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

Answer.4. Conservation of Energy

Explanation

Tellegen’s theorem is one of the most powerful theorems in network theory. The physical interpretation of Te|legen‘s theorem is the conservation of power. As per the theorem, the sum of powers delivered to or absorbed by all branches of a given lumped network is equal to 0 i.e. the power delivered by the active elements of a network is completely absorbed by the passive elements at each instant of time.

Tellegen’s theorem depends on KCL and KVL but not on the type of the elements. Tellegen theorem can be applied to any network linear or non-linear, active or passive, time-variant or time-invariant.

Answer.2. 30°

Explanation

The phase angle θ of a series RL circuit is given as

θ = tan⁻¹(X_L ⁄ R)

Where

X_L = Inductive Reactance = 37.53

R = Resistance = 65

θ = tan⁻¹(37.53 ⁄ 65)

θ = tan⁻¹(0.57)

θ = 30°

Answer.3. 15.91

Explanation

Given

Inductance L = 0.01 mH = 0.01 × 10⁻³

Capacitance C = 0.01 mF = 0.01 × 10⁻³

The resonance frequency of an L-C tank circuit is given as

[latex]\begin{array}{l}{f_{LC}} = \dfrac{1}{{2\pi \sqrt {LC} }}\\\\ = \dfrac{1}{{2\pi \sqrt {0.01 \times {{10}^{ – 3}} \times 0.01 \times {{10}^{ – 3}}} }}\\\\ = \dfrac{{{{10}^3}}}{{2\pi \times 0.01}}\\\\{f_{LC}} = 15.91kHz\end{array}[/latex]

Answer.1. 7.36

Explanation

Given

Inductance L = 1.8 H

Resistance R = 90 Ohms

Voltage V = 20 V

Time = 20 ms = 20 × 10⁻³

The current in an RL series circuit is given as

[latex display=”true”]{I_{\left( t \right)}} = \dfrac{V}{R}\left( {1 – {e^{ – Rt/L}}} \right)[/latex]

The voltage drop across the inductor is

V_L = L(di/dt)

The Circuit is shown in the figure

[latex]\begin{array}{l}{I_{\left( t \right)}} = \dfrac{V}{R}\left( {1 – {e^{ – Rt/L}}} \right)\\\\{V_L} = L \times \dfrac{{20}}{R}\left( {{e^{ – Rt/L}}} \right)\dfrac{R}{L}\\\\{V_L} = 20\left( {{e^{ – Rt/L}}} \right)\\\\{V_L} = 20\left( {{e^{ – \dfrac{{20 \times {{10}^{ – 3}} \times 90}}{{1.8}}}}} \right)\\\\{V_L} = 20 \times {e^{ – 1}}\\\\{V_L} = 20 \times 0.367\\\\{V_L} = 7.35V\end{array}[/latex]

Answer.1. 12

Explanation

Delta (Δ) to (Y) star conversion: We can convert the delta connection into its equivalent star connection with the following equations:

So, the equivalent the impedance of each branch of the star connection is obtained by the multiplication of the impedances of the two delta branches that meet at its end divided by the sum of three delta impedances.

R_A, R_B, R_C = Product of an adjacent resistor ⁄  Sum of all resistor in delta

Given

R_AB = R_BC = R_CA = 36Ω

Therefore the delta to star conversion will be

[latex]\begin{array}{l}{R_A} = \dfrac{{{R_{AB}} \times {R_{CA}}}}{{{R_{AB}} + {R_{BC}} + {R_{CA}}}}\\\\{R_A} = \dfrac{{36 \times 36}}{{36 + 36 + 36}}\\\\{R_A} = 12\Omega \\\\{R_B} = \dfrac{{{R_{BC}} \times {R_{AB}}}}{{{R_{AB}} + {R_{BC}} + {R_{CA}}}}\\\\{R_B} = \dfrac{{36 \times 36}}{{36 + 36 + 36}}\\\\{R_B} = 12\Omega \\\\{R_C} = \dfrac{{{R_{CA}} \times {R_{BC}}}}{{{R_{AB}} + {R_{BC}} + {R_{CA}}}}\\\\{R_C} = \dfrac{{36 \times 36}}{{36 + 36 + 36}}\\\\{R_C} = 12\Omega \end{array}[/latex]

So the value of per phase resistor in the equivalent star connection is 12 ohms.

Answer.2. 0.78

Explanation

Power consumption in an AC circuit is given as

P = VI cosφ

Given

Impedance Z = 16Ω

Current I = 4A

Power = 200 W

Power factor cosφ = ?

Voltage V = I × Z = 16 × 4 = 64 V

200  = 64 × 4 × cosφ

cosφ = 0.78

Power Factor = 0.78

Answer.3. 30

Explanation

The power triangle or Impedance triangle of the AC circuit is shown Below

Cosφ = Base ⁄ Hypotenuse = Active Power  ⁄ Apparent Power
Active Power = Apparent power × Cosφ
Sinφ = Perpendicular ⁄ Hypotenuse = Reactive Power  ⁄ Apparent Power

Reactive Power = Apparent power × Sinφ

or

Reactive Power = (Active Power × Sinφ) ⁄ Cosφ

cosφ = 0.8

Sin²φ = 1 − cos²φ

Sin²φ = 1 − (0.8)²

Sinφ = 0.6

Reactive Power =(40 × 0.6) ⁄ 0.8

Reactive Power = 30 VAR

Answer.2. Insulator

Explanation

When an electric field is applied to a conductor, there occurs a large scale physical movement of free electrons because these are available in large numbers in Conductor.

On the other hand, if an electric field is applied to an insulator, there is hardly any movement of free electrons because these are just not available in an insulator. Plastics, wood, and rubber are examples of good insulators. Pure water is also an insulator. Tap water, however, contains salts that form ions which can move through the liquid, making it a good conductor.

The insulator is also called the dielectric. There are practically no free electrons in the dielectric. The electrons in dielectric normally remain bounds to their respective molecules.

There are some materials, called semiconductors, which are intermediate between conductors and insulators.

Superconducting materials are the materials that conduct electricity without resistance below a certain temperature. Superconductivity is one of the most exciting phenomena in Physics, because of the peculiar nature and the wide application of this phenomenon. This phenomenon of superconductivity was first discovered by a Dutch physicist, H.K. Onnes. Superconducting materials are having very good electrical and magnetic properties. Before the discovery of superconductors, it is believed that the electrical resistivity of the material becomes zero, only at the absolute temperature.

Answer.3. 20

Explanation

Given

Resistance R = 5Ω

Current I = 2 A

Power dissipated by the resistor is

P = I²R

P = 2² × 5

P = 20 watts