√(5 )/2 or √(5/2)
Walk through it step by step -- and show the steps here. You are SO close!

I mean to say will it be √(5 )/2 = √2.5
or
√(5/2)=2.2/2?

No. This is wrong. Walk through it step by step, showing your work. We can't find where you went wrong when you won't show us what you did.

 

If we're to help find the error we need to see your work

For starters you should have
Req^2-2*Req-2Ω=0

This is dimensionally unsound.

There's not much point putting the units on one term if you're not going to put them, properly, on all terms. It's sorta like the Karate Kid -- either you do units yes, or you do units no. Do units 'guess so' and squish!

It needs to be

Req^2 - 2Ω*Req - 2Ω² = 0

If you track the units through to the end, you will see that they work out very nicely -- but only if they are correct to begin with.

 

This is dimensionally unsound.

There's not much point putting the units on one term if you're not going to put them, properly, on all terms. It's sorta like the Karate Kid -- either you do units yes, or you do units no. Do units 'guess so' and squish!

It needs to be

Req^2 - 2Ω*Req - 2Ω² = 0

If you track the units through to the end, you will see that they work out very nicely -- but only if they are correct to begin with.

Granted! -- Hence my preference for Req^2-2Req-2=0 --- While I do see your point - he has challenges enough at this point

 

Granted! -- Hence my preference for Req^2-2Req-2=0 --- While I do see your point - he has challenges enough at this point

It's a tough call. I have seen many, many students literally go from sub-C work to A quality work within a couple weeks just by forcing them to use units properly. When someone has the kind of math challenges that RRITESH has, it is even MORE valuable since most of the errors he makes would mess up the units. I wouldn't be surprised to discover that the mistake he is making right now in solving the quadratic would be detectable by tracking units. But we won't know until he shows his work.

 

Then too there is the (future) requirement of 'unlearning' bad habits --- In other words I agree in principle (at least) -- Even so - that 'square' on the unit symbol seems likely to 'spook' a 'novice'...?

With genuine deference to your experience!
HP

 

sorry, i have to go for litle bit.
Actually i am going in university to post my resume.

Well hey! -- In case I'm absent upon you return, I'll check the thread at my nearest convenience -- Should I find that you have solved the problem in the interim I will 'like' 10 of your posts as promised --- Otherwise we'll keep 'forging ahead' with it ...K?

Very best regards
HP

 

This is where the sanity checks that you did (or were supposed to do) a long time ago come in.

Looking at the first three resistors, the middle one is in parallel with another resistance, so the extremes that are possible are that it is in parallel with an open and with a short. This places bounds of

2 Ω ≤ Req ≤ 3 Ω

So this part is consistent with your answer. But if we go one more level, then we see that the parallel resistance is also subject to these same bounds, which means

2 Ω + (1Ω || 2 Ω) ≤ Req ≤ 2 Ω + (1 Ω || 3 Ω)

This reduces to

2.67 Ω ≤ Req ≤ 2.75 Ω

Thus we already know what the value of Req must be to within 1.5% before we even begin. And we know that 2.47213 Ω is definitely NOT the correct answer.

So go back and ask yourself whether you chose a, b, and c correctly.

I like this solution an remember it.
anyway, how i you get 1.5%?

 

Hello,
(1 Ω)(Req) = (Req)(1 Ω) + (Req)^2 - (2 Ω)(1 Ω) - (2 Ω)(Req)
(Req)(1 Ω) + (Req)^2 - (2 Ω²) - (2 Ω)(Req) - (1 Ω)(Req) = 0
(Req)^2 + (Req)(1 Ω) - (2 Ω)(Req) - (1 Ω)(Req) - (2 Ω²) = 0
(Req)^2 + (Req)[1 Ω - 2 Ω - 1 Ω] - (2 Ω²) = 0
(Req)^2 - (2 Ω)(Req) - (2 Ω²) = 0

Why are you Ω² the unit?

 

Well hey! -- In case I'm absent upon you return, I'll check the thread at my nearest convenience -- Should I find that you have solved the problem in the interim I will 'like' 10 of your posts as promised --- Otherwise we'll keep 'forging ahead' with it ...K?

Very best regards
HP

Here is the resume is there any mistake?

Ritesh Kakkar
24 Green Park,Niranjanpur, Dehradun(Utarakhand).

