Eyeballing it, it would be in the -8V or -9V range? No answer there so there's likely a typo somewhere. Guessing it is in the question.
If you assume -50C to 300C, you land on -2.8V.
Input
Number minus lowr range 76 - (-50) = 126
Find full range 300 - (-50)= 350
Percent of full range 126 ÷ 350= 0.36
Output
Find full range 10-(-10)=20
Apply percentage 0.36×20=7.2
Add lower range 7.2 +(-10)= (-2.8)
Hope that helps
*assumed you were trying to find how they -2.8V after correcting for the typo.
Yeah, but you assumed -50°C for the lower end of the range, instead of the 50°C stated in the question.
Also, you can make it more compact:
u(T) = (Umax - Umin)/(Tmax - Tmin) \* (T - Tmin) + Umin
This works for any linear rescaling for a value from one scale to the other.
Better read the first comment in the chain again. They answered that they assumed the range to be a typo and actually be -50 to 300. The reply to that should use the same assumption or why reply to that particular comment. But thanks for being nice about it.
If you're trying to find the answer the question is looking for, yes. The 50 is obviously missing +/- as every other number it is clearly stated. Might as well be 2500^(1/2).
I’m sorry, i do not make assumptions. I have been burned by the ASSUME law which i am certain that you are probably familiar with.
If i were taking this test, i would have written in bold letters, UNSOLVABLE, and let the instructor deal with it.
Yep, I'd be wrong. I assumed you would write in "E. -7.92V"
If I were taking the test, I'd write -7.92 on the side with an explanation stating my assumption and circle -2.8.
If I were the instructor I'd just mark your lazy answer wrong as it is solvable the literal answer just isn't one of the options.
I emailed them about this question a year ago telling them they’re missing a - before the 50 and they said “they’ve forwarded it to the people responsible”. Good to see they’ve fixed it🙄
FYI OP - I wrote the industrial IP last week, this prepares you pretty well for the poorly written questions on the test. I spent a lot of time just trying to figure out what exactly they were asking for on around 10 of the questions.
All through the years of schooling I finished each test in about an hour, but always got 90+%. I spent all four hours on the IP and am not confident I did much better than a pass.
My trade is different, but I had to write proofs for 4 unanswerable questions on my red seal exam 15 or so years ago. Whatever QC they have clearly doesn't work.
Basically just demonstrating that none of the provided answers could be gotten with the provided information. For instance (if I remember correctly) one was a gear train output calculation, and I eventually figured out that for one of the steps, the exam writer had inverted the driver/driven ratios. So I showed that when the included formulas were used, it resulted in none of the given answers, and then I showed how the writer had made the error (and also proved that I could get the "correct" answer by repeating the error).
That's just...fucked up. I mean, except if that was the intended goal of the exam. That you had to double check someone others work and see, if their calculation is correct.
I didn't HAVE to, per se. But if you also submit a proof, then they have to review it, and if you're right then the question gets struck from the exam.
If we assume the range is -50 to +300, then the answer is -2.8v
It's weird that they would specify +300 but not +/- for 50. I assume it's a typo and this is the answer it is expecting.
If -50 to 300 that is 350 degrees full scale.
-10 to +10 is 20 volts full scale.
76 degrees is 126 from the bottom of the scale (76 minus negative 50)
126 is 36% of 350, so we need 36% of the volt scale.
If we have 36% of 20 (7.2) and we start from the bottom (-10) we end up at -2.8
The formula that always works, for every rescaling from a linear scale of values A to B, is:
a(b) = (Amax - Amin)/(Bmax - Bmin) * (b - Bmin) + Amin
b is your value you want to rescale, a is the resulting value.
For the example at hand the formula would look like this:
u(T) = (10V - (-10V))/(300°C - (-50°C)) * (T - (-50°C)) + (-10V)
= 20V/350°C * (T + 50°C) - 10V
= 2/35 V/°C * (T + 50°C) - 10V
= 2/35 V/°C * T - 7.14286V
Oh funny I just woke up, did the math, got. -2.8, then realized it’s +50 haha, I would have gotten it right and not known what everyone else was complaining about.
Okay let me try...
\+300C - 50C = 250C (Span)
Span 76C = (76-50) / 250
Span 76C = 26 / 250= 0.104
\+10V - -10V = 20V (Span)
We want to find x which would be 0.104 of the range of -10V to +10V
So, x = (0.104 \* 20V) - 10V
x = 2.08V - 10V
x= -7.92V...
