SSC JE Basic Electrical Questions (2009 – 2018) Solved (Part-3)

Ques.11.  Examine the two statements ‘A’ and ‘R’ and select your answer.

Statement A:
Switching off a lamp in the house produces noise on a radio.

Statement R:
The switching operation produces an arc across separating contacts.

  1. Both A and R are true and R is a correct explanation of A
  2. Both A and R are true and R is not a correct explanation of A
  3. Both A and R are true and R is not a correct explanation of A
  4. A is false but R is true

Answer.1. Both A and R are true and R is a correct explanation of A

Explanation:

Statement A:

  • Usually, a radio will make a sharp ‘pop’ when another electrical item is switched off.
  • The cause of the noise is the arc that momentarily bridges the contacts of the lamp’s switch when it is opened (shut off).
  • The arc produces radio-frequency energy, which nearby radios (especially AM, but sometimes also FM) detect.
  • Since the radio can’t differentiate between intended and unintended signals, it will reproduce whatever comes its way

Statement R:

During the opening of the contact, the medium between the contacts gets highly ionized therefore low resistive path is created between the contacts and current tends to flow through that path.

Hence both the statement are true and  A is the correct explanation of Statement R.

 

 

Ques.12.  In electronic circuits, for blocking the DC component of a voltage signal, a/an ____________ is connected in series with the voltage source.

  1. Capacitor
  2. Diode
  3. Resistor
  4. Inductor

Answer.1. Capacitor

Explanation:

Why does a capacitor block DC but pass AC?

  • A capacitor is nothing but 2 metal plates separated by an insulator. The insulator can be anything like paper, air, rubber, etc.
  • When a capacitor is connected to a voltage source, positive charges from the positive terminal of the voltage source and negative charges(electrons) from the negative terminal of the voltage source, travel and GET ACCOMMODATED ON BOTH THE PLATES OF THE CAPACITOR.
  • Once the capacitor gets fully charged it will now act as the insulator and it doesn’t allow current to pass through it.
  • Now as we know that the voltage value of DC remains constant for all the time the power supply is ON. Hence capacitor gets charged to its total limit and thus doesn’t allow any current to flow hence DC BLOCKED.
  • Talking about AC it is the time-varying Voltage i.e the polarity of AC changes from time T1 to T2, therefore, at time T1, the capacitor gets charged such that the positive charges are accommodated on the left plate.
  • After time T1, the polarity changes (because the voltage source is AC) and now at the start of T2, voltage source V will attract the negative and positive charges from the respective plates to its terminals and current will flow in another direction. And because the capacitor is losing its charges – the Capacitor is getting discharged.

 

Ques.13. If the angular frequency of an alternating voltage is ω, then the angular frequency of instantaneous real power absorbed in an ac circuit is

  1. 2 ω
  2. ω
  3. 3 ω
  4. ω/2

Answer 1.

Instantaneous power p(t) is defined as the product of instantaneous voltage v(t) and instantaneous current i(t).

Assuming the sinusoids waveforms v(t) = Vm cos (ω t + θ v ) and i(t) = Im cos (ω t + θi ) represent the voltage and current, then it can be shown:

p(t) = P + P cos 2ω t – Q sin 2ω t

where P = ½ Vm Im cos (θ v- θi )

and Q = ½ Vm Im sin (θ v – θi )

The term P is a constant and represents the average of the instantaneous power p(t) (since the averages of cos 2ω t and Q sin 2ω t are both zero). The term P + P cos 2ω t represents the instantaneous real power (instantaneous active power). Q is called the reactive power and the term Q sin 2ω t represents the instantaneous imaginary power (instantaneous reactive power). The second term is a time-varying sinusoid whose frequency is equal to twice the angular frequency of the supply.

 

 

Ques.14. In the circuit, v is the input voltage applied across the capacitor of 2 F. Current through the capacitor is

Answer. D

The given triangle output can be changed into the square output

The instantaneous value of capacitor current is

i(t) = cdv/dt

where C is the capacitance in farad

dv/dt = = instantaneous rate of voltage change(Volt/second)

For charging rate of voltage change is 10 and for discharging rate of voltage change is -11 therefore

 

Ques.15. A geyser is operated from 230 V, 50 c/s mains. The frequency of the instantaneous power consumed by the geyser is

  1. 25 c/s
  2. 50 c/s
  3. 100 c/s
  4. 150 c/s

Answer 3. 100 c/s

Explanation:

Frequency of instantaneous power consumed = 2f = 2 x 50 = 100 c/s

 

Ques.16.  Ampere-second is the unit of

  1. E.M.F
  2. Power
  3. Electric Charge
  4. Energy

Answer.3. Electric Charge

Explanation:

The SI unit of charge, the coulomb, “is the quantity of electricity carried in 1 second by a current of 1 ampere

 

Ques.17.  Two lossy capacitors with equal capacitance values and power factor of 0.01 and 0.02 are in parallel, and the combination is supplied from a sinusoidal voltage source. The power factor of the combination is

  1. 0.03
  2. 0.015
  3. 0.01
  4. 0.0002

Answer.2. 0.015

Explanation: 

Since capacitor are connected in parallel therefore

PF = PF1 + PF2 / 2

= 0.01 + 0.02 /2

= 0.015

 

Ques.18. In a 3-phase 400 V, 4-wire system, two incandescent lamps, one having 230 V, 100 W specification and the other 230 V, 200 W are connected between R phase-neutral and Y phase-neutral respectively. If the neutral wire breaks

  1. 100 W lamp will fuse first
  2. 200 W lamp will fuse first
  3. Both the lamp will fuse together
  4. Both the lamp will glow

Answer.1. 100 W lamp will fuse first

Explanation:

  • The 100W lamp will fail first, Although the voltage across 100 w lamp will be higher (293V), it will fail because of effective current (0.489A) which is more than the 100 w lamp normal current (0.4A).
  • Increase in current cause more heat and heating element will fail.

 

Ques.19.  The peak value of the output voltage of half-wave rectifier is 100 V. The r.m.s value of the half-wave rectifier output voltage will be

  1. 100 V
  2. 50 V
  3. 70.7 V
  4. 35.35 V

Answer.2. 50V

Explanation:

RMS value of Half wave Rectifier is given as

VRms = Vm / 2

= 100 / 2

= 50 V

 

Ques.20.In a series, R-L circuit supplied from a sinusoidal voltage source, the voltage across R and L are 3V and 4V respectively. The supply voltage is then

  1. 7 V
  2. 1 V
  3. 3.5 V
  4. 5 V

Answer.4. 5V

Explanation:

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