# SSC JE Electrical Previous Year Question Paper With Solution and Explanation 2018-Set 5 | MES Electrical | SSC JE 2018 | SSC Junior Engineer Exam Paper “Evening shift”

Ques.1. The potential difference is measured in _____

1. Watt
2. Joule
3. Volts
4. Ampere

The difference between the initial electric potential and the final potential measurements is known as potential difference and is measured in joules per coulomb, or volts. Thus, the potential difference is commonly referred to as the voltage.

or

Electric Potential is the measure of electrostatic potential energy per unit charge for a given point in space.

or

The difference in electric charge creates a potential difference or voltage. The size of the voltage is directly proportional to the charge difference.

Ques.2. Which of the following statement is TRUE about the resistance of a conductor?

1. The resistance of a conductor does not depend upon the length
2. The resistance of a conductor does not depend upon the material
3. The resistance of a conductor does not depend upon the temperature
4. The resistance of a conductor does not depend upon the pressure

Factors Affecting the Resistance

The resistance R offered by a conductor depends on the following factors :

1. Length of the material (l): The resistance of a material is directly proportional to the length. The resistance of the longer wire is more.
2. Cross-Section Area (a): The resistance of a material is inversely proportional to the cross-sectional area of the material. The more cross-sectional area allowed the passage of more number of electrons offering less resistance. 3. Nature of Material: As discussed earlier the conductor has a large number of free electrons hence it offers less resistance whereas Inductor has less number of free electrons hence which offers more resistance.
4. Temperature: The temperature of the material affects the value of the resistance. In the General case, the resistance of the material increases as its temperature increases.

Ques.3. In the series combination of resistance, the current through each resistance is

1. Higher in largest resistance
2. Lower in largest resistance
3. Same in each resistance
4. Higher in smaller resistance

In a series circuit, there is only one path in which current can flow. There are no branches through which current can leave, or enter the series loop. Since no additional current can enter or leave the circuit. the current is the same everywhere in a series circuit. This means the current is the same through every load in series

Mathematically

IT = I1 = I2 = I3 ——— = IN

Ques.4. Siemens is the S.I unit of ______

1. Resistance
2. Conductance
3. Capacitance
4. Inductance

Conductance

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 and is the ease with which an electric current passes. Electrical conductance is measured in siemens (S) σ and denoted by (G). Ques.5. Determine the conductance (in Mho) of a conductor, when the value of current that flows through the conductor is 2 A and the potential difference between the ends of the conductor is 40 V.

1. 0.04
2. 0.05
3. 0.62
4. 0.24

Given

Current I = 2A

Voltage V = 40 V

Resistance R = V/I = 40/2 = 20

Hence Conductance

G = 1/R = 1/20

G = 0.05 (Mho)

Ques.6. 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?

1. 0.026
2. 0.032
3. 0.065
4. 0.011

The three capacitor C1, C2, C3 is connected in a series as shown in the figure Now the Equivalent capacitance connected in the series will be

1/Ceq = 1/C1 + 1/C2 + 1/C3

1/Ceq = 1/0.04 + 1/0.08 + 1/0.02

1/Ceq = 25 + 12.5 + 50

1/Ceq = 87.5

Ceq = 1/87.5

Ceq = 0.011

Ques.7. Determine the resistance (in ohms) of a 14 m long circular wire when the diameter and the conductivity of the wire are 0.6 m and 12 mho/meter respectively.

1. 1.79
2. 2.84
3. 3.64
4. 4.17✓

Given

Length =14 m

Radius = Diameter/2 = 0.6/2 =0.3m

Area of circle = πR2 = π × 0.32

Conductivity = 1/Resistivity

Resistivity = 1/12

The formula of resistivity is given as ∴ Resistance = (L × Ρ) ⁄ A

$.\begin{array}{l}R = \dfrac{1}{{12}} \times \dfrac{{14}}{{\pi \times {{0.3}^2}}}\\\\R = 14.17\Omega \end{array}$

Ques.8. What will be the color-coding of a resistor when the resistance of the resistor is 50 + 2% ohms?

1. Green-Black-Brown-Red
2. Green-Black-Black-Brown
3. Yellow-Brown-Black-Red
4. Green-Black-Black-Red

Resistor values can be known by the color band.

0 black

1 brown

2 red

3 orange

4 yellow

5 green

6 blue

7 violet

8 grey

9 white

First color = digit
Second color=digit
Third color= power of ten
Fourth color =tolerance

Now in the given question 50 × 100  ± 2%

5 = Green

0 = Black

100 = Black

± 2% = Red

Ques.9. What will be the value of current (in A) drawn from a 4 V battery when a wire of 20 ohms resistance is stretched to double its original length and then cut into two equal parts and these equal parts are connected in parallel with the battery?

1. 2
2. 4
3. 0.2
4. 0.4

Resistance R = p(L/A)………..(1)

p is the resistivity of the material of wire.

L is the length of wire.

A is the area of cross-section of wire.

We multiply and divide equation (1) by l, then

R=p(L2/LA)

We further know that,

Volume = Length × Cross-Sectional Area

We can assume that the material volume does not change (density of the material not affected)

∴ R ∝ L2.

Then, if we double the length, Resistance becomes R’=4R.

Interpretation can be drawn qualitatively as follows:

Due to an increase in the length of the wire, it will become thinner and longer. If it is thinner it will be harder for the charge to move through, so the resistance will increase. If it is longer, the charge has further to travel, so its resistance has increased.

Now in the given question, the 20Ω is resistance is stretched to double hence the new resistance will be

R’ = 4R = 4 x 20 = 80Ω

Now the resistance 80Ω is cut into two equal parts and are connected in the parallel circuit as shown in the figure R = 40 || 40 = (40 × 40) ⁄ (40 + 40)

R = 20 Ω

Hence current I = V/R = 4/20

I = 0.2 A

Ques.10. Determine the heat dissipated (in Joule) through a conductor of 10 ohms resistance, when 1 A of current is flowing through the conductor for 5 seconds.

1. 50
2. 40
3. 20
4. 60

If current flows for time t then the production of heat is given as

H = I2R.t

Where

I = current = 1 A

R = Resistance = 10 ohms

t = time = 5 sec

H = 12 × 10 × 5

H = 50 watt-sec or Joule

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