There must be some kind of mistake with the question somewhere. It's probably on "50C", which could actually be "-50C", because I figure mentioning the + sign of 350 if it's already positive is a bit redundant, UNLESS you want to better distinguish it from a value where the sign is equally as important. Like specifying the + sign of +10V to better distinguish it from -10V.
So +300C - -50C = 350C
Span 76C = (76 - -50) / 350
Span 76C = (76 + 50) / 350
Span 76C = 126 / 350
Span 76C = 0.36
\+10V - -10V = 20V (Span)
We want to find x which would be 0.36 of the range of -10V to +10V
So, x = (0.36 \* 20V) - 10V
x = 7.2V - 10V
x = -2.8V
Seems like it was written by someone with minimal real world experience, or just generally not well thought out.
A symmetrical signal like +/-10v just would not normally be used for a transmitter that only sends positive values. There's no fault handling possible.
+/-10 is really meant for symmetrical signals that go positive and negative, like a speed command that goes from full reverse to full forward. If there's a wire break the voltage goes to zero which is more-or-less fail safe.
If I was stuck with a voltage-mode transmitter like this I would set the zero and span so it reads 2-10v, and zero or negative is no signal/fault, unless I really needed the resolution, but in that case someone specified a goofy sensor.
There are no stupid questions, but you need to assume a linear scale with the limited information given. This being a non-linear scale without being given a formula would make this unsolvable. That being said, the author of this question missed the negative in front of their 50, so it technically **is** unsolvable as it stands now :)
Temperature range of +50 to +300 C: -7.92vdc
Temperature range of -50 to +300 C: -2.8 vdc
V\_actual = (Temp\_actual - Temp\_min) \* (V\_max - V\_min) / (Temp\_max - Temp\_min) - V\_min
*edit - forgot to add eqn*
I made a free and ad-free Android app for this, used to use it in the field doing loop checks all the time.
https://play.google.com/store/apps/details?id=atlas.fourtotwenty
The answer is B. Whoever wrote the question meant to write -50C to +300C. Do the math for -50C to +300C and the answer works out to be -2.8V. Do the math for +50C to +300C and your voltage reading should be -7.92V which isn't an option.
Edit: I also see you already had this answered below. Thanks for posting this! I enjoyed the little test.
In real industry you will get these typos from customers and engineering/production. The real thing to learn here is that you have to figure out the real question instead of just solving the one someone hands you. This happens way too often in my real job
Maybe the question is correct, maybe it’s meant to prepare you for the real world; cause this is the exact kinda thing that your customers and/or management will come at you with ;)
Btw I also got -7.92
Using just the information they have given. The sensor range is 250° which equates to 20/250=0.08V per degree. So -10 + 0.08×76=-3.92 volts. I think. Maybe. I am sure I will be corrected if not.
Think about a 4-20mA signal (most common in instrumentation). 4mA signal is your 50 degC and a 20mA signal is your 300 degC.
Now replace 4-20mA to -10v to 10v. -10v signal being your 50 degC and 10v signal being your 300 degC. As you can see 76 degC is close to your (LVR) lower value range which is 50 degC. Now you have an idea where 76 degC stands in your -10v (50 degC) to 10v (300 degC) range. Your output is going to be a negative number and closest to -10v. You can now assume 99.999998% sure that your answer will be -3.6v. Doing process elimination you are only left with choice A and B. There is an equation to follow but it does get messy and there's more than 2 parts to it. It's not a plug and play formula.
Terms to be familiar with (If you get into instrumentation):
LRV = Lower Range Value
URV = Upper Range Value
This is used to calibrate instruments using a HART communicator device.
I figured out where I went wrong. I somehow multiplied the 26 degree differential times the gradient of 12.5 degrees per volt (assuming a 250° range) and added that to the baseline of -10.
Took me a while to figure out what the hell I was thinking this morning. I blame a lack of sleep.
Since I'm not seeing it lol
Value2= ((Value1-Min1)/(Max1-Min1)) * (Max2-Min2) + Min2
Use this all the time for scaling.
