put solution (eee-201 april 2012)

8/2/2019 Put Solution (Eee-201 April 2012)

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Q4.(b) Bandwidth is defined as the band of frequencies which lie between two half power frequencieswhere the current falls to 1 / 2 times of the maximum value.B.W = F2 F1 , where F2 & F1 are the two half power frequencies.Quality factor is defined as the the ratio of maximum stored energy per cycle to the energydissipated per cycle. Q = 2 maximum stored energy per cycle / energy dissipated per cycle.

DERIVATION :


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Q4(c)

= 10.6 + 3.06+ 0 = 13.66

= 0 + 1.76 + 7.07 = 8.83


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= 16.26

= 28.50

Resultant voltage = 22.99 sin( -28.50)

Q5(a)


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Q5(c)

Deflecting torque/force:The defection of any instrument is determined by the combined effect of thedeflecting torque/force, control torque/force and damping torque/force. The value ofdeflecting torque must depend on the electrical signal to be measured; thistorque/force causes the instrument movement to rotate from its zero position.

Controlling torque/force:This torque/force must act in the opposite sense to the deflecting torque/force, andthe movement will take up an equilibrium or definite position when the deflecting


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and controlling torque are equal in magnitude. Spiral springs or gravity usuallyprovides the controlling torque.

Damping torque/force:A damping force is required to act in a direction opposite to the movement of themoving system. This brings the moving system to rest at the deflected positionreasonably quickly without any oscillation or very small oscillation. This is providedby i) air friction ii) fluid friction iii) eddy current. It should be pointed out that any

damping force shall not influence the steady state deflection produced by a givendeflecting force or torque. Damping force increases with the angular velocity of themoving system, so that its effect is greatest when the rotation is rapid and zerowhen the system rotation is zero.

NUMERICAL:

Instrument resistance = Rm = 50 x103 / 20 x 103 = 2.5Series resistance required for full scale deflection for V = 500 V

R = (Vm / Im) Rm= (500 / .02) 2.5 = 2497.5 `

Q6(a)

GRID: The interconnection of various power system enables economic ggeneration by optimum use o f highcapacity generating plants. It si being done using either HVDC or HVAC links. An interconnected powersystem covering a major portion of a countrys territory ( or state) is called GRID. The different grids may beinterconnected through transmission lines called tie lines to form regional grid. When the different regional gridsare interconnected, they form a national grid.

Layout of Typical power system:


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Q6(b) Saving in Conductor Material:

Weight of copper NIwhere, I = Current in the windingand N = Number of turns of the winding.

In two winding transformer,Weight of copper in primary N1I1Weight of copper in secondary N2I2

Wtw = N1 I1 + N2 I2 = 2 N1 I1 ------------------------(1)

In case of step down autotransformer


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Weight of copper in section AC (N1 N2) I1Weight of copper in section BC N2 (I2 I1)

Wat = (N1 N2) I1 + N2 (I2 I1)= 2 (N1 N2) I1 ---------------------------------(2)

Taking ratio of eqn (1) and (2) we may find the saving as:

Saving of conductor material = k x (Conductor weight in 2 winding transformer)

Applications of Auto transformer:

1. For interconnecting systems which are operating roughly at same voltage.

2. For starting rotating machines like induction motors, synchronous motors etc.

3. To give small boost to a distribution cable to correct for voltage drop.

4. As a furnace transformer for getting required supply voltage

5. As a variac to vary the voltage to the load smoothly from load to the rated value

6. As dimmer stat for dimming the lights in cinema hall.

Q6(C)

Max Flux = 0.6 Wb

Flux density = 0.6 / 9 x 10-1 = 0.667 T

Length of air gap lg = 1.5 mm = 1.5 x10-3 m

Length of iron path li = 2{25-(2x 1.5) +35-(2x1.5)} 0.15 = 1.0785 m

Total AT = Bm lg /o + Bm li /o r

= 796+ 152 = 948 AT

Maximum value of excitation required Im = AT max / N = 948/ 500 = 1.896 A.

