SSC JE 2012 Electrical question paper with solution

Ques 41. In the circuit forward resistance of the diode D is 2Ω and its reverse resistance is infinitely high.

numerical 41a

A list consists of meters (List-I) and another list shows, the meter readings (Lis-II)

Numerical 41 1

Which one of the options given here is correct to indicate the type of meter (List-I) and its reading (List-II)

  1. i → (a), (ii) → c
  2. (i) → b, (ii) → (d)
  3. (i) → (a), (ii) → (b)
  4. (i) → (b), (ii) → (a)

For PMMC and Hot wire

Given Vrms =100V

Peak voltage Vm = VRMS x √2

= 100 x √2 = 141.42 V

Im = Vm/R = 141.42/(8+2)

Im = 14.14 A

RMS current of half wave rectifier is

Irms = Im/2 = 14.14/2 = 7.07 A →Hot wire

In the most analogous meter movement which employs a PMMC deflection system the average current flows through the meter coil. And the average current for have wave rectified sine wave is

Iavg = Im/π =14.14/π = 4.5A→PMMC

 

Ques 42. In figure D in an ideal diode. If the RMS value of the input voltage is 50V, then the RMS current through 100 Ω is

numerical 42

  1. 0.5/√2A
  2. 0.25A
  3. 0.5A
  4. 0.5√2 A

In the given figure the circuit consists of a single diode, therefore, the Output will be rectified as half-wave, hence RMS current of half-wave rectifier is

Irms = Im/2

RMS voltage = 50 V

RMS Value of output (Vrms) of Half voltage rectifier

Vrms = Vm/√2

Vm = Vrms x √2 = 50 x √2

Im = Vm/R

= 50√2/100 = 1/√2

Irms = Im/2

= 1/2√2 = 0.5/√2

 

Ques 43. Which one of the following materials is the semiconductor?

  1. Chromium
  2. Selenium
  3. Bismuth
  4. Silica

Selenium is produced as a byproduct in the refining of metal sulfide ores. Selenium is a semiconductor and is used in photocells. Applications in electronics, once important, have been mostly replaced with silicon semiconductor devices.

 

Ques 44. In a rectifier circuit, the primary function of the filter is to

  1. Control the DC level of the output voltage
  2. Remove ripples from the rectified output
  3. Minimize AC input variations
  4. Suppress odd harmonics In the rectifier output

We know that rectifiers are used to convert AC to DC, but not a pure DC. The output that is obtained from a rectifier is pulsating in nature, which basically means that it has a certain amount of AC component called the ripple.

These ripple components are very much unwanted and undesirable in a rectifier circuit as they reduce the efficiency of AC to DC conversion. So, in order to remove these components, filters are used. A filter ( capacitor-based filter) in a circuit takes this mixed input and produces a pure DC output, bypassing the AC component to earth / neutral.

 

Ques 45. In an R-L series circuit R = 20 Ω, L = 0.056 H and the supply frequency in f = 50 Hz. The magnitude of the impedance of the circuit is

  1. 26.46Ω
  2. 20.0Ω
  3. 37.6Ω
  4. 20.056Ω

Impedance of RL circuit is

3

 

Ques 46. A pure sinusoidal current is being rectified. For the given maximum value of sinusoidal current if the RMS value of half-wave rectified current is 50 A, then the RMS value of full-wave rectification will be

  1. 50/πA
  2. 100/πA
  3. 100A
  4. 70.7 A

For half-wave rectifier RMS current

Irms = Im/2

Im = Irms x 2 = 50 x 2 = 100 A

For Full wave rectifier RMS current

Irms = Im/√2

= 100/√2 = 70.71 A

 

Ques 47. Under thermal equilibrium semi-conductor, the ratio of the number of holes to the number of conduction electrons is

  1. 1/2
  2. Infinity
  3. 1
  4. 2

Thermodynamic (thermal) equilibrium is present when a semiconductor is kept at the constant temperature without any outside bias excitation for a sufficient length of time. In thermal equilibrium, electrons and holes are generated by thermal ionization only. The same amount of carrier generated in any volume element must be recombined in the same volume amount i.e the ratio of the number of holes to the number of conduction electrons is 1.

 

Ques 48. As  per IE rules, the maximum allowable variation between the declared and actual voltage at consumer’s premised should be (Most asked question in SSC)

  1. ± 4%
  2. ±4.5%
  3. ± 3%
  4. ± 3.5%

According to Indian Electricity Rules, the permissible voltage drop from the supply terminal to any point on the wiring system should not exceed ±4% volt of the nominal supply voltage

 

Ques 49. The alternator used in hydel power station has more number of poles in it than used in thermal power station, because

  1. Power generated by the alternator is less
  2. The speed of the prime mover may be changed whenever required
  3. Power generated by the alternator may be changed according to the demand
  4. The speed of its prime mover is less

In hydraulic turbine in the hydroelectric plants operates at low speed, hence the speed of prime mover is low(as compared to steam in case of thermal plants). according to the relation, we have

f=(P x N)/120

P= no of poles

F= frequency

N= speed

So in order to have a fixed frequency product of ‘p’ and ‘n’ should be a constant. Hence P should be more(as N is less). In order to accommodate more number of poles, the diameter of the rotor should be large.

 

Ques 50. The connected load of a consumer is 2kW and the maximum demand is 1.5kW. The demand factor of the consumer is

  1. 1.33
  2. 0.75
  3. 0.375
  4. None of these

Demand factor = Maximum demand of a system / Total connected load on the system

1.5/2 = 0.75

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