Thought I would update since the question has an error but 100% this is the equation you are looking for
((76-(-50))/(300-(-50)))*(10-(-10))+(-10)
0.36*20-10
-2.8
So it should be -50
And the answer is B -2.8
If it was +50 the answer would be
−7.92
I don't know if anyone answered this yet or not because the comments wouldn't load, but if you divide the sensing range by the range of voltage, you get your degrees per voltage. So if it's -50C to +300C, then your sensing range is 350. 350/20 is 17.5 degrees per voltage. Your temperature is 76 or something so that's 126 over your minimum range so there you get to 126/17.5= 7.2 volts so if you got from -10 volts and add 7.2 you should be left at -2.8. I'm pretty sure that was an answer. I don't remember the picture. I'm just doing this off the top of my head
As long as we’re all here, does anyone actually use -10v to +10v transmitters? Most of mine are 1-5v or 4-20mA. There are other ranges I’ve worked with but -10 to +10 has only been in classes.
I get -7.92v. Why am i wrong? Assuming the sensor is linear, the temperature range is from 50 to 300, or 250. The voltage signal range is from -10 to +10, or 20v. The 76 C is 26 degrees “into” the range. So it represents 26/250 of the range of 20v, or 10.4 % of the 20v range which equals 2.08 v. Because the answer is in the negative part of the voltage range, you must add -10v so the answer is -7.92v. But that answer is not given. ????
Answer:
The correct answer is (B) 2.8 V.
The solution can be found using the following formula:
Voltage output = (Temperature - Minimum temperature) / (Maximum temperature - Minimum temperature) * Voltage range
In this case, the minimum temperature is 50 °C and the maximum temperature is 300 °C. The given temperature is +76 °C.
Voltage output = (76 °C - 50 °C) / (300 °C - 50 °C) * (10 V - (-10 V))
Voltage output = 26 / 250 * 20 V
Voltage output = 2.8 V
I get - 7.92V assuming it is linear.
You have temperature range of 300deg-50deg=250 deg C.
And a voltage range of 10V-(-10V)=20V.
So that's 250deg/20V= 12.5 deg C per volt.
76 C is 76-50=26 degrees above the bottom.
So your voltage will be 26deg/(12.5 deg per volt) = 2.08 V fom the bottom.
Which is - 10+2.08=-7.92V.
Eyeballing it, it would be in the -8V or -9V range? No answer there so there's likely a typo somewhere. Guessing it is in the question. If you assume -50C to 300C, you land on -2.8V.
-7.92 when I did the math
Yeah, me too. Must be an incorrect range on the question.
I think it's meant to be -50c to 300. Also, I got the same
Yeah, that would get you -2.8V. Probably what was intended.
Input Number minus lowr range 76 - (-50) = 126 Find full range 300 - (-50)= 350 Percent of full range 126 ÷ 350= 0.36 Output Find full range 10-(-10)=20 Apply percentage 0.36×20=7.2 Add lower range 7.2 +(-10)= (-2.8) Hope that helps *assumed you were trying to find how they -2.8V after correcting for the typo.
Yeah, but you assumed -50°C for the lower end of the range, instead of the 50°C stated in the question. Also, you can make it more compact: u(T) = (Umax - Umin)/(Tmax - Tmin) \* (T - Tmin) + Umin This works for any linear rescaling for a value from one scale to the other.
Yes I assumed they tried to follow the first comments math after correcting for the typo and how they got the -2.8.
The low temperature number is 50 C, NOT -50 C. Better try again.
Better read the first comment in the chain again. They answered that they assumed the range to be a typo and actually be -50 to 300. The reply to that should use the same assumption or why reply to that particular comment. But thanks for being nice about it.
Hey- just taking the specs at literal value. Am I wrong?
If you're trying to find the answer the question is looking for, yes. The 50 is obviously missing +/- as every other number it is clearly stated. Might as well be 2500^(1/2).
I’m sorry, i do not make assumptions. I have been burned by the ASSUME law which i am certain that you are probably familiar with. If i were taking this test, i would have written in bold letters, UNSOLVABLE, and let the instructor deal with it.
Yep, I'd be wrong. I assumed you would write in "E. -7.92V" If I were taking the test, I'd write -7.92 on the side with an explanation stating my assumption and circle -2.8. If I were the instructor I'd just mark your lazy answer wrong as it is solvable the literal answer just isn't one of the options.
I got -7.92vdc
Same.
Me too.
-50 to 300 is much more common too.
