XOR GATE USING NAND GATE

Hi guys.In any Electrical Interview(Engineering services Interview or GATE Interview) this question will be surely asked."Form XOR Gate using NAND Gate logic or NOR Gate logic (Universal Gates)."The answer for this question can be formed using the truth table of XOR Gate and truth table of NAND Gate.In Interviews they might ask you to form other logic gates using the Universal Gates(NAND Gate and NOR Gate). So be prepared to form other logic gates too.If you have any doubts please post your doubts in the comments section.

TRUTH TABLE FOR NAND GATE(UNIVERSAL GATE)

EXCLUSIVE OR GATE USING NAND GATE

 

TRUTH TABLE FOR EXCLUSIVE OR GATE

RESISTANCE COLOUR CODE FOR RESISTORS

In many interviews for the Electrical Engineers,they will give you a resistor in your hand ask you to tell the resistance value by seeing the colour codes on the resistor.In basic Electrical workshop, the first thing you learnt would be "How to identify the resistance value using colour code for the resistor ?".The way to remember the colour code is BB ROY of Great Britain has Very Good Wife.

Usually the resistors would be having 4-band code.Starting from the left the first 2 bands are the significant figure of the resistance value.The Third band is the decimal multiplier.There will be a much bigger gap between the third and fourth band.The fourth band indicates the tolerance value.Usually the colour code for tolerance is either Gold or Silver.

Gold indicates 5% tolerance level,Silver indicates 10% tolerance level and no fourth band indicates 20% tolerance level.Some resistors have five band colour.In that case first three bands are significant figures of the resistance value,fourth band is the decimal multiplier,fifth band is the tolerance level.

B    - Black

B(ROY)    - Brown
R    - Red
O    - Orange
Y    - Yellow
G(reat)    - Green
Britain    - Blue
Very    - Violet
Good    - Grey
Wife    - White 

ONLINE ELECTRICAL ENGINEERING TEST 2

Hi all.Please take up the Electrical Engineering Test-2.Before starting the test take a paper and pen for rough work.Donot use calculator.Questions have been prepared in such a way that you can gain knowledge of some concepts.Some of these questions were asked in competitive exams.Get ready and start the Test !!Your valuable comments are welcome to improve the quality of the tests.

TIME LEFT :

CLICK HERE TO START THE TEST

1. If the resistance of a material at 0 ℃ is 3.83 Ω and its temperature co-efficient,∝ =2.27 ×10ˉ3 ℃ˉ1, What is the resistance at 60 ℃

   A) 4.15 Ω

   B) 4.35 Ω

   C) 4.45 Ω

   D) 4.20 Ω

ANSWER :

B) 4.35 Ω
EXPLANATION :
Rt = R0(1+∝T) = 3.83(1+(2.27×10ˉ3)*60)
= 3.83(1.1362) = 4.35 Ω.

2. Three resistances 7 Ω,8 Ω,10 Ω are connected in series across a 100 V source. What is the current through 10 Ω resistor ?

   A) 5 A

   B) 20 A

   C) 4 A

   D) 40 A

ANSWER :

C) 4 A.
EXPLANATION :
Rtotal = 7+8+10 = 25 Ω
Is = Vs/Rtotal = 100 /25 = 4 A
Current is same across all the three resistors as they are connected in series.

3. Three resistances 10 Ω, 8 Ω,4Ω are connected in parallel across 100 V source.Find the power drawn by 8 Ω.

   A) 1.25 KW

   B) 1150 W

   C) 1000 W

   D) 1200 W

ANSWER :

A) 1.25 KW
EXPLANATION :
I8Ω = 100/8 = 12.5 A
P8Ω = (I8Ω)^2×R = 12.5×12.5×8 = 1250 W (or) 1.25 KW
P8Ω = V×I8Ω = 100×12.5 = 1250 W (or) 1.25 KW
Voltage is same across all the three resistors as they are connected in parallel.

4. Find Rab

   A) 20 Ω

   B) 18 Ω

   C) 19 Ω

   D) 17 Ω

ANSWER :

C) 19 Ω
EXPLANATION :
1/Rcd = (1/4)+(1/6)+(1/3) = (3+2+4)/12 =9/12
Rcd = 12/9 = 1.33 Ω
1/Rae = (1/(.77+.33+6))+(1/8) = (1/8)+(1/8) = 2/8
Rae = 8/2 = 4 Ω
Rab = 4+15 = 19 Ω

5. Determine magnitude & direction of I

   A) 9A into the node

   B) 8A awaythe node

   C) 9A awaythe node

   D) 8A into the node

ANSWER :

D) 8A into the node
EXPLANATION :
Assume inward direction as positive.
I-2+6+5-3-7+1-3 = 0
I = 8A
direction is into the node.

6. Two resistances R1 = 10 Ω & R2 = 30 Ω are connected in parallel and are supplied by a current sourcw I = 5 A. Find the power consumed by 10 Ω.

