SSC JE Electrical Previous Year Question Paper With Solution and Explanation 2018-Set 1 | SSC JE 2018

Ques.1. The S.I unit of Electric energy is

Watt

Volts

Ampere

Joule✓

Electrical Power

The rate at which the electric work is done in an electric circuit is called an electric power. The SI unit of power is joule per second or watt.

Note:-

Electrical Energy

Electric energy is the total amount of electrical work done in an electrical circuit. Electric energy can also be defined as the product of power and time. The S.I Unit of Electrical- Energy is joule or watt-sec.

Ampereis the Unit of Current

Volt is the S.I unit of Potential Difference.

Ques.2. What is the value of the current I_{4} (in A) for the given junction?

4

−4

6✓

−6

According to Kirchhoff’s Current Law: At any point in an electrical circuit, the sum of currents flowing towards that point is equal to the sum of currents flowing away from that point.

From the above Diagram

Current Flowing towards the Point: I_{2}, I_{6}, I_{4}

Current Flowing Away from the Point: I_{1}, I_{3}, I_{5}

Now from the above result, we can conclude that the equivalent resistance in the series combination is larger than the largest resistance in the combination (since the largest resistance was 3Ω).

Therefore, Option.1. is correct.

Ques.4. The resistivity of a conductor depends upon

Pressure

Temperature✓

Degree of Illumination

Shape of cross-section

The resistance of a material having unit length and the unit cross-sectional area is known as Specific Resistance or Resistivity.

In the S.I. system of units

Hence, the unit of resistivity is ohm-meter (Ω-m).

Specific Resistance depends only on the temperature and material of the conductor but not on the dimensions of the conductor, on which resistance depends, and mechanical deformation such as stretching, etc. As ρ depends only on the material of a conductor at a given temperature, hence it is a characteristic constant.

Material with the highest value of resistivity is the best insulator while the material with the low value of resistivity is a good conductor.

Ques.5. Determine the energy (in J) stored by a 0.4 H inductance, If the current flowing through it is 2A.

1.6

0.8✓

0.4

1.4

Energy in a Magnetic Field of inductance

The magnetic flux of the current in inductance is the electric energy supplied by the voltage source producing the current. The energy is stored in the magnetic field since it can do the work of producing the induced voltage when the flux moves. The amount of electric energy stored is

Energy = ½LI^{2} joule

Given

Inductance H = 0.4 Current I = 2A

E = ½ × 0.4 × 2^{2}

E = 0.8 Joule

h({});

Ques.6. What will be the voltage (in V) across an 8H inductor, when the rate of change of current flowing through it is 0.5A/sec.

2

6

4✓

8

Let “I” be the current through the circuit and (di/dt) be the rate of change of current through the circuit at that instant.

∴ Instantaneous voltage across Inductor L V = L(di/dt)

Given Inductance = 8H Rate of change of current di/dt = 0.5A

V = 8 × 0.5 = 4 V

Instantaneous voltage across Inductor V = 4 V

Ques.7. What will be the equivalent capacitance (in mF) of three capacitors connected in a series having the capacitance of 0.04 mF, 0.08 mF, and 0.02 mF respectively?

0.026

0.032

0.065

0.011✓

The three capacitor C_{1}, C_{2}, C_{3} is connected in a series as shown in the figure

Now the Equivalent capacitance connected in the series will be

1/C_{eq} = 1/C_{1} + 1/C_{2} + 1/C_{3}

1/C_{eq} = 1/0.04 + 1/0.08 + 1/0.02

1/C_{eq} = 25 + 12.5 + 50

1/C_{eq} = 87.5

C_{eq} = 1/87.5

C_{eq} = 0.011

Ques.8. Determine the voltage (in V) of a battery connected to a parallel plate capacitor (filled with air) when the area of the plate is 10 square centimeters, the separation between the plate is 5 mm and the charge stored on the plate is 20 pC.

12.3

10.3

11.3✓

14.3

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 = 10 × 10^{−4} meter ε_{ο} = Permittivity of free space = 8.85 × 10^{-12} d = distance between the parallel plate capacitor = 5mm = 5 × 10^{−3} meter
q = Charge stored in the capacitor = 20 pC = 20 × 10^{−12}

Ques.9. Determine the value of current (in A) through both the resistors of the given circuit.

−2, −1.5

2, 1.5✓

−2, 1.5

2, −1.5

Current through the 10Ω resistance

I_{1} = V/R = 20/10 = 2A

I_{1} = 2A

Now current through the 20Ω resistance

I_{2} = V − (-10)/R = (20 + 10)/20 = 1.5 A

I_{2} = 1.5 A

Ques 10. The relative permeability of the diamagnetic material is

Greater than 1

Greater than 10

Less than 1✓

Greater than 100

Permeability is a measure of how easy it is to establish the flux in a material. Ferromagnetic materials have high permeability and hence low Reluctance, while non-magnetic materials have low permeability and high Reluctance.

Diamagnetic materials are materials with relative permeabilities slightly smaller than 1 (μ_{r} < 1). This class includes important materials such as mercury, gold, silver, copper, lead, silicon, and water. The relative permeability of most diamagnetic materials varies between 0.9999 and 0.99999 (susceptibility varies between (-10^{-5} and -10^{–4}), and for most applications, they may be assumed to be nonmagnetic.

An interesting aspect of diamagnetism is the fact that the magnetic flux density inside the diamagnetic material is lower than the external magnetic field. If we place a piece of diamagnetic material over a permanent magnet, the magnet will repel the diamagnetic material, as shown in Figure.

Since the magnet and the equivalent magnetic field (due to the magnetization of the diamagnet material) oppose each other, the diamagnetic material is always repelled from the magnetic field in the same way that two magnets repel each other when their magnetic flux densities oppose each other. However, this force is extremely small for diamagnetic materials, except superconductors, in which it is very large. This repulsion is the reason why a permanent magnet floats above a superconducting material.