SSC JE 2013 electrical question paper with solution

Explanation:

• Steel is used as a magnetic core in a transformer, it acts as a good conductor and whenever a time-varying flux passes through the core, circulating currents are produced and known as “Eddy currents”.
• To reduce conductivity (eddy current losses) by not destroying magnetic properties, about 4 to 5% of silica impurity is added so that the conductivity of steel decreases.
• The high content of silica is not preferred the material becomes more brittle and it will destroy the mechanical properties.

Explanation:

Power consumed by wattmeter = CT ratio x PT ratio x VIcosΦ……..1

In the above question

Power consumed by wattmeter = 2000 watts

CT ratio = 15 / 30

PT ratio = 300 / 600

VI = 600 x 30

Putting all the value in equation number 1 we get

2000=(15/30) x (300/600) x 600 x 30 x cosΦ

CosΦ = 0.4444

Power = VI cosΦ = 600 x 30 x 0.4444

Power = 7999.2 ≅ 8000 watts

Explanation:

• A deliberate connection of resistance in parallel with the contact space (or arc) is called resistance switching.
• Resistance switching is employed in the circuit breaker having high post zero resistance of contact space (i.e air blast circuit breaker).
•  High restriking voltage appears across the contacts because of current chopping. If these voltages are not allowed to discharge, they may cause the breakdown of insulation of the circuit breaker or the neighboring equipment.
• The resistance discharges the heavy arc current through them this results in the decrease of arc current and an increase in the rate of deionization of the arc path.
• Thus the arc resistance increased leading to a further increase in current through the shunt resistance R.
• This build-up process continues until the current becomes so small that it fails to maintain the arc.

Answer 1. 2ω

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

Explanation:

• GLS ( general lighting service) lamps are the source of incandescent light.
• Acid etching creates a very smooth, glossy, and satin finish; the acid-etched lamp is maintenance-free as it does not show dirt marks or fingerprints.

Answer 2. 0.5

I_L = 12A

W1 = 11KW

W2 = 0

Power factor of the wattmeter is

tanΦ = √3(W1 – W2)(W1 + W2) = √3 (11 – 0)(11 + 0)

tanΦ = √3 = Φ = 60º

Cos 60º = 1/2 = 0.5

Explanation:

Plugging in DC Motor

• Plugging of Dc motor is the method of reconnecting the motor to the line with reverse polarity hence now the motor will produce torque in the opposite direction to that of rotation.
• Plugging in a DC motor means the reversing of either field or armature current. So either Eb or V gets reversed. Therefore, the voltage across the armature will be V+Eb which is almost twice the supply voltage.
• The rotor speed decreases until it becomes zero and then the rotor accelerates in the opposite direction. Therefore, the plugging is used to get a quick reversal and a rapid stop or Braking.

Explanation:

• Welding usually requires high current (over 80 amperes) and it can need above 12,000 amperes in spot welding.
• A transformer-style welding power supply converts the moderate voltage and moderate current electricity from the utility mains (typically 230 or 115 VAC) into a high current and low voltage supply, typically between 17 and 45 volts and 55 to 590 amperes.
  

Explanation:

Plant Capacity Factor

It is the product of the load factor and the utilization factor.

Capacity factor = Load factor × Utilization factor

Where load factor is the ratio of average demand and maximum demand

And Utilization Factor is the ratio of kWh generated to the product of the plant capacity and the number of hours for which the plant was in operation.