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

Total capacitance between point L and N

=2 + 1 = 3μF

Now total capacitance between L and O

= 1 + (3 x 2)/(3 + 2) = 2.2μF

Total capacitance Between L and M

= 1 + (2.2 x 2)/(2.2 + 2) = 2.05μF

The rate of change of flux linkage is directly proportional to the induced EMF in a coil. Hence, the EMF is maximum when the rate of change of flux linkage is maximum.

Time constant of RL series circuit is given by

τ = L/R

τ = 2/20 = 0.1 sec

When the rotor of a three-phase induction motor is blocked then the rotor speed “Nr” = 0

and

s = Ns – Nr/Ns

= Ns – 0/Ns = 1

For symmetrical fault

If = E/(Zi +Zn)

Where E = Pre fault voltage Which is equal to 1

Zi = 0.5j & Zn = 0.1 j

If = 1/(0.5j + 0.1j)

If = -j 1.67

Current Chopping in the circuit breaker is defined as a phenomenon in which current is forcibly interrupted before the natural current zero. Current Chopping is mainly observed in Vacuum Circuit breakers and Air Blast Circuit breakers.

For average values of load current, current chopping occurs more frequently in the Air circuit breaker.

In an air blast breaker, a high-velocity jet of air is made to flow through the arcing contacts that extinguish the arc much before natural ac current 0 is reached, the air blast breaker retains the same extinguishing power irrespective of the nature of the fault, consequently, the blast of air sweeps out the ionized particles in between the contacts thereby building the resistance and hence ultimately extinguishing the arc.

In the saturation mode, both the junctions of the transistor (emitter to base and collector to base) are forward biased. In other words, if we assume two p-n junctions as two p-n junction diodes, both the diodes are forward biased in saturation mode.

Mutual inductance

M = K√L1L2

For mutual inductance between two unity coupled coils, K =1

M = √L1L2 = √9 x 4 = 6H

I = V/R = 120/(100 + 50) = 120/150 = 4/5 A

V_CB = 50 x 4/5 = 40 V

Now voltage at ‘c’ w.r.t to ground = Vc

V_CB = Vc – Vb

Vc = V_CB – V_B

= 40 + 0 = 40 V

Maximum Power Transfer theorem

The maximum power is transferred when the load resistance RL is equal to internal resistance Rth or the equivalent resistance Rth.

RL = Rth

\({I_L} = \frac{{{V_{th}}}}{{{R_{th}} + {R_L}}}\)

\({P_L} = Load\;power = I_L^2{R_L}\)

\({P_L} = {\left( {\frac{{{V_{th}}}}{{{R_{th}} + {R_L}}}} \right)^2} \times {R_L}\)    –(1)

\({P_i} = Input\;power = {V_{th}} \times {I_L}\)

\({P_i} = {V_{th}}\left( {\frac{{{V_{th}}}}{{{R_{th}} + {R_L}}}} \right)\)    –(2)

For RL = Rth,

Equation-(1) becomes:

\({P_L} = {\left( {\frac{{{V_{th}}}}{{{R_{th}} + {R_{th}}}}} \right)^2} \times {R_L}\)

\(= \frac{{V_{th}^2}}{{4{R_{th}}}}\)

Similarly, Equation-(2) becomes:

\({P_i} = {V_{th}}\left( {\frac{{{V_{th}}}}{{{R_{th}} + {R_{th}}}}} \right)\)

\(= \frac{{V_{th}^2}}{{2{R_{th}}}}\)

Power Transfer Efficiency will be:

\(\frac{{{P_L}}}{{{P_i}}} = \frac{{V_{th}^2 \times 2{R_L}}}{{4{R_L} \times V_{th}^2}}\)

\(= \frac{1}{2} = 0.5 = 50\%\)

Under this condition, the same amount of power is dissipated in the internal resistance Rth and hence the efficiency is 50%.