.0Ω in series with the meter
(C) 0.04Ω in parallel with the meter 
(D) 0.05Ω in parallel with the meter



Ans: D



Explanatory answer :


In ammeter designs, external resistors added to extend the usable range of the movement are connected in parallel with the movement rather than in series as is the case for voltmeters. This is because we want to divide the measured current, not the measured voltage, going to the movement, and because current divider circuits are always formed by parallel resistances.

The shunt resistance can be calculated as follows

Let  Rm = internal resistance of coil=0.2 Î©   
 Rsh  = low shunt resistance going to add                               


Im = full scale deflection current=5A
Ish = shunt current    

As the two resistances are in parallel, the voltage drop across them is same

therefore    IshRsh  = IRm 

 Rsh  =  IRm /Ish 

Rsh   = 5* 0.2 / (25-5)

Rsh  = 0.05  Î©    is connected in parallel with the meter


Thursday, September 5, 2013

A bulb in a stair case has two switches, one switch being at ground floor and other one at first floor. The bulb can be turned on and also can be turned off by any one of the switches irrespective of the state of other switch.The logic of switching of bulb resembles


A bulb in a stair case has two switches, one switch being at ground floor and other one at first floor. The bulb can be turned on and also can be turned off by any one of the switches irrespective of the state of other switch.The logic of switching of bulb resembles
(GATE-EC-2013)


(a) an AND gate
(b) an OR gate
(c) an XOR gate
(d) an NAND gate



ANS: C


EXPLANATORY ANSWER:

The purpose of staircase wiring is, one should be able to switch on and switch off lights from more than one point and these can be achieved by 2-way switches.The position of switches A, B and the condition of the lamp can be tabulated as below.



(i.e)  lamp glows when  switch A & B positions are different.

This truth table simply resembles the logic of a XOR gate as shown below




From the truth table it can be verified that XOR logic is implemented.







Tuesday, September 3, 2013

ANNA UNIVERSITY INTERNAL ASSESMENT AND ATTEDANCE DETAILS FOR NOV/DEC2013&JAN 2014

Anna university internal and attendance report,NOV/DEC-2013

Anna university instructed to all colleges to enter their Attendance Details and Assessment Marks in the COE Web Portal  from  the academic year 2013 onwards. 


How to check  attendance report and internal mark :

1. click here

2.go to student log in, enter your registration number, date of birth & verification then click log in

3.Click assesment details tab to check your attendance report and internal marks.

A practical current source is usually represented by

 A practical current source is usually represented by                                                         (GATE-EE-1997)




 A.a resistance in series with a ideal current source 
 B.a resistance in parallel with a ideal current source 
 C.a resistance in parallel with a ideal voltage source 
 D.none of the above.

Ans: B

Explanatory Answer:

The diagrams shown below represents an ideal and practical current sources

Ideal current source

Practical current source

An ideal current source does not exists in practice because there is no current source which can maintain current supplied by it constant even when its terminals are open circuited.Hence a practical current source can be represented as an ideal current source in parallel with an resistance as shown above.This resistance Rs is called internal resistance of source. 


                                                                          


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Monday, September 2, 2013

GATE-EE-2014-SYLLABUS



GATE 2014 Syllabus for Electrical Engineering (EE)


ENGINEERING MATHEMATICS

Linear Algebra: Matrix Algebra, Systems of linear equations, Eigen values and eigen vectors.
Calculus: Mean value theorems, Theorems of integral calculus, Evaluation of definite and improper integrals, Partial Derivatives, Maxima and minima, Multiple integrals, Fourier series. Vector identities, Directional derivatives, Line, Surface and Volume integrals, Stokes, Gauss and Green’s theorems.
Differential equations: First order equation (linear and nonlinear), Higher order linear differential equations with constant coefficients, Method of variation of parameters, Cauchy’s and Euler’s equations, Initial and boundary value problems, Partial Differential Equations and variable separable method.
Complex variables: Analytic functions, Cauchy’s integral theorem and integral formula, Taylor’s and Laurent’ series, Residue theorem, solution integrals.
Probability and Statistics: Sampling theorems, Conditional probability, Mean, median, mode and standard deviation, Random variables, Discrete and continuous distributions, Poisson,Normal and Binomial distribution, Correlation and regression analysis.
Numerical Methods: Solutions of non-linear algebraic equations, single and multi-step methods for differential equations.
Transform Theory: Fourier transform,Laplace transform, Z-transform.

ELECTRICAL ENGINEERING

Electric Circuits and FieldsNetwork graph, KCL, KVL, node and mesh analysis, transient response of dc and ac networks; sinusoidal steady-state analysis, resonance, basic filter concepts; ideal current and voltage sources, Thevenin’s, Norton’s and Superposition and Maximum Power Transfer theorems, two-port networks, three phase circuits; Gauss Theorem, electric field and potential due to point, line, plane and spherical charge distributions; Ampere’s and Biot-Savart’s laws; inductance; dielectrics; capacitance.

Signals and SystemsRepresentation of continuous and discrete-time signals; shifting and scaling operations; linear, time-invariant and causal systems; Fourier series representation of continuous periodic signals; sampling theorem; Fourier, Laplace and Z transforms.

Electrical MachinesSingle phase transformer – equivalent circuit, phasor diagram, tests, regulation and efficiency; three phase transformers – connections, parallel operation; auto-transformer; energy conversion principles; DC machines – types, windings, generator characteristics, armature reaction and commutation, starting and speed control of motors; three phase induction motors – principles, types, performance characteristics, starting and speed control; single phase induction motors; synchronous machines – performance, regulation and parallel operation of generators, motor starting, characteristics and applications; servo and stepper motors.

Power SystemsBasic power generation concepts; transmission line models and performance; cable performance, insulation; corona and radio interference; distribution systems; per-unit quantities; bus impedance and admittance matrices; load flow; voltage control; power factor correction; economic operation; symmetrical components; fault analysis; principles of over-current, differential and distance protection; solid state relays and digital protection; circuit breakers; system stability concepts, swing curves and equal area criterion; HVDC transmission and FACTS concepts.

Control SystemsPrinciples of feedback; transfer function; block diagrams; steady-state errors; Routh and Niquist techniques; Bode plots; root loci; lag, lead and lead-lag compensation; state space model; state transition matrix, controllability and observability.

Electrical and Electronic MeasurementsBridges and potentiometers; PMMC, moving iron, dynamometer and induction type instruments; measurement of voltage, current, power, energy and power factor; instrument transformers; digital voltmeters and multimeters; phase, time and frequency measurement; Q-meters; oscilloscopes; potentiometric recorders; error analysis.

Analog and Digital ElectronicsCharacteristics of diodes, BJT, FET; amplifiers – biasing, equivalent circuit and frequency response; oscillators and feedback amplifiers; operational amplifiers – characteristics and applications; simple active filters; VCOs and timers; combinational and sequential logic circuits; multiplexer; Schmitt trigger; multi-vibrators; sample and hold circuits; A/D and D/A converters; 8-bit microprocessor basics, architecture, programming and interfacing.

Power Electronics and DrivesSemiconductor power diodes, transistors, thyristors, triacs, GTOs, MOSFETs and IGBTs – static characteristics and principles of operation; triggering circuits; phase control rectifiers; bridge converters – fully controlled and half controlled; principles of choppers and inverters; basis concepts of adjustable speed dc and ac drives.