There is a great deal to be answered immediately through the drafting of questions. The subject matter has grown repetitive. The nature of today’s presentation will be a little different from previous ones. I’ve come to offer you the sensation of sitting in front of a virtual viva board.

Consider the following scenario: you pass the written test for a corporation. It’s time for you to introduce yourself to the other brothers now. When you get up on the morning of your viva, you’re both excited and nervous. Because this is going to be the first brother of your employment, you should be prepared. Because to the strain, breakfast was not provided properly. Arrived at the company’s headquarters on time and in good spirits. I went to see how many job hopefuls were waiting for interviews, and I noticed that you were one of them. Everyone appears to be apprehensive. Brother, let’s start at the beginning. We’ll be going in one by one. Some people are leaving with a smile on their face, while others are dejected. Those who had previously come out with their brothers returned to the scene, and the interview began. The remainder of the group is quite interested in finding out what the question is. It’s finally your turn after a long wait. You were allowed to enter the building and sit in the chair with permission. On the viva board, there were four individuals present. They are senior executives in that organization. The episode of self-introduction has come to a conclusion. Then four officers came up to you and asked you four questions.

## Question 1 : What will be the problem if the motor is driven first by Delta then Star?

If you start by running it in delta mode, the coil of the motor will receive full line voltage, and the starting current will be significantly higher because the rotor is steady at the beginning of the operation. As a result, the motor may be damaged or destroyed. When using a star connection, a device receives significantly less current than when using a delta connection. Because of this, the motor with Star Delta connection has been launched into the industry. Your statement was then documented on a sheet of paper by the officer, who instructed you to provide a mathematical proof of your claim here.

Proof based on mathematics

Consider the following scenario: I have a 15 KW motor. I will provide a voltage of 400 volts. Under these circumstances, I will compute the amount of current you will consume in both the Star and Delta connections.

At the beginning, I was able to make use of the star connection. Knowing that the line voltage is 1.732 times the phase voltage when a star connection is made, we may proceed to the next step.

As a result, the line voltage VL equals 1.732 x 220 = 400 volts (approx). The load current IL = 15000 / (1.732 x 400 x 0.8) = 27 Amps is calculated as follows:

The line voltage will be equal to the phase voltage if a delta connection is made to it at this point in time. In other words, VL = Vp = 400 volts. In this situation, the load current IL = 15000 divided by (400 divided by 0.8) equals 46 Amps. In other words, it can be shown that the motor in the delta connection consumes 1.7 times the amount of current that the motor in the star connection does. After that, the officer was really pleased with your response.

## Question 2 : Why inductor bank is not used for power factor improvement?

The majority of the loads utilized in the industry are inductive. It has a resistive and capacitive component. As a result, the power factor is lagging far behind the competition. The use of an inductor bank will result in additional lagging. As a result, the inductor bank is not utilized.

### The officer asked again what is the problem if lagging?

The power factor is defined as the cosine of the angle formed by the voltage and the current. Because of this fact, the power factor will decrease in value in direct proportion to how little the value of this intermediate angle is. The use of an inductor bank, on the other hand, increases the magnitude of the angle between voltage and current, resulting in a reduction in power factor. The questioner was happy and inquired, “What are the drawbacks of having a low power factor?” again and again.

- The loss of line will increase.
- It will require a substantial amount of space.
- A fine will be levied against the electric utility.
- It will be necessary to use instruments with greater kVA ratings.

The officer inquired once more as to how one might explain things mathematically.

The following describes the relationship between electric potential (P), voltage (V), and electric current (I) in alternating current systems:

In other words, P = V × I × cosθ ……. (1)

The angle between the vectors showing the voltage and current flow in the phasor domain is represented by the symbol. (In a DC system, if = 0, then P = V IP = V I is obtained. The power factor of a system is represented by the cosine. The apparent power, also known as the kVA rating, is the sum of the amounts V and I. It is denoted by the letter S. Typically, the kVA rating of an instrument is used to determine the capability of the device.

In essence, In essence, V × I = S and cosθ = P.F.

Once you determine that your system’s power factor is low, you will need to purchase an electrical device with a larger kVA rating in order to obtain the same amount of power (P). Because the first equation states that

S = P / P.F

The greater the kVA rating of the device, the higher the cost of the gadget is.. As a result, the price will increase.

If you continue to use the same device (in this case, S is no longer variable), your active power will be lowered owing to the low power factor (P). Because,

P = S × P.F.

It is possible that you will be fined by your power supply provider if the value of the power factor in your house is too low. This is due to the fact that when electricity with a low quality power factor is delivered, the power flow through the supply line is increased. How?

Return to Equation 1 for a moment. Electricity is always delivered to your home at a certain voltage by your supplier. Its voltage is 230 volts, which is appropriate for our subcontinent. The equipment we employ is specifically designed to maximize this potential. As a result, V must always be fixed. Assuming that you have a given quantity of power (P), remember that the lower your power factor, the greater your current.

I = PV × P.F.

Furthermore, if the current flow is large, there will be system loss, which is defined as the loss of power in the supply line (Power loss = I2 R). In order to transmit more power, the wire thickness must be raised as well. As a result, the price goes up. That is why the electric distribution company has the authority to charge you.

As a result, we strive to retain the power factor at a high level.

## Question 4 : Armature is made of iron and commutator is made of copper. Any special reason?

Because of its high magnetic permeability, iron is commonly employed in the armature. Copper is utilized in commutators because it has high conductivity and is therefore ideal for this use.

## Question 4: What is the main purpose of your job desire?

My professional interests are in the field of electricity. Furthermore, money is required in the worldly sphere. At the same time, I have vowed to devote my time and energy to the advancement of mankind.The four officers were satisfied with your performance, and you walked out of the Vaiba board with a smile on your face.