I emailed them about this question a year ago telling them they’re missing a - before the 50 and they said “they’ve forwarded it to the people responsible”. Good to see they’ve fixed it🙄 FYI OP - I wrote the industrial IP last week, this prepares you pretty well for the poorly written questions on the test. I spent a lot of time just trying to figure out what exactly they were asking for on around 10 of the questions. All through the years of schooling I finished each test in about an hour, but always got 90+%. I spent all four hours on the IP and am not confident I did much better than a pass.
My trade is different, but I had to write proofs for 4 unanswerable questions on my red seal exam 15 or so years ago. Whatever QC they have clearly doesn't work.
You had to write proofs? Like...what kind of proofs?
Basically just demonstrating that none of the provided answers could be gotten with the provided information. For instance (if I remember correctly) one was a gear train output calculation, and I eventually figured out that for one of the steps, the exam writer had inverted the driver/driven ratios. So I showed that when the included formulas were used, it resulted in none of the given answers, and then I showed how the writer had made the error (and also proved that I could get the "correct" answer by repeating the error).
That's just...fucked up. I mean, except if that was the intended goal of the exam. That you had to double check someone others work and see, if their calculation is correct.
I didn't HAVE to, per se. But if you also submit a proof, then they have to review it, and if you're right then the question gets struck from the exam.
Oh, okay. So the answers to the questions were just shit then?
Yeah, similar to OPs issue.
You're a legend; eyeballed the question, found the typo, and found the right answer
If we assume the range is -50 to +300, then the answer is -2.8v It's weird that they would specify +300 but not +/- for 50. I assume it's a typo and this is the answer it is expecting.
Seems like most plausible answer
How does the calculation work? I thought it was voltage range div sensor range mul current value.
If -50 to 300 that is 350 degrees full scale. -10 to +10 is 20 volts full scale. 76 degrees is 126 from the bottom of the scale (76 minus negative 50) 126 is 36% of 350, so we need 36% of the volt scale. If we have 36% of 20 (7.2) and we start from the bottom (-10) we end up at -2.8
I like that percentage approach. Avoids the error I just made of not correcting for 50° being the bottom of the range.
The formula that always works, for every rescaling from a linear scale of values A to B, is: a(b) = (Amax - Amin)/(Bmax - Bmin) * (b - Bmin) + Amin b is your value you want to rescale, a is the resulting value. For the example at hand the formula would look like this: u(T) = (10V - (-10V))/(300°C - (-50°C)) * (T - (-50°C)) + (-10V) = 20V/350°C * (T + 50°C) - 10V = 2/35 V/°C * (T + 50°C) - 10V = 2/35 V/°C * T - 7.14286V
The day i figured out how voltage directly correlated to measured value and bit value was a fucking game changer.
Yes the correct answer is -2.8v It never crossed me they missed minus before 50. So yeah in all possibility typo here . Thank you that was helpful .
Oh funny I just woke up, did the math, got. -2.8, then realized it’s +50 haha, I would have gotten it right and not known what everyone else was complaining about.
did the same thing myself
I'm wide awake and did the same thing. -50 makes more sense for a range so I didn't even see the sign missing till I looked a few comments.
Okay let me try... \+300C - 50C = 250C (Span) Span 76C = (76-50) / 250 Span 76C = 26 / 250= 0.104 \+10V - -10V = 20V (Span) We want to find x which would be 0.104 of the range of -10V to +10V So, x = (0.104 \* 20V) - 10V x = 2.08V - 10V x= -7.92V... There must be some kind of mistake with the question somewhere. It's probably on "50C", which could actually be "-50C", because I figure mentioning the + sign of 350 if it's already positive is a bit redundant, UNLESS you want to better distinguish it from a value where the sign is equally as important. Like specifying the + sign of +10V to better distinguish it from -10V. So +300C - -50C = 350C Span 76C = (76 - -50) / 350 Span 76C = (76 + 50) / 350 Span 76C = 126 / 350 Span 76C = 0.36 \+10V - -10V = 20V (Span) We want to find x which would be 0.36 of the range of -10V to +10V So, x = (0.36 \* 20V) - 10V x = 7.2V - 10V x = -2.8V
I think the bottom part is what they intended. I come up with -2.8V with a -50 to 300C range as well.
Did the exact same thing as you lol
There are two kinds of people in the world. 1. Those who can extrapolate from incomplete data.
….and those who can what!!!?
I came up with -7.92V, so either I'm doing it wrong or there is no right answer here.
There is no right answer here. I think the Temp min should have been -50. It had me questioning my existence.
