SSC JE Basic Electrical Questions (2009 – 2018) Solved | MES Electrical

The bandwidth is related to the resonant frequency by the quality factor:

Q = Resonance frequency ⁄ Bandwidth

But in the given question only the bandwidth is given but not the quality factor hence data is insufficient.

Answer.1.

Explanation

Quality Factor Q

In a practical circuit R is essentially resistance of the coil since practical capacitors have very low loss in comparison to practical inductor. Hence Q is a measure of the energy storage property (LI²) in relation to the energy dissipation property (I²R) of a coil or a circuit. The Q is, therefore, defined as

[latex display=”true”]Q = 2\pi \dfrac{{Maximum{\text{ }}energy{\text{ }}stored}}{{Energy{\text{ }}dissipated{\text{ }}per{\text{ }}cycle}}[/latex]

In electric circuit energy is stored in the form of electromagnetic field in the inductance where as in electrostatic form of energy across a capacitance.

Maximum energy stored = Electromagnetic energy in Inductor or Electrostatic Energy in Capacitor

[latex]=\dfrac{1}{2}L{{\mathop{\rm I}\nolimits} _{\max }}^2{\text{ or }}\dfrac{1}{2}C{V_{\max }}^2[/latex]

We know that average power P

P = I²R = Where I = I_m ⁄ √2 =  1⁄2 I²_mR

Energy dissipated per cycle = P ⁄ f

Therefore,

Where ω = Resonance frequency = 1/√LC

Thus, Q is inversely proportional to R. Hence, for series RLC circuit, a high value of quality factor implies low losses and a low value of Q implies high losses.

Also, the quality factor for a circuit is defined as

[latex display=”true”]Q = \dfrac{{Resonant{\text{ }}frequency}}{{Bandwidth}} = \dfrac{{{\omega _o}}}{{{\omega _1} – {\omega _2}}}[/latex]

The selectivityof the circuit is defined as the reciprocal of quality factor, i.e

[latex display=”true”]selectivity = \dfrac{1}{Q} = \frac{{Bandwidth}}{{Resonant{\text{ }}frequency}} = \dfrac{{{\omega _1} – {\omega _2}}}{{{\omega _o}}}[/latex]

Therefore, a circuit will be highly selective if it has a high value of Q. For a series RLC circuit, a high value of the quality factor implies a narrow resonant peak and a low value of Q implies broad resonant peak. The variations of magnitude and phase angle of current in an RLC series circuit for different values of quality factor (Q).

Answer.4. V_rms = 0.707V_peak

Explanation

The effective or r.m.s. value of alternating current is that steady current (d.c.) which when flowing through a given resistance for a given time produces the same amount of heat as produced by the alternating current when flowing through the same resistance at the same time. It is also called virtual value of a.c. and is represented by I_rms or I_eff or I_v.

For example, when we say that r.m.s. or effective value of an alternating current is 5A, it means that the alternating current will do work (or produce heat) at the same rate as 5A direct current under similar conditions.

To determine r.m.s. value of a sinusoidal alternating current, we have to first square it, then take its mean over one cycle or half cycle and then take the square root (note that r.m.s. value is calculated by making reverse operation, that is, the first square, then take mean and then take square root).

Square of current i = I_m sinθ = I²_m sin²θ

Its mean over one cycle is calculated by integrating it from 0 to 2π and dividing by the time period of 2π.

Mean value of alternating current

[latex]\begin{array}{l}{\text{Mean of square}}\\\\ = 1/2\pi \int\limits_0^{2\pi } {{I^2}_m} {\sin ^2}\theta d\theta \\\\{\text{RMS Value}}\\\\{\rm{I = }}\sqrt {1/2\pi \int\limits_0^{2\pi } {{I^2}_m} {{\sin }^2}\theta d\theta } \\\\{\sin ^2}\theta = \dfrac{{1 – \cos 2\theta }}{2}\\\\{\rm{I = }}\sqrt {\dfrac{{{I^2}_m}}{{4\pi }}\int\limits_0^{2\pi } {(1 – \cos 2\theta } )d\theta } \\\\ = \sqrt {\frac{{{I^2}_m}}{{4\pi }}\left[ {\theta – \dfrac{{\sin 2\theta }}{2}} \right]_0^{2\pi }} \\\\ = \sqrt {\dfrac{{{I^2}_m}}{{4\pi }}} \times 2\pi \\\\ = \sqrt {\dfrac{{{I^2}_m}}{2}} \\\\I = \frac{{{I_m}}}{{\sqrt 2 }} = 0.707\end{array}[/latex]

