# 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?

1. M−1 L−2 T3 A2
2. M L3 T3 A2
3. M2 L2 T3 A2
4. M L2 T3 A2

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.

IR

⇒ 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,

[VLMLT2 ⁄ 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 L2 T−3 A−1/ A

Dimensional formula for R = [M L2 T −3 A−2]

Now the Dimensional formula of conductance will be

G = 1 ⁄ [M L2 T −3 A−2]

G = [M−1 L−2 T3 A2]

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

1. εA/d
2. εA2/d
3. εA2d2
4. A2d2

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/εo, where σ is the magnitude of the charge per unit area on each plate and εο = 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:

$\begin{array}{l}C = \dfrac{Q}{{\Delta V}} = \frac{{\sigma A}}{{Ed}} = \dfrac{{\sigma A}}{{\left( {\sigma /{\varepsilon _o}} \right)d}}\\\\C = {\varepsilon _o}\dfrac{A}{d}\end{array}$

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

1. Equal to the largest inductance of the combination
2. Lower than the largest inductance of the combination
3. Lower than the smallest inductance of the combination
4. Greater than the largest inductance of the combination

Let us Suppose that the three Resistance L1, L2, L3 of 1 Henry, 2 Henry, and 3 Henry are connected in series respectively. Now add all the inductance of the circuit we get

Leq = L1 + L2 + L3
Leq = 1 + 2 + 3 = 5H

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, Option.4. is correct.

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

1. Reluctivity
2. Susceptibility
3. Conductivity
4. 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 a high value of conductivity is a good conductor of electricity while the material having a 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.

1. 4L
2. L/4
3. 4/L
4. 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 parallel to a source voltage V as shown in Fig. In these 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 I1, I2, I3, I4 through the corresponding inductors L1, L2, L3 & L4 can be written as

I1 = V ⁄ L1

I2 = V ⁄ L2

I3 = V ⁄ L3

I4 = V ⁄ L4

Applying KCL to the circuit we get

I = I1 + I2 + I3 + I4

$\begin{array}{l}I = \dfrac{V}{{{L_1}}} + \dfrac{V}{{{L_2}}} + \dfrac{V}{{{L_3}}} + \dfrac{V}{{{L_4}}}\\\\I = V\left( {\dfrac{1}{{{L_1}}} + \dfrac{1}{{{L_2}}} + \dfrac{1}{{{L_3}}} + \dfrac{1}{{{L_4}}}} \right)\\\\{L_1} + {L_2} + {L_3} + {L_4} = L\\\\\dfrac{I}{V} = \left( {\dfrac{{1 + 1 + 1 + 1}}{L}} \right)\\\\R = \dfrac{L}{4}\end{array}$

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.

1. 0.1
2. 1.1
3. 2.4
4. 4.2

Conductance “G” is measured in Siemens and it is the reciprocal of 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 resistor, when the current through the resistor is 3A?

1. 30
2. 40
3. 60
4. 90

Given

Resistance R = 10Ω

Current I = 3 A

Power dissipated by the resistor is

P = I2R

P = 32 × 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.

1. 20
2. 23
3. 24
4. 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 Ro at 0°c and Rat t°c. From the above discussion, we can conclude that the change in the resistance i.e (Rt – Ro) is

1. Directly proportional to the initial resistance Ro
2. Directly proportional to the rise in temperature t°c.
3. Depends on the nature of the material for conductor metals and alloy

Hence

(Rt – Ro) ∝ Rot

(Rt – Ro) = αRot

Rt  = Ro(1 + αot)

Where αo is constant and called the temperature coefficient of resistance at 0°c and its value depends upon the nature of material and temperature.

Ro = 10 ohms

T = 40°

α =.04

RT = ?

Hence the value of resistance at 40°C is

RT = 10(1 + .04 × 40)

RT = 26 Ω

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

1. 94
2. 90
3. 60
4. 40

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

W = 1/2 L I2

where

W = energy stored (joules, J)

L = inductance (henrys, H) = 5 H

I = current (amps, A) = 6A

W = (5 × 62) ⁄ 2

W = 90 Joules

Ques.10. Determine the equivalent resistance (in ohms) for the circuit given below. 1. 2
2. 4
3. 6
4. 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 ohm each share two common colors i.e yellow and green hence this two resistance are 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 these 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 the same two colors i.e green and sky blue, therefore, these 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 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 green and blue hence these 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

R = 2 + 1.96 = 3.96 ≅

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