square root calculation for dp flow transmitter? give me a explanation with ex

square root calculation for dp flow transmitter?
give me a explanation with example

Question Posted / arun_greaves

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square root calculation for dp flow transmitter? give me a explanation with example..

Answer / ginoop .k.m

ma (Linear) =(% *
16)/100 +4
example - ma=
(25*16)/100 +4
linear ma of 25%
=400/100 +4
linear ma of 25%=
4+4
linear ma of 25% =
8ma

ma (sq rt) = (sqrt
of %*16)/10 +4
example -
sqrt ma of 25% =
(sqrt
of25*16)/10 + 4
sqrt ma of 25% =
(5*16)/10 +4
sqrt ma of 25% =
8+4
sqrt ma of 25%
=12ma

Is This Answer Correct ?

45 Yes

7 No

square root calculation for dp flow transmitter? give me a explanation with example..

Answer / sampath

Sq.Root { x / 100 } * 16 + 4 = ____ mA
where, x is the % ( percenteage value )

For eg, 25% Sq.Root { 25 / 100 } * 16 + 4 = 12 mA

50% Sq.Root { 50 / 100 } * 16 + 4 = 15.31 mA

75% Sq.Root { 75 / 100 } * 16 + 4 = 17.86 mA

Is This Answer Correct ?

51 Yes

23 No

square root calculation for dp flow transmitter? give me a explanation with example..

Answer / jackass

So, the way you would go about doing square root extraction is by using an analog multipler. An analog multiplier has three inputs Vx, Vy & Vz. The output is given by Vo = (Vx*Vy)/Vz; If we set Vy = 1V and Vo = Vz through a feedback loop, we get Vo.^2 = Vx => Vo = sqrt(Vx).

If the interviewer further asks about an analog multiplier, you could mention that it can be designed by a combination of logarithmic/anti-logarithmic amplifier and an adder.

Hope that helps!

Is This Answer Correct ?

2 Yes

14 No

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