# Top 15 Examination Question And Answer On Electrical Engineering Examination

Electrical engineering is a professional engineering discipline that generally deals with the study and application of electricity, electronics, and electromagnetism. In this article, we are going to consider the top 15 examination question and answer on electrical engineering examination.

## 1.Question:

What is a solenoid?

A solenoid is an electromagnet placed in an electronic device (like a television or a computer) whose purpose is to either conduct or resist electrical flow. When a solenoid conducts electricity, it passes power on so that a device can function; when it resists electricity, it will turn off the flow of power so that the device does not function. This means that the function of solenoids are changed by on-off switches and are located in parts of electronics that control the flow of electricity.

## 2.Question:

What is the full form of ARC in arc welding?

The word ‘arc’ is not an acronym. Arc welding refers to the process of fusing metal together through the controlled use of an electric arc. Electricity is generated by an electrode and then crosses the gap between the two metallic surfaces. The ensuing heat of the electrical energy bonds the surfaces together. Arc welding can be done using either alternating current (AC) or direct current (DC).

## 3.Question:

What were Nikola Tesla’s hobbies?

Not only was he smart, Nikola Tesla was also creative. Tesla loved to write poetry, read, and obviously, invent things.

## 4.Question:

How to find equivalent capacitance of a network of capacitors

A network can have one of two kinds of circuits. In parallel circuits, the wires with their components run next to each other. To find its total capacitance, you add the value of each capacitor. In a series circuit, the wire and its components are in a straight line. In this case, you add the inverse of each capacitance. Using these methods makes it easy to find the total capacitance of any network.

A capacitor, C, holds an electric charge, Q, across two plates. Its capacitance, C, equals the charge it has divided by the voltage, V, between the plates, or C = Q/V.

When capacitors are in parallel, the current will go by different paths. But this circuit had only one total voltage drop and one total charge. In effect, the individual capacitors can be viewed as being a single, bigger one. So to find the total capacitance, all that is needed is to add each capacitor, or:

C Total = C1 + C2 +…+ Cn.

For example, three capacitors are in parallel, with 10, 20, and 40 microfarads. This would be the same as one capacitor of 70 µF.

When doing this calculation, you must be sure the units for all the capacitors are the same! If some are in pF and others in µF you have to convert one set of units to the other. Otherwise, you will be adding numbers that have a 1000 times difference.

When capacitors are in series, the same current goes through all of them, so each has the same charge. So while C = Q/V still works and Q is the same for all the capacitors, each one its V will be only a fraction of the whole. This gives the following formula for total capacitance:

1/C Total = 1/C1 + 1/C2 +…+ 1/Cn.

To solve such an equation it is necessary to have a common denominator for the fractions.

For example, to find the total capacitance of the same capacitors that were given for the parallel circuit of 10, 20, and 40 µF, you would calculate them like this:

1/10 + 1/20 + 1/40 = 1/C. Using 40 as the common denominator for the fractions, we now have: 4/40 + 2/40 + 1/40 = 7/40.

Because the equation gives its answer as 1/C we and want to find the total capacitance, we must take the inverse of the final number. ”Flipping” the fraction, we find that: C = 40/7, which is the same as 5.714 µF.

A handy check on your work is that in series circuits, total capacitance will always be less than that of the smallest capacitor.

If you have just two capacitors in series, you have a few tricks you can use to make this work easier. If the capacitors are equal, the total capacitance of the circuit is equal to half that of either one of them. So two capacitors in series each of 10 µF is the same as one of 5 µF.

If the capacitors are the same or different, you can find the answer by:

C Total = (C1 x C2) / (C1 + C2).

To use the last example, (10 x 10) / (10 + 10) = 100/20 = 5 µF.

In a complex circuit with both series and parallel components, the easiest way to calculate the total capacitance is to take each part separately and then add them together. For example, when several capacitors in series are along one branch of a parallel circuit, calculate their total C first. Then treat that number as though it was from a single capacitor in parallel with those on the other branches.

Note that the laws for capacitors in series and parallel are exactly the reverse of those for resistors.

## 5.Question:

Who was Alexander Graham Bell?

Alexander Graham Bell (1847-1922) was the inventor of the telephone. He founded Bell Telephone Company in 1877. He was born in Scotland and educated at the University of Edinburgh and University College London.

## 6.Question:

What is nanoengineering?

Nanoengineering refers to designing and manufacturing electronic devices or circuits on a nanometer scale. This could be anywhere from 0.1 to 100 nanometers in size, and for reference, a nanometer is one billionth of a meter.

## 7.Question:

Solid copper wire is a good conductor because

A solid copper wire is a good conductor because it allows electrons to freely move through it without resistance since it can ”give up” its valence electrons easily. Also, copper conserves the electrical energy that flows through the wire. Because of this, copper is often used for electrical circuits in homes, cars, businesses, etc.

The copper wire also has a few other properties that make it one of the best metals to use for electricity. It is not only very durable, but it can also be bent easily. Copper is also resistant to corrosion, which is one downfall of some conductive metals

## 8.Question:

Why back emf used for a DC motor?

The induced emf developed when the rotating conductors of the armature between the poles of the magnet in a DC motor, cut the magnetic flux, opposes the current flowing through the conductor, when the armature rotates, is called back emf. Its value depends upon the speed of rotation of the armature conductors. In starting the value of back emf is zero.

## 9.Question:

What is Slip in an induction motor?

Slip can be defined as the difference between the flux speed and the rotor speed. Speed of the rotor of an induction motor is always less than its synchronous speed. It is usually expressed as a percentage of synchronous speed and represented by the symbol “S”.

## 10.Question:

Define Real power, Apparent power, Reactive power

Real power – It is the product of voltage current and power factor.

Apparent power – It is the product of voltage and current.

Reactive power – It is the product of voltage, current, and sine of the angle between the voltage and current

## 11.Question:

Benefits of AC systems over DC system

Due to the following factors, AC systems are recommended over DC systems:

a. Maintenance is easy and can change the voltage of AC power for transmission and distribution.

b. Plant cost for AC transmission (circuit breakers, transformers etc) is much lower than the equivalent DC transmission

c. AC is produced from the power stations so AC is better for users than DC instead of converting it.

## 12.Question

How can you compare power engineering with electrical engineering?

Power engineering is a sub-department of electrical engineering. It deals with the generation, transmission, and distribution of power in electric form. Design of all power accessories also comes under power engineering. Power engineers focus on maintenance and design of the power grid i.e. called on grid systems and they might perform on off-grid systems that are disconnected to the program.

## 13.Question:

What is the various kind of wires used for transmission?

• Super tension wires can transmit voltage 66 kV to 132 kV
• High-tension wires can transmit voltage up to 23000 v.
•  Low-tension wires, which can transfer voltage up to 1000 v.

## Question:

Why back emf used for a dc motor? emphasize its importance.

The emf which is induced developed when the rotating conductors of the armature between the magnetic poles, in a DC motor, cut the magnetic flux, is a contradiction to the current flowing through the conductor, when the armature is rotating, is known as back emf. Its value depends upon the rotational speed of the armature conductors. The initial value of back emf is zero.

## Question:

Explain the application of storage battery power.