Hi, pls I need understand how to solve this instrument linear equation detaily

[nyame]

Hi, please I wish to understand how to solve this analogue instruments linear equation detaily, solving it step by step in a basic way

Will appreciate any help.
Thanks

[nyame]

Hi, please I wish to understand how this instrument linear equation was solved in a simplified method. How the answer was achieved

Will appreciate any help
Thanks

[TheHolyHorse]

This is some really simple math and nothing fancy at all.

20mA - 4mA give you 16mA, which means you have a range of 16mA to represent 0-100%. So 16 divided by a 100 gives you 0.16mA per percent of signal. So you multiply this with the percentage of the signal which in this case is 34.7 which will get you 5.552. And since the range is 4-20mA you just add 4 to offset it back to the 4-20mA range.

Just take some time and look at your problems and you'll be fine don't give up because you didn't get it the first time you looked at it.

[soldar]

Hi, please I wish to understand how to solve this analogue instruments linear equation detaily, solving it step by step in a basic way

Will appreciate any help.
Thanks

Please crop and edit your images. Some of us have slower connections and do not need to download huge files of poor graphics. In fact, if it is just text you could transcribe it.

[Zero999]

Simply converting to a monochrome PNG saves a lot of bandwidth, with no loss of clarity.

[GerryR]

The last statement "The slope-intercept formula...." is the key to what has been done; y + mx +b where m is the slope (16/100) and b is the y-intercept (4).  As previously stated, a basic formula for linear applications.

[Mechatrommer]

 

[GerryR]

@ Mechatrommer:
I guess I was 2 minutes late; hey, great minds work together.   

[Mechatrommer]

@ Mechatrommer:
I guess I was 2 minutes late; hey, great minds work together.   

i got sleepy when bending my neck 90 degree, but what i'm aware of is your image is up when i hit the post button so i guess i'm the 2 minutes late one. but nevermind i wont bother to correct my graph "de-offsetted" y range. i think yours is more corrected graph. and more detaily explanation will be needed if OP havent attend secondary class topic like ratio, percentage and how area of trapezium is derived.

i should have make it clearer
(x - a) / (y% - 0%) = (b - a) / (100% - 0%) ... slope or ratio formula... but well...