SSC JE Basic Electrical Questions (2009 – 2018) Solved (Part-3)

Answer.1. Negative twice, zero four times

Explanation:

• Instantaneous power is the power of the object at any instant of time, in general, it is defined as p(t) = v(t) x  i(t)
• In a single-phase ac circuit, the instantaneous power to a load is of pulsating nature. Even at unity power factor, the instantaneous power is always less than unity
• For a sinusoidal signal,  the voltage swings through zero volts twice a cycle, so the instantaneous power is zero two times per cycle.
• If the load has both resistance and reactance, then the current, and voltage swings through zero twice a cycle, with different phase angle and thus the instantaneous power will swing through zero 4 times a cycle.
• Instantaneous power is positive for one part of the cycle and negative for another part of the cycle if there is both resistance and reactance in the circuit. then there will be one negative for voltage and one negative for current thus instantaneous power will have two negative per cycle for single-phase R-L series circuit.

Answer.1. 1/R√L/C

Explanation:

• Q factor is a parameter that describes the resonance behavior of an underdamped harmonic oscillator (resonator).
• In an ideal series RLC circuit, and in a tuned radio frequency receiver (TRF) the Q factor is: 1/R√L/C
• The larger the series resistance, the lower the Q factor.

Answer.2. 6400 W, 800 VAR capacitive

Explanation:

Active power = 200 x 30 + 40 x10 =6400 W

Reactive Power = 200 x 10 – 40 x 30 = 800 VAR 

Answer.1. A current source and impedance in parallel

Explanation:

Norton’s theorem illustrates that a network consists of  several voltage sources, current sources and resistors with two terminals, which is electrically equivalent to an ideal current source ” I_NO” and a single parallel  resistor, R_NO

Answer 4. 14 Ω

From the circuit diagram 

e = V – 10i

r = e/i 

= V/i -10

On differentiating

dr = dv/di -10

= 12/0.5 – 10 = 14 Ω

Answer.1. 50 Ω at a power factor of 0.6 lead

Explanation:

For maximum power transfer

Z_L = Zs* = 30 – j40

Apparent power

Zs² = 30² + 40²

Zs = 50 Ω

Power factor = Real power/ Apparent power

CosΦ = 30/50 = 0.6 leading

 A leading power factor signifies that the load is capacitive, as the load “supplies” reactive power, and therefore the reactive component  Q is negative as reactive power is being supplied to the circuit.

Answer.3. Power consumed will be 1/3 rd of the previous value

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

Answer.1. 110 V

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

Answer.2. 20 mH

Explanation:

For equivalent star connection

Z_Y = 1/3 Z_Δ

= 60/3

= 20 mH