SSC JE Electrical Previous Year Question Paper 2018-SET 6|MES Electrical

SSC JE Electrical Previous Year Question Paper With Solution and Explanation 2018-Set 6 | MES Electrical | SSC JE 2018 | SSC Junior Engineer Exam Paper “Morning shift”

Ques.1. Why are the same types of cells connected in parallel?

  1. To decrease the voltage rating
  2. To increase the voltage rating
  3. To decrease the current rating
  4. To increase the current rating

As we know that in the series circuit the current remain the same and in the parallel circuit the voltage remains the same.

WHEN THE CELL ARE CONNECTED IN SERIES

Now consider the series circuit in which all the cell are connected in series. If there is “n” number of the cell than the total EMF is nE.

Now all the battery has total internal resistance therefore if n cell is connected in series that the total internal resistance will be “Rn”

Now total current in the circuit is I = nE/nR = E/R

WHEN THE CELL ARE CONNECTED IN PARALLEL

When n number of cell are connected in parallel then total EMF is equal to E (since in parallel circuit voltage remain the same).

Internal resistance in parallel = R’ = R/n

Total current in parallel circuit = I = nE/R

Now consider 3 cell each of 2V having the internal resistance of 0.1 ohm

Current in series = I = E/R = 2/0.1 = 20A

Current in parallel = I = nE/R = 60A

Hence when the individual cells are connected in parallel, the voltage is the same as that across one cell. However, the current rating of the combination equals the sum of the individual current-rating values. Only cells that have the same voltage should be connected in parallel. Whereas when the cells are connected in a series-aiding manner. The current rating of the series-aiding cells is the same as that for the cell with the lowest current rating.

 

Ques.2. There are N resistances, each are connected in parallel having value R with the equivalent resistance of X. What will be the total resistance when these N resistances are connected in series?

  1. NX
  2. RNX
  3. X/N
  4. N2X

Let the equivalent resistance when they are connected in parallel be RP

1/RP = 1/R + 1/R ———– N times

1/RP = N/R

RP = R/N

Since the equivalent resistance, RP is given as X

∴ X = R/N

or

R = NX

Now the equivalent resistance when they are connected in series be Rs

Rs = R + R———- N times

Rs = NR

Rs = N2X-————————–(since R = NX)

Hence the total resistance when these N resistances are connected in series is Rs = N2X

 

Ques.3. Which of the following is equivalent to 0.5 kWh?

  1. 1800000 W
  2. 1800000 J
  3. 18000000 J
  4. 36000000 J

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.

Power

∴ 1 watt is defined as 1 joule/second.  i.e
1 watt – sec = 1 joule

So 1kW is 1000W, and an hour is 60 seconds x 60 minutes.

So 1kWh = 3600kWs = 3,600,000Ws = 3,600,000J

0.5 kWh = 0.5 × 3,600,000Ws = 1800000 joules

 

Ques.4. What is the conductivity (in Mhos/m) of a 2 Ohm circular wire, when the length and the diameter of the wire are 10 m and 0.8 m respectively?

  1. 10
  2. 1
  3. 0.1
  4. 5

Given

Length =10 m

Radius = Diameter/2 = 0.8/2 =0.4m

Area of circle = πR2 = π × 0.42

Resistance = 2Ω

The formula of resistivity is given as

Unit-of-Resisitivity

∴ 

\begin{array}{l}\rho = \dfrac{{\pi \times {{0.4}^2} \times 2}}{{10}}\\\\\rho = 0.1\Omega - m\end{array}

Conductivity = 1/Resistivity

G = 1/0.1 = 10 Mhos/m

 

Ques.5. ‘Erg’ is a unit of measurement for

  1. Energy
  2. Power
  3. Voltage
  4. Impedance

Erg is the unit of energy or work in the centimeter-gram-second system. It is equivalent to 1.0 × 10−7 Joules or 2.4 × 10−8 calories.

 

Ques.6. What is the equivalent capacitance (in μF) between the terminal A and B in the circuit given below?

  1. 4.56
  2. 5.67
  3. 18.58
  4. 51

Let us mark the capacitance as shown in the figure

The capacitance C2, C3, C4 are in parallel with capacitance C6, C7

1/C = (20 + 20 + 10) || (30 + 20)

C = (1/20 + 1/20 + 1/10) || (1/30 + 1/20)

C = 5 + 12 = 17 μF

 

Now all the three capacitors are connected in series as shown in the figure.

