EE MCQ

SSC JE Electrical 2019 with solution SET-1
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If dΦ is the luminous flux incident normally on an area dA, then Illuminance is given as

Luminous flux measures the total radiant power reaching a surface, but without regard to surface position: the power might be concentrated on only part of the surface or spread evenly across it.

Irradiance (Ee) or illuminance (Ev) is the density of incident radiant flux or luminous flux at a point on a surface and is defined as radiant flux or luminous flux per unit area, as given by

E = dφ/dA

where (dφ) is the radiant flux or luminous flux incident on an element dA of the surface containing the point. The unit of irradiance is Wm−2, and that of illuminance is lux

Related Question

SSC JE Electrical 2019 with solution SET-1
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The output of the logic circuit given below represents the gate.

In the given diagram all are NOR Gate . The final output is shown in the figure.

At stage 1 the output will be \overline A \& \overline B

At stage 2 the output will be \overline {\overline A + \overline B } = A.B

And the final output will be \overline {A.B}

Hence for input A & B the output is \overline {AB} in case of Nand gate.

SSC JE Electrical 2019 with solution SET-1
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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

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.

SSC JE Electrical 2019 with solution SET-1
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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.

SSC JE Electrical 2019 with solution SET-1
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Please check out below figure.

From the figure it can be concluded that the voltmeter reads 5 volts as shown in the figure below. Based on the voltmeter and ammeter readings in the measuring network, determine the value of the resistor R Here Current I = 1/2 A = 0.5 A Voltage V = 5 V R = V/I = 0.5/5 R = 10Ω