SSC JE Electrical Previous Year Question Paper With Solution and Explanation 2018-Set 4 | MES Electrical | SSC JE 2018 | SSC Junior Engineer Exam Paper “Afternoon Shift “

Ques.1. Which property of an electrical conductor which opposes the flow of current through it?

Conductance

Resistance✓

Reluctance

Permeance

Electrical Resistance:– The property of a conductor which opposes the flow of electric current through it is known as resistance.

R = V/I

Conductance:- The property of a conductor which conducts the flow of current through it is called conductance. In other words, conductance is the reciprocal of resistance. Its symbol is G (G= I /R) and its unit is mho represented by ℧.

Reluctance:- Flux in a magnetic circuit also depends on the opposition that the circuit presents to it. Reluctance (R_{M}), is the opposition a magnetic circuit offers to the formation of magnetic flux.

Permeance:- The reciprocal of reluctance is called permeance and is a measure of the readiness with which the magnetic flux is developed. it is analogous to conductance in an electric circuit and is measured by Weber/ampere-turn.

Permeance = 1/Reluctance

∴ Permeance = flux ⁄ m.m.f

Ques.2. Ohm is the S.I unit of____

Capacitance

Inductance

Resistance✓

Conductance

Ohm is the S.I unit of electric resistance.

Ques.3. What will be the resistance (in Ohms) of a resistor, when the dissipated power and the current flowing through the resistor is 12 W and 2 A respectively?

6

24

2

3✓

Given

Resistance P = 12 W

Current I = 2 A

Power dissipated by the resistor is

P = I^{2}R

12 = 2^{2} × R

R = 3 Ω

Ques.4. What will be the resistance (in Ohms) of a cylindrical wire, if the length, diameter and the resistivity of the wire is 4 m, 0.2 m, and 0.4 Ohms-m respectively?

49

56

54

51✓

The resistivity of the conductor is defined as

Where A = cross-section area of the conductor = πr^{2} = π × 0.1^{2}

R = Resistance of the conductor = ?

L = Length of the conductor = 4 m

ρ = Resistivity = 0.4 Ω-m

∴ 0.4 = π × 0.1^{2 }× R ⁄ 4

R = 50.92 ≅ 51 Ω

Ques.5. Which of the following material has the highest resistivity?

Insulator✓

Semiconductor

Conductor

Superconductor

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. The resistivity of the conductor is 10^{−8} to 10^{−6}.

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. The resistivity of the insulator is 10^{7} to 10^{23}.

Note:-

There are some materials, called semiconductors, which are intermediate between conductors and insulators. The resistivity of the conductor is 10^{−6} to 10^{7}.

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. Every superconductor has zero resistivity, i.e. infinite conductivity, for a small-amplitude of dc current at any temperature below T_{c}

Ques.6. What will be the equivalent inductance (in mH) for the circuit given below?

40

50✓

30

60

To identify whether the Inductance is connected in series or in parallel consider the following method

Use one color for each continuous wire

DO NOT cross any circuit elements

Any element that shares the same two colors are in parallel

As you can see from the above figure the three inductance L_{1}, L_{2}, L_{3} each share two common colors i.e Red and Purple hence this three inductance are connected in parallel.

L_{1} || L_{2} || L_{3} = 20 || 30 || 50

= ( 20 × 30) ⁄ (20 + 30) || 50

= 12 || 50 = ( 12 × 50) ⁄ (12 + 50)

L = 9.67

Now the inductance L and L_{4} are connected in series

Leq = L + L_{4} = 9.67 + 40 = 49.68 ≅ 50 mH

Ques.7. Determine the power (in W) of a lamp of 220 V, when the resistance of the lamp is 100 Ohms.

448

446

484✓

488

Given

Voltage V = 220 V

Resistance R = 100 ohms

Power consumed by the lamp

P = V^{2}/R = 220^{2}/100

P = 484 Watt

Ques.8. What will be the voltage (in V) of a battery connected to a parallel plate capacitor with air as the dielectric, having plate area of 6 sq. cm and separation between the plates are 2 mm, which stores a charge of 8.0 pC on the plates?

3✓

9

6

5

Consider the parallel plate capacitor connected to the battery as shown in the figure

The capacitance “C” of the parallel plate capacitor is given as

C = ε_{ο}A/d

Where A = Area = 10 square centimeter = 6 × 10^{−4} meter ε_{ο} = Permittivity of free space = 8.85 × 10^{-12} d = distance between the parallel plate capacitor = 2mm = 2 × 10^{−3} meter
q = Charge stored in the capacitor = 8 pC = 8 × 10^{−12}

C = (8.85 × 10^{-12} × 6 × 10^{−4}) ⁄ 2 × 10^{−3}

C = 26.55 × 10^{−13 }F

The self-capacitance of a conductor is defined by

C = q ⁄ V

Hence the voltage across the capacitor

V = q ⁄ C = (8 × 10^{−12}) ⁄ (26.55 × 10^{−13})

V = 3.013 Volts

Ques.9. What will be the value of current in the given circuit?

8

4✓

2

6

In the given circuit the resistor of 8Ω & 8Ω are connected in parallel with each other

R_{eq} = (8 × 8) ⁄ (8 + 8)

R_{eq} = 4 Ω

Now all the three resistance i.e 4Ω, 4Ω, & 2Ω are in series therefore there equivalent resistance will be

R_{eq} = 4 + 4 + 2 = 10Ω

Now the value of current in the circuit

I = V/R = 40 ⁄ 10 = 4 A

Ques.10. Find the power delivered (in W) by the current source in the circuit diagram given below?

300

400✓

500

600

To identify whether the resistance is connected in series or in parallel consider the following method

Use one color for each continuous wire

DO NOT cross any circuit elements

Any element that shares the same two colors are in parallel

Now the circuit will look as shown below

As you can see from the above figure the resistance R_{1} & R_{2} share two common colors i.e Green and Blue hence this two resistance are in parallel.

Similarly, the resistance R_{3} & R_{4} share two common colors i.e Red and Blue hence this two resistance are in parallel.

Therefore equivalent resistance for R_{1} and R_{2} is

R = (4 × 6) ⁄ (4 + 6) = 2.4 Ω

Similarly, the equivalent resistance for R_{3} and R_{4} is

R = (8 × 2) ⁄ (8 + 2) = 1.6 Ω

Now the final circuit will look like this

Since the resistance don’t share a common node hence they are connected in series

R_{eq} = 2.4 + 1.6 = 4Ω

Now the power delivered in the circuit is given as

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