# SSC JE electrical 2019 question paper with solution SET-1

Ques.51. The law that explains about the economic size of a conductor is:

1. Newton’s law
2. Lamberts law
4. Kelvin’s law

Explanation:-

In transmission lines, a large amount of power is transmitted over a long distance. So, voltage regulation is not important because in some lines, 40% regulation is considered satisfactory. In such cases, only the economy is important. The cost of the conductor is the main part which decides the total cost of the transmission line. Hence, the selection of the proper size of the conductor for a particular line is most important.

The most economical area of the conductor is that for which the total annual cost of the transmission line is minimum. This is known as Kelvin’s law and was given by Lord Kelvin in the year 1881.

It states that the most economical cross-section of a conductor is the value at which the annual cost of the electric energy wasted in the conductor, and annual cost of the interest and depreciation on the capital cost of the conductor are equal. Thus, the total annual charge on an overhead transmission line can be expressed as :

Total annual charge =P1 + P2α or Variable part of the energy charge = Annual cost of energy wasted.

where

P1 and P2 are constants

α is the area of the X-section of the conductor.

Ques.52. Z1 and Z2 are connected in series to form a load. A wattmeter’s current coil is connected in series with the load, whereas its pressure coil is connected across Z2. The wattmeter reads

1. Power consumed by Z1 and Z2
2. Power consumed by Z1
3. Power consumed by Z2
4. Zero always

Explanation:-

As you can see from the below figure in load Z1 is connected with the only current coil. In Load Z2 both current from the current coil (CC) and voltage from voltage coil (PC) are present (Power = V × I). Hence the  Wattmeter will read power consumed by Z2.

Ques.53. _______is a tube that brings water to turbines.

1. Spillway
2. Trash rack
3. Forebay
4. Penstock

Explanation:-

Penstock

• Penstock pipe is used to bring water from the dam to the hydraulic turbine.
• Penstock pipes are made up of steel or reinforced concrete.
• The turbine is installed at a lower level from the dam.
• The penstock is provided with a gate valve at the inlet to completely close the water supply.
• It has a control valve to control the water flow rate into the turbine.

Ques.54. Find the net capacitance of the combination in which ten capacitors of 10 μF are connected in parallel.

1. 50 μF
2. 100 μF
3. 1 μF
4. 0.1 μF

Explanation:-

When the capacitance is connected in series the total capacitance is given by

Ceq = C1 + C2 + C3………CN

Ceq = 10n

Ceq =10 × 10 = 100 μF

Ques.55. Hysteresis loss is NOT a function of

1. Frequency
2. The volume of a material
3. Steinmetz co-efficient of material
4. Ambient temperature

Explanation:-

Hysteresis Loss = Kh × BM1.67 × f × v watts

where
Kh = Hysteresis constant depends upon the material
Bm = Maximum flux density
f = frequency
v = Volume of the core

Hence the hysteresis loss does not depend upon the ambient temperature.

Ques.56. ________ is used to manufacture stay wire, earth wire, and structural components.

2. Hard drawn copper
3. Nichrome
4. Galvanized steel

Explanation:-

Galvanized steel conductors do not corrode, and possess high resistance. Hence such Wires are used in telecommunications circuits, earth wires, guard wire, stray wire, etc.

Ques.57. The voltmeter is shown in the circuit reads:

1. 24 V
2. 2.4 V
3. 1.2 V
4. 12 V

Explanation:-

Total resistance in the given circuit

R = (250 + 250)MΩ = 500 MΩ

Current I = V/R = 24/(500 × 103)

Now the Voltage in the voltmeter

[latex]= \dfrac{{24}}{{500 \times {{10}^3}}} \times 250 \times {10^3}\$

V = 12 V

Ques.58. In the two wattmeter method, the readings of the two wattmeters are 500W, 500W respectively. The load power factor in a balanced 3-phase 3-wire circuit is:

1. 0.8
2. 0.9
3. 1
4. 0

Explanation:-

In two wattmeter method the phase angle is

tanφ = √3(W1 − W2)/(W1 + W2)

tanφ = √3(500 − 500)/(500 + 500)

tanφ = 0°

φ = tan−10° = 0°

Power factor = cosφ

PF = cos0° = 1

Ques.59. In a single-phase capacitor start-and-run motor, the minimum number of capacitors to be used in it is:

1. 4
2. 2
3. 1
4. 3

Explanation:-

Capacitor start and capacitor run motor: Two capacitors are used for starting, but one of them is cut out when speed reaches 70 percent of the synchronous speed. The capacitor start-and-run motor starts with a high value and a low-value capacitor connected in parallel with each other but in series with the starting winding. This provides a very high starting torque. The centrifugal switch disconnects the high-value capacitor at 80 percent speed, but the lower value capacitor remains in the circuit.

Ques.60. The meter constant of an energy meter will be given in:

1. kW/Revolutions
2. kWh/Revolutions
3. Revolutions/kW
4. Revolutions/kWh