EE MCQ

SSC JE Electrical 2018 with solution SET-2
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Which of the following material does not allow the current to flow in it?

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 which 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.

Related Question

SSC JE Electrical 2018 with solution SET-2
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Which of the following law is based on the conservation of energy?

Kirchhoff’s Voltage Law (KVL,) or Kirchhoff’s Loop Rule. This law is based on the conservation of energy and may be stated as under:

In any closed electrical circuit or loop, the algebraic sum of all the electromotive force (e.m.f s) and voltage drops in resistors is equal to zero, i.e., in any closed circuit or loop.

The algebraic sum of e.m.f s + Algebraic sum of voltage drops = 0

The validity of Kirchhoff’s voltage law can be easily established by referring to the loop ABCDA shown in Fig.

If we start from any point (say point A) in this closed circuit and go back to this point (i.e., point A) after going around the circuit, then there is no increase or decrease in potential. This means that algebraic sum of the e.m.f.s of all the sources (here only one e.m.f. source is considered) met on the way plus the algebraic sum of the voltage drops in the resistances must be zero. Kirchhoff’s voltage law is based on the law of conservation of energy, i.e., the net change in the energy of a charge alter completing the closed path is zero.

V1 + V2 − V = 0

or

Kirchhoff’s voltage law is also called as loop rule.

SSC JE Electrical 2018 with solution SET-2
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How much power (in W) will be dissipated by a 5 Ohm resistor in which the value of current is 2 A?

Given

Resistance R = 5Ω

Current I = 2 A

Power dissipated by the resistor is

P = I2R

P = 22 × 5

P = 20 watts

SSC JE Electrical 2018 with solution SET-2
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What will be the resistance (in ohms) of bulb A for the circuit given below?

The power can be defined as

P = I2R

Let the Power dissipated by Bulb A be

P = I2RA = 100 = I2RA

And power dissipated by Bulb B be

P = I2RB = 10 = I2RB

As we Know that the current flow in the series circuit is same

RA ⁄ 100 = RB ⁄ 10

RA ⁄ RB = 10

or

RA = 10RB

Power can also be defined as the

P = V2/R

Total Power consumption = 100 Watt + 10 Watt = 110 Watt

Applied voltage = 20 Volt

∴ 110 = 202 ⁄ R

or

R = 202 ⁄ 110 = 40 ⁄ 11

Now in series connection, the equivalent resistance is the sum of the individual resistance

∴ R = RA + RB

40 ⁄ 11 = 10RB + RB

RB = 0.33 Ω

Hence RA = 10RB

= 10 × 0.33 =3.3

RA = 3.3Ω

SSC JE Electrical 2018 with solution SET-2
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Determine the Norton’s current (in A) and Norton’s resistance (in ohms) respectively, for the given electrical circuit across the load resistance RL.

Determine the resistance RN of the network as seen from the network terminals. (Its value is the same as that of Rth).

RN = (4Ω || 8Ω) + 5Ω = (4 × 8)/(4 + 8) + 5

RN = 7.66

The value I for the current used in Norton’s Theorem is found by determining the open circuit voltage at the terminals AB and dividing it by the Norton resistance r.

According to voltage Division Rule

VAB = V1R3 ⁄ (R1 + R3)

= 24 × 8 ⁄ (4 + 8)

VAB = 16 V

Now Norton Current IN is

IN = VAB ⁄ RN

IN = 16 ⁄ 7.66

IN = 2.08

SSC JE Electrical 2018 with solution SET-2
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Determine the value of current (in A) through both the resistor of the given circuit

Current through the 10Ω resistance

I1 = V/R = 20/10 = 2A

I1 = 2A

Now current through the 20Ω resistance

I2 = V − (-10)/R = 20 + 10/30 = 1.5 A

I2 = 1.5 A