SSC JE Electrical Previous Year Question Paper 2018-SET 3|MES Electrical

SSC JE Electrical Previous Year Question Paper With Solution and Explanation 2018-Set 3 | MES Electrical | SSC JE 2018 | SSC Junior Engineer Exam Paper “held on 25 Jan 2018”

Ques.1. Which of the following is the dimensional formula for conductance or conductivity?

M^{−1} L^{−2} T^{3} A^{2}✓

M L^{3} T^{−}^{3} A^{−}^{2}

M^{2} L^{2} T^{−}^{3} A^{−}^{2}

M L^{2} T^{3} A^{−}^{2}

The conductance is given as

Conductance = 1 ⁄ Resistance

Firstly consider the resistance.

Ohm’s law states that electric current flowing through the conductor is directly proportional to the potential difference between its two ends when the temperature and other physical parameters of the conductor remain unchanged.

V = IR

⇒ R = VI

Now V has units of (electric field)*(distance).

But the electric field has units (force)/(charge).

Also, charge has dimensions of (current)(time) and force has dimensions (mass)(length)/(time)^{2}.

Thus, dimensions of V is,

[V] = LMLT^{−}^{2} ⁄ AT

⇒ [V] = M L² T ⁻³ A⁻¹

Dimensional formula for I = A

Put these values in equation (1), we get:

The dimensional formula for R = M L^{2} T^{−3} A^{−1}/ A

Dimensional formula for R = [M L^{2} T ^{−3} A^{−2}]

Now the Dimensional formula of conductance will be

G = 1 ⁄ [M L^{2} T ^{−3} A^{−2}]

G = [M^{−1} L^{−2} T^{3} A^{2}]

Ques.2. Which of the following is the CORRECT expression for the capacitance of a parallel plate capacitor?

εA/d✓

εA^{2}/d

εA^{2}d^{2}

A^{2}d^{2}/ε

The capacitance C of a capacitor is the ratio of the magnitude of the charge on either conductor (plate) to the magnitude of the potential c^lif^rerence between the conductors (plates).

C = Q/V

THE PARALLEL-PLATE CAPACITOR

The figure shows the structure of a parallel plate capacitor. It is made up of two parallel plates of the area of cross-section A, kept in an insulating medium and separated from each other by a distance of d metros. Then, if the two plates are applied with a potential, one of the plates will be positively charged and the other negatively charged. Let the charge on the plates be given by Q coulombs.

The capacitance of a device depends on the geometric arrangement of the conductors. The capacitance of a parallel-plate capacitor with plates separated air can be easily calculated from three facts.

The magnitude of the electric field between two plates is given E = σ/ε_{o}, where σ is the magnitude of the charge per unit area on each plate and ε_{ο} = Permittivity of free space = 8.85 × 10^{-12}

Second, the potential difference between the two plates is ΔV = Ed, where d is the distance between the plates.

Third, the charge one plate is given by q = σA, where A is the area of the plate. Substituting the three facts into the definition of capacitance gives the desired result:

Ques.3. In a series combination of several inductors, the equivalent inductance is______

Equal to the largest inductance of the combination

Lower than the largest inductance of the combination

Lower than the smallest inductance of the combination

Greater than the largest inductance of the combination✓

Let us Suppose that the three Resistance L_{1}, L_{2}, L_{3} of 1 Henry, 2 Henry, and 3 Henry are connected in series respectively.

Now from the above result, we can conclude that the equivalent Inductance in the series combination is greater than the largest Inductance in the combination (since the largest Inductance was 3H).

Therefore, the Option.4. is correct.

Ques.4. Which of the following is the reciprocal of the resistivity?

Reluctivity

Susceptibility

Conductivity✓

Permittivity

Resistivity:-

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

Conductivity

The electrical resistance of an electrical conductor is a measure of the difficulty to pass an electric current through that conductor. The inverse quantity is electrical conductance or conductivity and is the ease with which an electric current passes. Electrical conductance is measured in siemens (S) σ and denoted by (G).

The material having high value of conductivity is a good conductor of electricity while the material having the low value of conductivity is a good insulator.

Ques.5. Determine the value of equivalent inductance, if 4 inductors having inductance L are connected in parallel.

4L

L/4✓

4/L

8L

Inductors are sometimes connected in parallel in a circuit. Two or more circuit elements are said to be in parallel when they share the same pair of nodes. In a parallel circuit, the voltage across each of the parallel circuit elements is the same, but the current through each of the elements might be different. Let us consider four inductors that are connected in a parallel to a source voltage V as shown in the Fig. In this cases, the voltage across each of these resistors will be V.

