SSC JE 2015 Electrical question paper with solution

Ques 31. To operate properly a transistor’s base-emitter junction must be forward biased with the reverse bias applied to which junction?

  1. Base-collector
  2. Collector-emitter
  3. Base-emitter
  4. Collector-base

Biasing of the transistor means setting up a fixed level of current which should flow on the transistor with the fixed level of the voltage drop across the transistor.

For proper working of the transistor, the base-emitter junction is forward biased and the collector-base junction is reversed biased.

By making the base-emitter junction forward biased the resistance of the base-emitter become very low as compared with the resistance of the collector-base junction.

Transistor Biasing

A forward-biased semiconductor junction has low resistance while the reversed biased junction has high resistance. And the large difference in junction resistance makes the transistor capable of power gain.

 

Ques 32. There are 3 lamps 40 W, 100 W, and 60 W. To realize the full rated power of the lamps they are to be connected in

  1. Parallel Only
  2. Series or Parallel
  3. Series Only
  4. Series-parallel
By connecting the 3 lamps in parallel the total resistance of a parallel is less than the resistance of the smallest resistance. Hence the overall resistance is decreased.

The power dissipated in a circuit is equal to

P = V2/R

So by reducing the resistance, we increase the overall power. Therefore to realize the full rated power of the lamps they are to be connected in parallel.

 

Ques 33. In a three-phase system, the volt-ampere rating is given by?

  1. 3 VL IL
  2. Vph Iph
  3. VL IL
  4. √3VL IcosΦ
The 3-phase power is given by

P = 3VICosΦ

For star connection VL = √3V

For Delta connection IL = √3I

(3VLILCosΦ)/√3

= √3VL IL  cosΦ

 

Ques 34. With the positive probe on an NPN base, an ohmmeter reading between the other transistor terminal should be?

  1. Infinite
  2. Open
  3. Low resistance
  4. High Resistance
Since the base is connected to +ve probe it has a higher potential than other terminals so it makes the transistor in forward bias mode for both terminals and as we know in forward bias resistance is very low.

 

Ques 35. The primary and secondary windings of a transformer are wound on the top of each other in order to reduce ___

  1. Iron losses
  2. Leakage reactance
  3. Copper losses
  4. Winding Resistance

A secondary coil is wound over the primary coil in a transformer to have maximum flux linkages such that the maximum efficiency is obtained.By doing this, the distance between the coils is reduced and thus lower flux leakages.

 

Ques 36. If a dynamometer-type wattmeter is connected to an AC circuit the power indicated by the wattmeter will be?

  1. Instantaneous Power
  2. Peak Power
  3. Volt-ampere product
  4. Average power
Wattmeter measure the active power P in the circuit. In AC circuit the active power at any instant is given by

P = VI

And the average power is given by which can be measured by wattmeter only is given as

P = VICosΦ

Where V and I are RMS values of voltage and current

 

Ques 37.  A Lissajous pattern on an oscilloscope has 5 horizontal tangencies and 2 vertical tangencies. The frequency of the horizontal input is 1000 Hz. The frequency of the vertical input will be

  1. 400 Hz
  2. 2500 Hz
  3. 4000 Hz
  4. 5000 Hz
Lissajous Pattern with different Ratios is given as

Fy/fx =hx/hy

Fy: unknown frequency

Fx: known frequency

hy: tangencies at the y-axis

hx: tangencies at the x-axis

Fy/1000 = 5/2

Fy = 2500 Hz

 

Ques 38. Three wattmeter method of power measurement can be used to measure power in

  1. Balanced Circuits
  2. Unbalanced circuits
  3. Both Balanced and Unbalanced circuits
  4. None of the above

3 wattmeter method

In case if the supply is 3-phase, 4 wire system, then for the unbalanced system or loads the two wattmeters can’t be used as there is current flowing through neutral. The three-wattmeter method of power measurement will work regardless of whether the load is balanced or unbalanced, delta or wyne connected.

 

Ques 39.  If the stator voltage of a squirrel cage induction motor is reduced to 50 percent of its rated value, the torque developed is reduced by how many percentages of its full load value?

  1. 50%
  2. 25%
  3. 75%
  4. 57.7%

The squirrel cage induction motor has a very rugged construction, the three-phase squirrel-cage motor is capable of handling the starting current without any damage to winding. However very large motors may cause line drops, which may affect other equipment operating from the same system For such installations, reduced voltage starters a used. The reduced voltage limits the starting current to a lower value.

Starting a squirrel cage induction motor under reduced voltage may be reduced the starting voltage considerably. From the torque equation of a three-phase induction motor, the starting torque is approximately proportional to the square of the applied voltage i.e T ∝ V2

By reducing the voltage by 50% ( reducing voltage by 1/2) than

T = (1/2)2

T = 1/4 = 25%

Hence by reducing the voltage by 50%, the torque is reduced to 25% and 75% of the full load value.

 

Ques 40. A short shunt compound generator supplies a load current of 100 A at 250 V. The generator has the following winding resistances:

shunt field = 130Ω, Armature 0.1 Ω and series field = 0.1 Ω . Find the Emf generated if the brush drop is  1 V per brush.

  1. 272.2 V
  2. 262.2 V
  3. 272.0 V
  4. 262.0 V
Voltage = 250 V

Load current (IL) = 100 A

Shunt field (Rf) = 130Ω

Armature field (Ra) = 0.1Ω

Series field  (Rse) = 0.1Ω

Brush voltage Drop = 1V

Now voltage drop in the field winding = ILRse

= 100 x 0.1 = 10 V

For short shunt compound generator, the voltage drop across series field winding gets added with voltage drop across armature winding

Voltage drop across field winding  + Voltage drop across armature winding

V1 = 250 + 10 = 260 V

Field current If = V1/Rf = 260/130 = 2 A

So armature current

Ia = IL + If = 100 + 2 = 102 A

Generated EMF

E = V1 + IaRa

= 260 + 102×0.1

E = 270.2 Volt

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