SSC JE 2014 Electrical question paper with solution(Morning-Shift)

• When the synchronous motor operates at the leading power factor then Rotor is overexcited in such a way that back emf (Eb which is generated in stator due to dc excitation of the rotor ) is greater than the supply voltage (V).
• At this time resultant flux is greater than that is required for the unity power factor,  then this extra flux will generate reactive power so the motor will generate additional reactive power.  And it will also use active power for mechanical work.
• Therefore synchronous motor working on leading PF will work as a synchronous condenser or phase advance

For DC motor voltage is given as

V = Eb + IaRa

Ia = (V – Eb)/Ra

= (150 -142)/0.4 = 20A

The peak inverse voltage (PIV) of four diode bridge is one half that of full wave rectifier using two diodes. Therefore four bridge diode rectifier required low peak inverse voltage

Let’s take a 200 volts – 100  watts bulb and calculate the resistance of its filament :

We know,

P=V²R

or

R=V²P

When V=200 volts and P = 100 watts,

R = (200*200)/100 = 400 ohms.

Since now the two bulbs are in series, the total resistance offered by them is :

R’ = 400 Ω + 400 Ω = 800 Ω

Let them be powered by a 200 volts DC source.

Now the current that flows through either of them is :

I=V/R

=200/800=0.25A

Now total power consumed by each lamp

P=I² x R

P=(0.25 x 0.25)400

P = 25 watts

Biot-Savart’s law describes the three different types of currents, which is nothing but the filament or line current, surface current and volume current. Hence, this law is a general modification of Ampere’s circuital law.

Active Power P = 60 W

Reactive Power Q = 80 VAR

S =√P² + Q² = √60² + 80²

= 100 VA

Power factor = cos Φ = P/S = 60/100 =0.6

Second Method

When the load on the supply system changes, the terminal voltage of the alternator also varies due to the changed voltage drop in the synchronous reactance of the armature.

The voltage of the alternator can be kept constant by changing the field current of the alternator in accordance with the load. This is known as an excitation control method. The excitation of the alternator can be controlled by the use of an automatic or hand-operated regulator acting in the field circuit of the alternator.

The excitation control method is satisfactory only for relatively short lines. However, it is not suitable for long lines as the voltage at the alternator terminals will have to be varied too much in order that the voltage at the far end of the line may be constant.

Unit Protection Scheme:

Unit type protection is also called main protection or primary protection it protects a specific area of the system such as a  transmission line, transformer,  generator, or bus bar. The unit protection scheme is based on Kirchhoff’s Current Law – the sum of the currents entering an area of the system must be zero Any deviation from this must indicate an abnormal current path.

This protection scheme is based on the principle of selectivity i.e the effects of a disturbance or operating condition outside the area of interest are totally ignored and the protection must be designed to be stable above the maximum possible fault current that could flow through the protected area.

Loss load factor

Several empirical formulae are available to determine the Loss load factor

Loss load factor is the ratio of average power losses to the power losses at peak demand

Loss load factor = Average power losses/Power losses at peak demand

Loss load factor can also be determined by

Loss load factor = 0.25 x (load factor) + 0.75 x (load factor)2

In the above question, load factor is given as F

Loss load factor = 0.25 F² + 0.85 F