If we calculate the r.m.s. value for half cycle, it can bc seen that we will get the same value by calculating as

[latex]\begin{array}{l}{\text{Mean of square}}\\\\ = 1/\pi \int\limits_0^\pi {{I^2}_m} {\sin ^2}\theta d\theta \\\\{\text{RMS Value}}\\\\{\rm{I = }}\sqrt {1/\pi \int\limits_0^\pi {{I^2}_m} {{\sin }^2}\theta d\theta } \\\\{\sin ^2}\theta = \dfrac{{1 – \cos 2\theta }}{2}\\\\{\rm{I = }}\sqrt {\frac{{{I^2}_m}}{{2\pi }}\int\limits_0^\pi {(1 – \cos 2\theta } )d\theta } \\\\ = \sqrt {\dfrac{{{I^2}_m}}{{2\pi }}\left[ {\theta – \dfrac{{\sin 2\theta }}{2}} \right]_0^\pi } \\\\ = \sqrt {\dfrac{{{I^2}_m}}{{2\pi }}} \times \pi \\\\ = \sqrt {\dfrac{{{I^2}_m}}{2}} \\\\I = \dfrac{{{I_m}}}{{\sqrt 2 }} = 0.707\end{array}[/latex]

Hence the r.m.s. value or effective value or virtual value of alternating current is 0.707 times the peak value of alternating current for the half-cycle as well as for full cycle.

Answer.40

Explanation

Frequency is inversely proportional to the time period

F = 1/T = 1/(25 × 10⁻³)

F = 40 Hz

Answer.3. 6.37

Explanation

The average value of alternating current (sinusoidal) is zero over one cycle. It is because of the positive area exactly cancels the negative area. However, the half-cycle average value is not zero.

Therefore if an average value of alternating current and voltage is asked, it is understood for the half cycle.

I = I_msinωt

And the average value of an alternating current is given by

I_avg = 2I_m ⁄ π = 0.637I_m

Where

I_m = peak value of alternating current = 10A

∴ I_avg = 0.637 × 10 = 6.37A

I_avg = 6.37A

Answer.2. 10.77

Explanation

In RLC series circuit the total impedance of the series LCR circuit is given as

Z² = R² +  (X_L – X_C)²

where

R = Resistance = 10Ω

X_L is inductive reactance = 6Ω

and X_C is capacitive reactance = 2Ω

Z² = 10² + (6 − 2)²

Z² = 116

Z = 10.77Ω

Answer.4. 5

Explanation

In a delta connected system, the phase current is 1/√3 times the line current.and the phase voltage is equal to the line voltage of the system.

I_PH = Line/√3

I_PH = 8.7/√3 = 5.02 A

I_PH = 5.02 A

Answer.4. 7

Explanation

As explained in question no 32 the quality of the RLC circuit is given as

Q = ω_oL ⁄ R ———— (1)

or

Q = 1 ⁄ ω_oC.R ———–(2)

Now the given quantities are

Quality factor Q = 2

Resonance Frequency = ω_o = 6 rad/Sec

Capacitance C = 4 F

Resistance R =?

Inductance L = ?

Now from equation Number 2, we can find the value of resistance

Q = 1 ⁄ ω_oC.R

2 = 1 ⁄ 6 ×4 × R

R = 1 ⁄ 48 Ω

Now again To find the value of an inductance we will use equation number 2

Q = ω_oL ⁄ R

2 = 6 × L ⁄ 1 ⁄ 48

L = 0.0069 ≅ 0.007

L = 7mH

Answer.1. 50

Explanation

Let the expression be as follows:-

I(t) = I_m sin(ωt ± φ)

Where φ represent the concerned phase-shift

I_m = Maximum value of the current = 10 A or IRMS = I_m/√2 = 10/√2

φ = 60°

Similarly

V(t) = V_m sin(ωt ± φ)

Vm = 20 V or V_RMS = 20/√2

Average power consumed is

P_AVG = V_RMS. I_RMS. Cosφ

P_AVG = (20/√2) × (10/√2) × Cos60°

P_AVG = 200/4 = 50 watt

P_AVG = 50 watts

Answer.4. 17.93

Explanation

The power consumed by the three-phase circuit is

P = 3 V_ph I_ph cosφ_ph

In delta connected system

Line voltage is equal to the phase voltage i.e

V_L = V_PH = 230 V

Phase current I_PH = 30 A

phase angle φ = 30°

∴ P = 3 × 230 × 30 × cos30°

P = 17926.2 watts or 17.93 kW