   A) 130.525 Watts

   B) 145.625 Watts

   C) 140.625 Watts

   D) 150 Watts

ANSWER :

C) 140.625 Watts
EXPLANATION :
P10Ω =
(I10Ω)^2×R1 = [(I×R2)/(R1+R2)]^2×R1
= [(5×5×30×30)/(40×40)]×10 = (25×900)/(4×40)
= 5625/40 = 140.625 Watts

7. Ra = 5 Ω , Rb = 8 Ω , Rc = 7 Ω , R2 = ?

   A) 4 Ω

   B) 3 Ω

   C) 2 Ω

   D) 1.5 Ω

ANSWER :

C) 2 Ω
EXPLANATION :
R2 = (RaRb)/(Ra+Rb+Rc)
= (5×8)/(5+8+7) = 40/20 = 2 Ω

8. Find Rth

   A) 55 Ω

   B) 50 Ω

   C) 65 Ω

   D) 60 Ω

ANSWER :

D) 60 Ω
EXPLANATION :
To find the Thevenin's Resistance short circuit the voltage source

Rth = [1/((2+3)/60)]+48 = [60/5]+48 = 12+48 = 60 Ω

9. The value of R required for maximum power transfer in the network

   A) 2 Ω

   B) 4 Ω

   C) 8 Ω

   D) 16 Ω

ANSWER :

C) 8 Ω.
EXPLANATION :
To find the Thevenin's Resistance short circuit the voltage source,open circuit the current source and the load resistance.
For maximum power transfer RL = Rth

Rth = [(20×5)/(20+5)]+4 = [100/25]+4 =4+4 = 8 Ω

10. The current Source given in the figure can be represented by equivalent voltage source of

   A) 20V with 4Ω in series

   B) 20V with 4Ω in parallel

   C) 5V with 4Ω in series

   D) 5V with 4Ω in parallel

ANSWER :

A) 20V with 4Ω in series

SUBMIT TEST

ELECTRICAL QUESTION AND ANSWERS PART2

So here is the Electrical Question and it has been asked by Bagi .His doubt is "What is No Load condition?Can anyone explain what No Load condition exactly mean?" .The answer for this question is very simple.

Let me give an analogy.If you want to carry a lap top or a luggage , it is just a load for you in addition to your weight.Imagine a train (Engine A) running with 2 coaches(Engine A) and another train with 20 coaches(Engine B) .Obviously the Engine A has less load than Engine B to pull.

Similarly when it comes to electrical,we say it as Load current.As the supply voltage is constant, the current consumption will be different for different appliances.If you take a mixer-grinder catalog,there they would have mentioned the rated current for the motor used in it.So whenever there is a appliance running at your home you will be consuming electricity and where does the electricity comes from ? .Can you guess now? .Yes,it comes from the power plant and the appliances we are running at home now act as load for these power plants.

But do our home appliance come in direct contact with the power plants.No is the answer,for supplying the generated electricity we have substations and distribution stations. So the electrical load is in substation as well.Now you will be having clear idea about load.

Whenever we connect/disconnect an appliance to the plug point we ensure that switch is in Off state.If you connect/disconnect keeping the switch in On state,you can see the sparks coming from the plug point.When the appliance is in off state there is no load for the appliance.Same thing applies to isolator in substation.Before doing any maintenance work,the load will be disconnected using circuit breakers.Then the isolators will be opened to perform the maintenance/repair work and closed when the work is over.And then close the circuit breaker to resume the operation.

Now you know why there is announced power cuts in your area.Its for maintenance work or to manage the more demand of electricity.

HVDC IN INDIA TALCHER KOLAR LINK

The Video describes about how power is transferred from Talcher to Kolar using HVDC Link.This is one of the largest HVDC Projects in India.Hope this video will be useful for the Electrical Engineers.Online Electrical Engineering does not own the rigths for this video.It has been shared on knowledge sharing basis, as it would be helpful for the budding engineers.

HVDC LINES IN INDIA

Here is the list of the HVDC Lines in India.These data might be useful for the electrical engineers in answering the questions during interview.Cheers !!!

HVDC TRANSMISSION LINES IN INDIA
HVDC LINK NAME	DESCRIPTION ABOUT HVDC LINK
Vindhyachal Back-To-Back link	connects Northern and Western regions with 2 lines.Transmission Power is 250 MW and the Operational Voltage is 70 KV
Chandrapur Back-To-Back link	connects southern and western regions with 2 lines.Transmission Power is 500 MW and the Operational Voltage is 140 KV
Sasaram Back-To-Back link	connects northern and eastern regions.Transmission Power is 500 MW and the Operational Voltage is 140 KV
Vizag Back-To-Back link	connects southern and eastern regions.Transmission Power is 500 MW and the Operational Voltage is 140 KV
Rihand-Dadri	Maximum Transmission Power is 1500 MW and the Operational Voltage is 500 KV.It is a Bipolar Link of length 814 KM in Uttar Pradesh.
Chandrapur - Padghe	Maximum Transmission Power is 1500 MW and the Operational Voltage is 500 KV.It is a Bipolar Link of length 753 KM in Maharastra.
Sileru-Barsoor	The Transmission length is about 196 KM and the Transmission Power is 400 MW and the Operational Voltage is 200 KV
Biswanath Chariyali -Agra	It is a bipolar Transmission Line of length 1825 KM from Assam to Uttar Pradesh.Maximum Transmission Power 6000 MW and the Transmission Voltage is 800 KV.
Talcher- Kolar	It is a bipolar Transmission Line of Length 1376 KM connecting Eastern and Southern regions . Maximum Transmission Power is 2000 MW  and Opertional Voltage is 500 KV.