C = 17 + 17 + 17

1/C = 1/17 + 1/17 + 1/17

C = 5.67μF

 

Ques.7. What is the equivalent inductance (in H) between the terminal A and B in the circuit given below?

  1. 1
  2. 1.42
  3. 3.2
  4. 7

Let us mark the capacitance as shown in the figure

The inductance L2, L3, L4 are in series therefore equivalent inductance

L = 1 + 1 + 1 = 3H

Similarly the inductance L6, L5 are connected in series 

L = 1 + 1 = 2 H

Now the inductance 3H and 2H are connected in parallel with each other therefore

L = 3 || 2 = (3 × 2) ⁄ (3 + 2)

L = 1.2 H

Now the all the three inductance are connected in series hence

L = 1 + 1.2 + 1

L = 3.2 H

 

Ques.8. Which of the following quantity will remain the same, when a layer of Teflon is inserted between the plates of a charged parallel plate capacitor?

  1. Capacitance
  2. Charge
  3. The energy of the capacitor
  4. Potential

Capacitors with Dielectrics

A dielectric is an insulating material such as rubber, glass, Teflon or waxed paper. Let us assume the capacitor is charged but not connected to anything.

Now, if the air space between the plates is filled with the dielectric medium such as Teflon the magnitude of electric field falls down. Weakening in E means decrement in V, since E = V/d. Q remaining constant, because there is no wire to carry the charge so it does not have a path to flow. Now the decrement in V will mean increment in the value of C (using Q = CV).

Note:-  If the capacitor remains connected to the battery, the voltage across the capacitor necessarily remains the same. If you disconnect the capacitor from the battery before making any modifications to the capacitor, the capacitor is an isolated system and its charge remains the same because the capacitor discharges when a conducting path is provided across the plates without any applied voltage connected.

 

Ques.9. What will be the value of resistance (in kilo-ohms) of a carbon composition resistor having color-coding of brown-black-brown-black?

  1. 400
  2. 200
  3. 300
  4. 100

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, the color of the resistor is given as brown-black-brown-black

1 = Brown

0 = Black

101 = Brown

± 0% = black

We get Resistance value from the color code (Brown,Black,Brown,Black) is 10x(101) = 100 Kiloohm

Here we are not considering the last color (tolerance-black) Because in the question there is no specification of minimum or maximum current.

 

Ques.10. Which of the following is NOT a type of capacitor

  1. Ceramic
  2. Electrolytic
  3. Film
  4. Wire Wound

Electrolytic Capacitors are generally used in DC power supply circuits due to their large capacitance’s and small size to help reduce the ripple voltage or for coupling and decoupling applications. Electrolytic’s generally come in two basic forms; Aluminium Electrolytic Capacitors and Tantalum Electrolytic Capacitors. The tantalum and electrolytic capacitors have ratings of 6, 10. 12. 15. 20. 25, 35, 50. 75, I00V and higher.

Paper capacitors: Paper capacitors use paper as the dielectric. They are made flat strips of metal foil plates separated by a dielectric, which is usually waxed paper. Paper capacitor have values in the picofarad and low microfarad, ranges. The voltage ratings are usually less than 6 V. Paper capacitors arc usually sealed with wax to prevent moisture problems.

The voltage rating of capacitors is very important. A typical set of values marked on a capacity might be “10 μF, 50 DCWV.” Such a capacitor h a capacitance of 10 μF and a “dc working voltage” of 50V. This means that a voltage in excess of 50 could damage the plates of the capacitors.

Ceramic capacitors: Ceramic capacitors make use of ceramic as the dielectric material. Basically, ceramic materials are fanned from titanium dioxide. Some of the dielectric materials used are barium titanate, magnesium titanic, and zirconium titanate.

Note:-

Wire Wound Resistors:- The wire wound resistors have a resistive wire wound on a central ceramic core. It is one of the oldest technologies, wire wounds provide the best-known characteristics of high-temperature stability and power handling ability. Nichrome, Manganin, and Evanohm are the three most widely used wires for wire wound resistor. Wirewound resistors have a significant amount of inductance due to the construction technique.

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