Career Objective: Seeking the position of an Teacher in an organization that will enhance my horizon of knowledge and give me the chance to prove myself has the best candidate in the organization.

Personal Qualities :- Using self ideology for a better result in this field.
- Making use of deferent technique for attaining the goal of the organization.

Academic Qualification:
- Bachelor of technology (B.Tech) Shivalik College of Engineering from Uttarakhand Technical University in 2009-13 with first division.
- Intermediate (10+2) from G.R.D. Niranjanpur I.S.C. Board in 2008 with 70%.
- High School from G.R.D. Niranjanpur Dehradun I.C.S.E. Board in 2006 with 62%.

Skills
-Ability to work with children of different attitudes and skills
-Excellent communication and ability to mix with children effortlessly
-Ability to create an effective teaching method for the children
-Ability to interact with the children effectively so they can learn quickly i.e. rhyming games, story telling, etc.
-Ability to motivate the children, so that they will consider themselves a special personality

Achievement Certificates:
- Indian Institute of Technology -Roorkee, College of Engineering Roorkee, Dev Bhoomi Institute of Technology, Graphic Era University Dehradun ,Tula's Institute Dehradun, Shivalik College of Engineering-Dehradun, NATIONAL Robotics event ROBOCON 2012 in PUNE .

Project included:
Robotics Technology , DESIGNING.
Worked LED TECHNOLOGY Semiconductor, Processor and Mermory EEPROM etc
-Embedded System (MICROCONTROLLER)
-LED Display Technologies

Project Minor Project UPS 500VA : In This project AIM was to build less expensive Square wave based UPS using iron core transformer with Feed back to control voltage requlation of power supply.
* PWM generator section. * AC to DC for Battery Charging.
*PUSH-PULL Terminology based.
Major Projects LED Display: The AIM of this project was to Dispaly info on Screen of LED pixel the Code was Written on C langauge in Microcontroller.

Training : For 45 days Dhakrani (3X11.25 MW) Power Station is located on the downstream of the. Dakpathar


Department of Engineering - Civil, Mechanical, Electrical and Electronics, Computer Science

Post: Professor, Associate Professor, Assistant Professor
Qualification: As per AICTE/UGC norms
http://www.himgirizeeuniversity.edu.in/career

What this mean Associate Professor, Assistant Professor?


Mod Edit:
Please don't show the Phone No. or Email, it will bring the email robot and spammer come to the forum.

 

ax^2- 2bx-2=0
so, a=1,b=-2, c=-2
For ax^2+bx+c=0

x=(-b±√(b^2-4ac))/2a
x=(-*-2±√((-2)^2-4*1*-2))/2*1
x=(2±√(4- (-8)))/2*1
x=(2±√(4+8))/2
x=(2±√(12))/2
x=(2±3.46)/2
x=2.73 an -0.73

 

ax^2- 2bx-2=0
so, a=1,b=-2, c=-2
For ax^2+bx+c=0

x=(-b±√(b^2-4ac))/2a
x=(-*-2±√((-2)^2-4*1*-2))/2*1
x=(2±√(4- (-8)))/2*1
x=(2±√(4+8))/2
x=(2±√(12))/2
x=(2±3.46)/2
x=2.73 an -0.73

Yes! But for accuracy let's leave it in radical form:

Hence the solution set for x (and, hence, Req) is: (√3+1) and (-√3+1)

Now you've only to identify the applicable root?

 

Hint -- We're talking resistors -- not specially biased tunnel diodes

 

@RRITESH KAKKAR

You've only to identify which of the solutions applies to the problem

(√3+1) or (-√3+1) ????

Don't give up now!

 

(√3+1) or (-√3+1) ????

2.73 an -.73

 

(√3+1) or (-√3+1) ????

2.73 an -.73

One of them is the correct solution -- which is it? -- Please give your reasons for your choice...

 

2.73 an -.73

2.73 because the Resistance will be in between 2<=R<=3

 

From where you got this value?
(√3+1) or (-√3+1) ????

 

2.73 an -.73

2.73 because the Resistance will be in between 2<=R<=3

Irrational values cannot be accurately expressed via fractions/expansions so Yes! the correct answer is (√3+1) --- because non-negative values are consistent with 'resistors'

Watch your alerts -- 10 likes coming your way!!!

 

Watch your alerts -- 10 likes coming your way!!!

how?

 

how?

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