EE MCQ

SSC JE Electrical 2019 with solution SET-1
######
The law that explains about the economic size of a conductor is:

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.

Related Question

SSC JE Electrical 2019 with solution SET-1
######
Which of the following fuel has the least calorific value?

In the above option, the wood has the lowest calorific value. Its value is around 17000 – 20000 kilojoule/kg.

The calorific value of LPG is 55000 kilojoule/kg.

The calorific value of Kerosene oil is 45000 kilojoule/kg.

The calorific value of hydrogen oil is 150000 kilojoule/kg.

SSC JE Electrical 2019 with solution SET-1
######
In the below circuit, X = ?

The circuit shown in the question is an AND GATE. In an AND gate has two or more inputs but it has only one output. An input signal applied to a gate has only two stable states, either 1 (HIGH) or 0 (LOW). In AND gate for any input A&B the output is A.B.

SSC JE Electrical 2019 with solution SET-1
######
In an AC network, the load connected is (10 + j10). The phase relation between the voltage applied and the current through the load is:

L0ad impedance Z = R + jX

z = 10 + 10j

Phase angle θ= tan−1(IL/IR)

θ= tan−1(10/10)

tanθ = 1

tanθ =45°