SSC JE 2012 Electrical question paper with solution

2 Comments / SSC previous year solved electrical papers / April 11, 2022

Ques 11. A current of i = 6 + 10sin(100π t) 20sin(200 π t) is flowing through a series combination of a PMMC and moving iron instrument. The ratio of the two currents as registered by the M.I. and PMMC meter is

  1. 1.81
  2. 3.11
  3. 2.82✔
  4. 2.63

i = 6 + 10sin(100π t) 20sin(200 π t)

PMMC read-only DC value

(I)PMMC = 6

For M.I instrument

solution 12

 

Ques 12. Three resistance 5Ω each are connected in star, values of equivalent delta resistance are

  1. 1.5 Ω each
  2. 2.5 Ω each
  3. 5/3 Ω each
  4. 15 Ω each✔

If three resistance each of value R is connected in delta formation. Then the equivalent delta connection will have three resistance of equal value which is 3R.

In the above question, Individual resistance in Star connection = 5Ω

Individual resistance in Delta connection = 3 x 5 = 15Ω

 

Ques 13. A 120 V, 60 W incandescent lamp has to be operated from 220 V, 50c/s, 1 Phase AC supply. In order to do, this a circuit element has to be connected in series with the lamp. Which one of the following elements is preferable?

  1. Pure Capacitance✔
  2. Pure Inductance or capacitance
  3. Resistance
  4. Pure Inductance

Series capacitors are generally employed to improve the stability of the system and to increase the voltage rating.

We know that the reactance of a capacitor is given as

Rc = 1/2πfc

For an A.C. source, frequency, f > 0 i.e Rc = 1/2πfc > 0 which means that a capacitor offers a constant resistance in the A.C circuit i.e. it allows the lamp to glow continuously.

In the case of Inductance, the brightness of the lamp will depend upon the number of turns in the inductor

 

Ques 14. The bandwidth of an AC series circuit consisting of R, L and C is

  1. L/R
  2. R/2πL✔
  3. L/RC
  4. RC/L

Bandwidth is defined as the frequencies at which the power in the circuit is half of its maximum value.

The bandwidth of  series RLC circuit

Bandwidth =(f2 – f1) = R/2πL

Where f1 is called lower half frequency

f2 is called as upper half frequency

 

Ques 15. For a balanced 3-phase supply system, the phasor sum of the line current is NOT zero if the load is

  1. Balanced Delta connected
  2. Unbalanced Delta connected
  3. Balanced star connected
  4. Unbalanced star connected✔

In the case of delta connection, whether the load is balanced or unbalanced the phasor sum of the line current is always zero.

But in the case of star connection, the phasor sum of the line current is not zero for unbalanced load but is equal to neutral current.

 

Ques 16. At series resonance of an RLC circuit, the impressed voltage is

  1. Equal to the resistive drop✔
  2. Equal to the capacity drop
  3. Greater than the resistive drop
  4. Equal to the inductive drop

At a particular frequency (resonant frequency),  X1=X2 because resonance of a series RLC circuit occurs when the inductive and capacitive reactances are equal in magnitude but cancel each other because they are 180 degrees apart in phase.
Thus, at the resonant frequency, the net reactance is zero because of X1=X2.

The circuit impedance Z becomes minimum and is equal to the resistance R. Impressed voltage is also called applied voltage therefore at a series resonance RLC  circuit the impressed voltage or applied voltage is equal to the voltage across R.

 

Ques 17. In a series RLC circuit R 20Ω, XL = 30Ω, and Xc =30Ω. If the supply voltage across the combination is v = 100sin(100πt + 30° ) volts, the instantaneous current and the power factor of the circuit are respectively

  1. I = 3.536 sin(100πt + 30° ) Amp, p.f =0.866
  2. I = 5 sin(100πt + 30° ) Amps, p.f = unity✔
  3. I = 3.536 sin(100πt + 30° ) Amps, p.f = unity
  4. I = 5 sin(100πt + 30° ) Amps, p.f = 0.866

Given R = 20Ω

XL = 30Ω

Xc = 30Ω

At series RLC circuit

tanΦ = (XL – Xc)/R

= 30 – 30/20 = 0

or cosΦ =1

Impedance of RLC circuit

sol1

Z = 20Ω

I = V/Z

I =  100sin(100πt + 30° )/20

I = 5 sin(100πt + 30° ) Amp at unity power factor

 

Ques 18. The RMS of the alternating current given by the equation

i = 50 sin(314t – 10°) + 30 sin (314 t – 20°)

  1. 41.23 A✔
  2. 58.31 A
  3. 38.73 A
  4. 77. 43 A

RMS value of alternating current is given as

1

= 41.23 A

 

Ques 19.  A series RLC  circuit will have a unity P.F is operated at a frequency of

  1. 1/LC
  2. LC
  3. 1/2π√LC✔
  4. 1/2LC 

Series RLC circuit will have unity power factor when resonant will occur.

For resonant, XC = XL

So, 1 / (2π fC) = 2π fL.

f2 = 1/4π2fL

f = 1/2π√LC

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