SSC JE 2012 Electrical question paper with solution

The direction of rotation of a DC shunt motor can be reversed by interchanging leads of either the field winding or the Armature Winding.

Generally changing the direction of the field is easier, because it carries lesser current as compared to armature current. However, the reversal should not be done while the armature is excited.

Why an AC series Motor are Built with few turns?

The AC series motor operates on either AC or DC circuits. When an ordinary DC series motor is connected to an AC supply, the current drawn by the motor is low due to the high series-field impedance. The result is low running torque. To reduce the field reactance to a minimum, AC series motors are built with as few turns as possible.

The decrease in the number of turns of the field winding reduces the load torque, i.e., if the field turns to decrease, its MMF decreases and then flux, which will increase the speed, and hence the torque will decrease. But in order to maintain constant load torque, it is necessary to increase the armature turns proportionately.

If the armature turns to increase, the inductive reactance of the armature would increase, which can be neutralized by providing the compensating winding.

Forward field slip Speed

S_F = (Ns -Nr)/Ns

Synchronous speed Ns = 120f/p = 120×50/4 = 1500

(1500 – 1300)/1500 = 0.133Hz

Rotor frequency due to forward field = S_F x F = 0.133 x 50 = 6.67 Hz

Backward slip of an induction motor

Sb = (2 – S_f) = 2 – 0.133 = 1.87 Hz

Rotor frequency due to Backward field = Sb x F = 1.87 x 50 = 93.35Hz

If the Key 2 is closed while the Key 3 and Key 1 is open, the circuit is closed for 100 W lamp and the lamp will glow with maximum brightness

If key 1 is closed while the other 2 are open then 100W and 40 W lamps will glow and In a series connection, the current flowing across each element is the same. So when a 40W bulb and 100W bulb are connected in series, the same current will flow through them. Since both bulbs are rated at the same voltage, we can say that the resistance of each bulb is inversely proportional to its rated power. Hence 40 W bulb will glow brighter.

The expression for mutual inductance

M = K√L1L2

Now the value of the coefficient of coupling K lies between 0 and 1 i.e Mutual inductance is maximum when K = 1 and mutual inductance is zero when K = 0

So K ≤ 1

∴ M ≤ √L1L2

The reluctance of a uniform magnetic circuit can be calculated as:

S = L/μA

S2/S1 = A1 L2/A2 L1

Since length l2 of the bar is increased by 20% = 120/100 = 1.2  i.e

L2 = 1.2L1

And area is decreased by 20% = 80/100 = 0.8 i.e

A2 = 0.8A1

The expression for mutual inductance

The self-inductance of a solenoid is given as

L = μN²A/I = L1

where

N is the number of turns of the solenoid

A is the area of each turn of the coil

l is the length of the solenoid

and μ is the permeability constant

so, if the number of turns was to be doubled the self-inductance would be

L2 = u (2N)²A/l

or

L2 = 4L1

it would be quadrupled or increased fourfold.

Both hysteresis loss and eddy current loss give rise to heat in a magnetic circuit. The two losses are usually taken together and are called ‘iron loss’.

Pi = P_E + P_H

Pi = f²B²max + f B^1.6max

Iron losses thus vary with both frequency and magnetic flux density. The power transformer is likely to be fed at the constant frequency and at the constant voltage, which means that magnetic flux density is almost constant. Thus iron losses in a transformer are constant from no load to full load.

The expression for mutual inductance

∴ M1 = M2 = 1H