SSC JE 2013 electrical question paper with solution

Explanation:

The diagram of the above question will be

Now in capacitor connected in parallel, their equivalent capacitance will be

C = 1/C1 + 1/C2 + 1/C3

= 1/3 +1/3 + 1/3

= 1μF

= 1μF

when the capacitors are connected in series their equivalent capacitance will be

C = C1 + C2

= 1 + 6

=7 μF 

Explanation:

Here

V = 400 V , I₁ = 20 A, N₁ = 1100

R_a = 0.06 ohm, R_se = 0.08 ohm

I₂ = 50 A, N₂ =?

Φ₂ = 130 % of Φ₁ = 1.3 Φ₁

While drawing line current of 20 A

E_b1 = V – I₁(R_a + R_se)

= 400 – 20(0.06 + 0.08)

= 397. 2 V

While drawing line current of 50 A

E_b2 = V – I₂(R_a + R_se)

= 400 – 50(0.06 + 0.08)

= 393 V

We know that

Explanation:

For series RLC circuit the maximum applied voltage is given as

On putting the value

V = 5 v 

Explanation:

The phase difference between applied voltage and current is

tanΦ = XL / R = Lω / R

= 4 Ω / 4Ω

or Φ = π/4 = 45°

Hence the current lags the voltage by 45°

Explanation:

• Superposition is used for circuit analysis methods when we have a circuit with multiple inputs or multiple power sources at least two sources are required which can be a voltage source or current source.

To solve the given circuit using the superposition theorem the contribution of each individual source, all of the other sources first must be “turned off” (set to zero) by

• Replacing all other independent voltage sources with a short circuit (thereby eliminating the difference of potential i.e. V=0; internal impedance of the ideal voltage source is zero (short circuit)).
• Replacing all other independent current sources with an open circuit (thereby eliminating current i.e. I=0; internal impedance of the ideal current source is infinite (open circuit).

Explanation:

• When 3 phase supply is given to the stator of a 3 phase induction motor, rotating stator flux will be produced, which will induce emf in rotor windings, according to Faraday’s law of electromagnetic induction.
• Now if the rotor windings are kept open-circuited, no current will flow in these windings.
• Copper bars are short-circuited at both ends via end rings.
• Due to short-circuited rotor windings, short circuit rotor currents will flow in them, giving rise to their own magnetic field linking with the stator flux hence the motor start running.

Explanation:

• Damper windings are windings that are wound to the rotor poles of the machine (winding is similar to that of an induction machine) which helps in two ways.
• We all know that a synchronous machine is not self-starting. Thus providing damper windings helps synchronous machines act as an induction motor ( only at starting). Which helps the machine to self-start.
• Hunting is a persistent phenomenon when it comes to synchronous machines.
• We can reduce hunting to a great extent by damping it. They don’t let the motor oscillate abruptly, they dampen the oscillations thus increasing the stability of the machine.

Answer.3. Hysteresis loss will Increase but eddy current loss will remain unchanged

Explanation:

• Eddy Current Loss in the transformer is given as

P_e = K_ef²B²m

Where Ke = eddy current constant 

Bm = Maximum flux density and Bm α (V/f)

• For any given voltage, if frequency decreases, Bm increases and if the frequency is increased Bm decreases correspondingly.
• Hence the eddy current loss P_e at any given voltage is independent of frequency.

For the fixed magnitude of applied voltage Hysteresis loss is given as

P_h = K_h V^1.6 f^0.6.

Hence from the above equation, it is clear that with the increase in the frequency Hysteresis losses will increase.

Explanation:

P = VIcosΦ

For motor A

2000 = 1000 × I × 0.8

I = 2.5 A

For motor B

1000 = 1000 × I × 0.5

I = 2 A

Total current drawn 2.5 + 2 = 4.5 A