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'

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Hello Hp,
The question was nothing but it has taken my mind, i was thinking we have to do lot of thinking as wb suggestion to make it open an close.
i don't believe i am talking to pilot.

 

@RRITESH KAKKAR

I wish to congratulate you on your industry and initiative on this -- you have proven your sincerity in my eyes!!!

Most sincerely
HP

 

I wish to congratulate you on your industry

industry?

 

industry?

Industry (in this context) ~ effort

 

Hello Hp,
The question was nothing but it has taken my mind, i was thinking we have to do lot of thinking as wb suggestion to make it open an close.
i don't believe i am talking to pilot.

I don't understand -- please rephrase

 

I don't understand -- please rephrase

which part.

 

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

Don't "remember" it, understand the concepts behind it so that you can apply it to problems in general.

The 1.5% comes as follows:

We know that the bounds are 2.67 Ω and 2.75 Ω. So our "best" estimate is the midpoint. You could use either the arithmetic mean (the average) or the geometric mean. The arithmetic mean is

\(
R_{arith} \; = \; \frac{R_{min}+R_{max}}{2} \; = \; 2.708333 \Omega \; = \; 2.71 \Omega
\)

while the geometric mean is

\(
R_{geo} \; = \; \sqrt{R_{min} \cdot R_{max}} \; = \; 2.708012 \Omega \; = \; 2.71 \Omega
\)

As you can see, we have the bounds so tight already that there is no practical difference between the two means.

Now for the tolerance we just take the difference between one of the bounds and our best estimate and divide by that best estimate.

\(
\frac{\Delta R}{R} \; = \; \frac{2.71 \Omega \; - \; 2.67 \Omega}{2.71 \Omega} \; = \; 1.599 %
\)

If you don't do any rounding you get 1.538%.[/tex]

 

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?

Think of area. If you have a rectangle that is 5 m on one side and 10 m on the other, the area is 50 square meters, or 50 m².

Mathematically it is all consistent because you are multiplying

A = (5 m)·(10 m) = 5·10·m·m = 50 m²

 

@RRITESH KAKKAR
Re: Post #547

Understand, however, that while approximation is useful in determination of the correct root -- it is not a substitute for solving the problem ---- Got it?

 

We know that the bounds are 2.67 Ω and 2.75 Ω.
from where we got this term?
i have studied arithmetic mean (the average) or the geometric mean but in real i on't get it.

from where you study so deep?

 

A = (5 m)·(10 m) = 5·10·m·m = 50 m²

but Ω² does not make sense

 

We know that the bounds are 2.67 Ω and 2.75 Ω.
from where we got this term?
i have studied arithmetic mean (the average) or the geometric mean but in real i on't get it.

from where you study so deep?

I walked through how to get those bounds earlier:
http://forum.allaboutcircuits.com/threads/need-help-in-aptitude-question.118081/page-26#post-934552

 

@RRITESH KAKKAR

Are you familiar with the process of integration?

 

but Ω² does not make sense

It's merely upshot of tracking your units (a good habit and quite helpful - especially in more complex problems where "automatic integrity checking" is essential)-- don't let it confuse you!

 

Are you familiar with the process of integration?

yes i have one in 12th

 

yes i have one in 12th

12'th what?

 

I walked through how to get those bounds earlier:
http://forum.allaboutcircuits.com/threads/need-help-in-aptitude-question.118081/page-26#post-934552

ok ok.

2.67 Ω ≤ Req ≤ 2.75 Ω

 

A = (5 m)·(10 m) = 5·10·m·m = 50 m²

but Ω² does not make sense

Remember that this is an intermediate, mathematical expression. Intermediate terms seldom have any direct physical meaning.

But let's look at how the units work out

a = 1
b = -2 Ω
c = -2 Ω²

\(
R_{eq} \; = \; \frac{-b \; \pm \; \sqrt{b^2 \; - \; 4ac} }{2}
\.
R_{eq} \; = \; \frac{- \( -2 \Omega \) \; \pm \; \sqrt{ \( -2 \Omega \) ^2 \; - \; 4\(1\)\(-2 \Omega ^2\)} }{2}
\.
R_{eq} \; = \; \frac{2 \Omega \; \pm \; \sqrt{ 4 \Omega^2 \; + \; 8 \Omega ^2} }{2}
\.
R_{eq} \; = \; \frac{2 \Omega \; \pm \; 2 \Omega \sqrt{ 1 \; + \; 2} }{2}
\.
R_{eq} \; = \; \( 1 \; \pm \; \sqrt{3} \) \Omega
\)

You treat units just like any other factor.

 

12'th what?

12th class 1st to 12th then 4year program on btech

 

12'th what?

I'm guessing 12th grade -- when he learned (grabs slab of salt) integration.