Note:- Inductor connected in parallel combine same way as the resistor in parallel.

According to Ohm’s law, the currents the current I_{1}, I_{2}, I_{3}, I_{4} through the corresponding inductors L_{1}, L_{2}, L_{3} & L_{4} can be written as

I_{1} = V ⁄ L_{1}

I_{2} = V ⁄ L_{2}

I_{3} = V ⁄ L_{3}

I_{4} = V ⁄ L_{4}

Applying KCL to the circuit we get

I = I_{1} + I_{2} + I_{3} + I_{4}

Ques.6. Determine the conductance (in Siemens) of a conductor, when the potential difference between the ends of the conductor is 30 V and the current flowing through the conductor is 3A.

0.1✓

1.1

2.4

4.2

Conductance “G” is measured in Siemens and it is the reciprocal the resistance

G = 1/R = G = 1/V/I

G = I/V

Where

I = current = 3A

V = voltage = 30V

G = 3/30 =0.1 Siemens

Ques.7. How much power will be dissipated by a 10 ohms resistors, when the current through the resistor is 3A?

30

40

60

90✓

Given

Resistance R = 10Ω

Current I = 3 A

Power dissipated by the resistor is

P = I^{2}R

P = 3^{2} × 10

P = 90 watts

Ques.8. Determine the value of resistance (in ohms) of a resistor at 40 degrees Celsius, when the resistance of 10 ohms at 0 degree Celsius and the temperature coefficient at 0 degree Celsius is 0.04.

20

23

24

26✓

Resistance Temperature Coefficient:

The change in resistance of a material with the increase in temperature can be expressed b means of the temperature coefficient of resistance. Consider a conductor having resistance R_{o} at 0°c and R_{t }at t°c. From the above discussion, we can conclude that the change in the resistance i.e (R_{t} – R_{o}) is

Directly proportional to the initial resistance R_{o}

Directly proportional to the rise in temperature t°c.

Depends on the nature of the material for conductor metals and alloy

Hence

(R_{t} – R_{o}) ∝ R_{o}t

(R_{t} – R_{o}) = αR_{o}t

R_{t} = R_{o}(1 + α_{o}t)

Where α_{o} is constant and called as the temperature coefficient of resistance at 0°c and its value depends upon the nature of material and temperature.

R_{o} = 10 ohms

T = 40°

α =.04

R_{T} = ?

Hence the value of resistance at 40°C is

R_{T} = 10(1 + .04 × 40)

R_{T} = 26 Ω

Ques.9. Determine the energy stored (in J) by a 5 H inductor, when the current flowing through the inductor is 6A.

94

90

60

40

The energy stored in the magnetic field of an inductor can be expressed as

W = 1/2 L I^{2}

where

W = energy stored (joules, J)

L = inductance (henrys, H) = 5 H

I = current (amps, A) = 6A

W = (5 × 6^{2}) ⁄ 2

W = 90 Joules

Ques.10. Determine the equivalent resistance (in ohms) for the circuit given below.

2

4✓

6

9

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 two resistance of 1 ohms each share two common colors i.e yellow and green hence this two resistance are in parallel.

1Ω II 1Ω = (1 × 1) ⁄ (1 + 1) = 0.5 Ω

Again redraw the circuit as shown in figure

Now the yellow wire is connected to only the resistance of 0.5 Ω & 1 Ω hence this two resistance are in series,

R = 0.5 + 1 = 1.5 Ω

Now again redrawing the circuit we get

From the above figure, it is clear that the resistance 1.5 Ω and resistance 2 Ω share same two colors i.e green and sky blue therefore this two resistance are connected in the series.

R = 1.5 || 2 = ( 1.5 × 2) ⁄ (1.5 + 2)

R = 6/7 Ω

Again redrawing the circuit we get

The sky blue wire is connected to the only resistance 6/7 Ω and 3Ω hence they both are connected in series, therefore,

R = 6/7 + 3 = 27 ⁄ 7 Ω

Redraw the given circuit

The resistance 4 Ω and the resistance 27/7 Ω is connected to the same wire i.e greenand bluehence this two resistance are connected in parallel

R = 4 || 27/7 = ( 4 × 27/7) ⁄ (4 + 27/7) = 1.96 Ω

Now the final circuit is shown below

Now the green wire is connected to the only resistance of 2 Ω & 1.96 Ω

please give ques no 30 answer