SSC JE 2013 electrical question paper with solution

Explanation:

Let us suppose Vs be the supply voltage per phase.

So the line voltage of the supply will be 3Vs.

Now assume any type of load; for simplicity, let’s assume it a resistive load which is R per phase

For Delta connected load:

Calculation of per phase power; P_D= I²R

Where I is load current (per phase)
And,

I = √3Vs/R {as line voltage of the supply is directly applied to the phase of the delta load}

So,

Pd = (√3Vs/R )²R = 3Vs²/R watts per phase.

For 3 phases:

P_3D = 3Pd = 3*3Vs²/R = 9Vs²/R watts.

Now for Star connected load:

P_S = I²R = (Vs/R)²R = Vs²/R watts

For 3 phases: P_3S = 3PS = 3 Vs²/R watts

Conclusion:

P_3S / P_3D = 3Vs²/R / 9Vs²/R = 1/3

i.e The power ratio between Star to Delta is 1:3

Explanation:

The average value of the second term is zero therefore answer is 110 V

Answer.3. 45 μH

Explanation:

The diagram of the above question can be constructed as

Explanation:

• The universal motor is so named because it is a type of electric motor that can operate on AC or DC power.
• It is a commutated series-wound motor where the stator’s field coils are connected in series with the rotor windings through a commutator.
• It is often referred to as an AC series motor. The universal motor is very similar to a DC series motor in construction but is modified slightly to allow the motor to operate properly on AC power.
• This type of electric motor can operate well on AC because the current in both the field coils and the armature (and the resultant magnetic fields) will alternate (reverse polarity) synchronously with the supply.

Explanation:

If the centrifugal switch is failed to open then the starting winding will draw very high current and it may burn the winding

Explanation:

• In an alternating current electric power system, synchronization is the process of matching the speed and frequency of a generator or other source to a running network. An AC generator cannot deliver power to an electrical grid unless it is running at the same frequency as the network. If two segments of a grid are disconnected, they cannot exchange AC power again until they are brought back into exact synchronization.
• Hence alternators are usually designed with the fixed frequency of AC voltage.

Explanation:

For equivalent star connection

Z_Y = 1/3 Z_Δ

= 60/3

= 20 mH

Explanation:
For a quarter part of wave pulse energy going down to the maximum value to minimum value. i,e
T/4 = 5ms
T =20ms
f = 1/T  =1/20ms = 50Hz

Answer.3. 2 A in parallel with 75 Ω

Explanation:

Voltage = 150V

R = 75 Ω

I = V/R = 2A

• If we remove the load resistance R_L then it becomes an open circuit. In this case, the terminal voltage between points A and B is V.
• In the next figure, if we remove the load resistance R_L then all the current goes through the internal resistance Ri. The terminal voltage between A and B is the same as the internal resistance Ri voltage for the open circuit.
• The voltage across the internal resistance R is V = I R. As open-circuit voltage between two circuits is electrically equal i.e V. So the circuits are equivalent.
• Hence the equivalent current source is 2 A in parallel with 75 Ω