RMS value of excitation current I =Im / 2 = 1.896 / 2 = 1.34 A

RMS value of induced emf in the coil = E = 4.44 fNm = 66.6 V

Q7(b)


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The set of torque-slip characteristics with variation in rotor circuit resistance is shown in Fig.. Thecharacteristics shown are for increasing rotor circuit resistances .The point to note that, the maximumtorque remains same for all the characteristics. As the maximum torque depends on rotor reactance only,but not on rotor circuit resistance, only the slip at maximum torque increases with the increase in rotorcircuit resistance. So, for constant load torque operation, the slip increases or the speed decreases with the

increase in rotor circuit resistance. The motor efficiency decreases, as the rotor copper loss increases withthe increase in slip. The load torque remains same, but the output power decreases, as the speed decreases.Also, it may be observed that the starting torque increases with the increase in rotor circuit resistance, withthe total rotor circuit resistance lower than rotor reactance. The starting torque is equal to the maximumtorque, when the total rotor circuit resistance is equal to rotor reactance. If the rotor circuit resistance ismore than rotor reactance, the starting torque decreases.


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Q7(c) DOUBLE REVOLVING FIELD THEORY:When the stator winding carries a sinusoidal current (being fed from a single-phase supply), a sinusoidalspace distributed mmf, whose peak or maximum value alternates with time, is produced in the air gap.This sinusoidally varying flux ( ) is the sum of two rotating fluxes or fields, the magnitude of which isequal to half the value of the alternating flux ( / 2), and both the fluxes rotating synchronously at thespeed, (n s = (2 f) / P) in opposite directions. This is shown in Fig. The first set of figures show theresultant sum of the two rotating fluxes or fields, as the time axis (angle) is changing from = 0 to (180) . In second fig.it shows the alternating or pulsating flux (resultant) varying with time or angle.


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The flux or field rotating at synchronous speed, say, in the anticlockwise direction, i.e. the same direction, as thatof the motor (rotor) taken as positive induces emf (voltage) in the rotor conductors. The rotor is a squirrel cageone, with bars short circuited via end rings. The current flows in the rotor conductors, and the electromagnetictorque is produced in the same direction as given above, which is termed as positive (+ve). The other part of fluxor field rotates at the same speed in the opposite (clockwise) direction, taken as negative. So, the torque producedby this field is negative (-ve), as it is in the clockwise direction, same as that of the direction of rotation of thisfield. Two torques are in the opposite direction, and the resultant (total) torque is the difference of the twotorques produced . The two torques are equal and opposite, and the resultant torque is zero. So, there is no startingtorque in a single-phase IM.But, if the rotor is started or rotated somehow, say in the anticlockwise (forward) direction, the forward torque ismore than the backward torque, with the resultant torque now being positive. The motor accelerates in the forward

direction, with the forward torque being more than the backward torque. The resultant torque is thus positive asthe motor rotates in the forward direction.

Various methods of starting 1- phase induction motor are:(i) Resistance split phase(ii) Capacitor start(iii)Capacitor start Capacitor run(iv) Shaded pole


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Resistance Split-phase Motor:

The schematic (circuit) diagram of this motor is given in Fig. above. As seen, another (auxiliary) winding

with a high resistance in series is to be added along with the main winding in the stator. This winding hashigher resistance to reactanceR /Xratio as compared to that in the main winding, and is placed at a spaceangle of 90 from the main winding as given earlier. The phasor diagram of the currents in two windingsand the input voltage is shown in Fig. b. The current (Ia ) in the auxiliary winding lags the voltage (V) byan angle,a , which is small, whereas the current (Im) in the main winding lags the voltage (V) by anangle, m , which is nearly 90. The phase angle between the two currents is (90 a), which should beat least 30.This results in a small amount of starting torque. The switch, S (centrifugal switch) is in serieswith the auxiliary winding. It automatically cuts out the auxiliary or starting winding, when the motorattains a speed close to full load speed. The motor has a starting torque of 100200% of full load torque,with the starting current as 5-7 times the full load current. The torque-speed characteristics of the motorwith/without auxiliary winding are shown in Fig. c. The change over occurs, when the auxiliary winding is

switched off as given earlier. The direction of rotation is reversed by reversing the terminals of any one oftwo windings, but not both, before connecting the motor to the supply terminals. This motor is used inapplications, such as fan, saw, small lathe, centrifugal pump, blower, office equipment, washing machine,etc.

put solution (eee-201 april 2012)

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