# SSC JE Basic Electrical Questions (2009 – 2018) Solved (Part-2)

Ques.130. After closing the switch ‘s’ at t = 0, the current i(t) at any instant ‘t’ in the network shown in the figure: (SSC-2014, M-Shift)

1. 10 – 10 e-100t
2. 10 + 10 e100t
3. 10 – 10 e100t
4. 10 + 10 e-100t

Explanation

Time constant τ  in series LR circuit is

τ =L/R = 0.01/1 = 0.01

Under steady state condition, L is short-circuited then

Io = V/R =10A

Now the instantaneous current i(t) is

Ques.131. The voltage across 5-H inductor is

Find the energy stored at t = 5 s. Assume zero initial current (SSC-2014, M-Shift)

1. 312.5 kJ
2. 0.625 kJ
3. 3.125 kJ
4. 156 .25 kJ

Explanation:-

Energy stored in inductor “E” = 1/2(LI2)

Voltage across inductor E(l) = L di(t)/dt

or  i(t) = 1/L di(t)/dt

= 30/5 x 125/3 = 250A

Energy stored in an inductor = (5 x 2502)/2 = 156.25 kJ

Ques.132. Find the current through 5Ω resistor: (SSC-2014, M-Shift)

1. 3.5 A
2. 7.15 A
3. 5 A
4. 2.85 A

Explanation:-

By applying current division Rule

Ques.133. An isolator is used in series with air blast circuit Breaker employed at UHV lines because (SSC-2014, M-Shift)

1. CB life is enhanced with the use of isolator
2. Current to be interrupted will be large
3. Gap between CB contacts is small so an isolator is used to switch off voltage
4. Gap between CB poles is small

Answer.3. Gap between CB contacts is small so an isolator is used to switch off voltage

Explanation:-

An air-blast circuit breaker is small in size, because of the growth of dielectric strength is so rapid (which final contact gap needed for arc extinction is very small) so isolator is used to switch off the voltage.

Ques.134. Regulation of an alternator supplying resistive or inductive load is (SSC-2014, M-Shift)

1. Infinity
2. Always Negative
3. Always Positive
4. Zero

Explanation:-

The voltage regulation of an alternator is defined as the change in its terminal voltage when the full load is removed, keeping field excitation and speed constant, to the rated terminal voltage.

Where Vph = Rated terminal voltage

An increase in the load current in a pure resistive load causes a decrease in the output voltage. For an inductive load, an increase in the load current causes a greater voltage drop as compared to the resistive load. Therefore for inductive and resistive load conditions, there is always a drop in the terminal voltage hence regulation values are always positive.

In the case of leading load that means capacitive load, the effect of armature flux on main field flux is magnetizing i.e, the armature flux is adding up with the main field flux. Since it is adding up, the total induced emf(Vph) will also be more than the induced emf at no load(Eph).Hence the regulation is negative

Ques.135.  If a 10 μF capacitor is connected to a voltage source with v(t) 50sin2000tV, then the current through the capacitor is _________  (SSC-2014, M-Shift)

1. 106 cos 2000t
2. 5 x 10-4 cos 2000 t
3. cos 2000t
4. 500 cos 2000 t

Explanation:-

The current flowing through the capacitor is given by

I(t) =Cdv(t)/dt

Where I(t) is instantaneous current

dv/dt =instanaeous rate of voltage changed

Ques.136. In a series resonance circuit, the impedance at half power frequencies is: (SSC-2014, M-Shift)

1. 2R
2. R/√2
3. √2R
4. R/2

Explanation:-

In series resonance circuit  Z = R = V/Im

Half power frequency is the frequency when the magnitude of voltage or current is decreased by the factor of 1/√2 from its maximum value. (Also known as cut-off frequency)

Therefore at half power frequency

I = Im/√2

Now impedance at half power frequency

= V/Im/√2 = √2V/Im =√2R

Ques.137. A circuit with a resistor, inductor, and capacitor in series is resonant of fo Hz. If all the component values are now doubled the new resonant frequency is (SSC-2014, M-Shift)

1. fo/4
2. 2fo
3. fo
4. fo/2

Explanation:-

The resonant frequency of LC is:

fo = 1/2π√LC

New resonant frequency

fnew = 1/2π√2L2C

fnew = f0/2

Ques.138. The rated voltage of a 3-phase power system is given as: (SSC-2014, M-Shift)

1. Peak line to line voltage
2. RMS phase voltage
3. Peak Phase Voltage
4. RMS line to line voltage

Answer.4. RMS line to line voltage

Explanation:-

In the AC system, we can’t take the average value because the average value is equal to zero over a period in AC.

So now to calculate different parameters like voltage, current, etc we need RMS value. This is the reason why we use RMS quantities while dealing with AC.

Now In a 3-phase supply, there exists 3 lines and a common neutral line. The voltage between two lines (for example ‘L1‘ and ‘L2‘) is called the line to line (or phase to phase) voltage.

The line to line voltage is the vector sum of the phase to phase voltage across each winding. This is not the same as the arithmetic sum and is given by the following equation VLL = √3 VLN

Ques.139. For a half-wave rectified sine wave the ripple factor is: (SSC-2014, M-Shift)

1. 1.00
2. 1.65
3. 1.45
4. 1.21

Explanation:-

Ripple Factor (r): It is the ratio of root mean square (RMS) value of the AC component to the DC component in the output and is given by

For half-wave rectifier

Irms = Im/2

Idc = Im/π

For the half-wave rectifier, RMS Current is given as Im/2 and DC current Im/π which results in the ripple factor of 1.21

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