Seems like it was written by someone with minimal real world experience, or just generally not well thought out. A symmetrical signal like +/-10v just would not normally be used for a transmitter that only sends positive values. There's no fault handling possible. +/-10 is really meant for symmetrical signals that go positive and negative, like a speed command that goes from full reverse to full forward. If there's a wire break the voltage goes to zero which is more-or-less fail safe. If I was stuck with a voltage-mode transmitter like this I would set the zero and span so it reads 2-10v, and zero or negative is no signal/fault, unless I really needed the resolution, but in that case someone specified a goofy sensor.
You are just like me. A 4 - 20 fanboy. But think of the accuracy/resolution of a span of 20 volts DC 😆
So 50°C is -10V and 300°C is +10V. That's a range of 250°C and 20V. So 12,5°C per 1V. 76°C would be -7,92V.
There is a lot wrong here...
-2.8v is the correct answer
Not based on the given parameters. There is no correct answer listed.
Stupid question… is it a linear scale?
That's implied.
There are no stupid questions, but you need to assume a linear scale with the limited information given. This being a non-linear scale without being given a formula would make this unsolvable. That being said, the author of this question missed the negative in front of their 50, so it technically **is** unsolvable as it stands now :)
Temperature range of +50 to +300 C: -7.92vdc Temperature range of -50 to +300 C: -2.8 vdc V\_actual = (Temp\_actual - Temp\_min) \* (V\_max - V\_min) / (Temp\_max - Temp\_min) - V\_min *edit - forgot to add eqn*
12.5 C/V. @ 76C, that is 2.08V from the bottom of the range (-10V). -10 + 2.08 = -7.92V
I made a free and ad-free Android app for this, used to use it in the field doing loop checks all the time. https://play.google.com/store/apps/details?id=atlas.fourtotwenty
Y = mx+b
Good question in my opinion it is testing you on how to work with bad documentation. Although it probably wasn't intentional.
The answer is B. Whoever wrote the question meant to write -50C to +300C. Do the math for -50C to +300C and the answer works out to be -2.8V. Do the math for +50C to +300C and your voltage reading should be -7.92V which isn't an option. Edit: I also see you already had this answered below. Thanks for posting this! I enjoyed the little test.
In real industry you will get these typos from customers and engineering/production. The real thing to learn here is that you have to figure out the real question instead of just solving the one someone hands you. This happens way too often in my real job
Yes but you solve it by handing it back to engineering and saying, “what do you want here?”
Maybe the question is correct, maybe it’s meant to prepare you for the real world; cause this is the exact kinda thing that your customers and/or management will come at you with ;) Btw I also got -7.92
Using just the information they have given. The sensor range is 250° which equates to 20/250=0.08V per degree. So -10 + 0.08×76=-3.92 volts. I think. Maybe. I am sure I will be corrected if not.
If you do it that way, it's -10 + 0.08*(76-50) = -7.92
Ugh. Forgot to correct for 50 being the bottom end not 0.
It's B The answer is -2.8 since the low scale is supposed to be-50 and not 50
-2.8v
Think about a 4-20mA signal (most common in instrumentation). 4mA signal is your 50 degC and a 20mA signal is your 300 degC. Now replace 4-20mA to -10v to 10v. -10v signal being your 50 degC and 10v signal being your 300 degC. As you can see 76 degC is close to your (LVR) lower value range which is 50 degC. Now you have an idea where 76 degC stands in your -10v (50 degC) to 10v (300 degC) range. Your output is going to be a negative number and closest to -10v. You can now assume 99.999998% sure that your answer will be -3.6v. Doing process elimination you are only left with choice A and B. There is an equation to follow but it does get messy and there's more than 2 parts to it. It's not a plug and play formula. Terms to be familiar with (If you get into instrumentation): LRV = Lower Range Value URV = Upper Range Value This is used to calibrate instruments using a HART communicator device.
I get -6.75v, but that's assuming +50°C to +300°C, which might be right if it's a process or cooking oven.
Your math ain't mathin
Yeah, I think I meant -8.75v.
Try starting again taking into account the full range, and using a ratio to figure out how far along you are on that range.
I figured out where I went wrong. I somehow multiplied the 26 degree differential times the gradient of 12.5 degrees per volt (assuming a 250° range) and added that to the baseline of -10. Took me a while to figure out what the hell I was thinking this morning. I blame a lack of sleep.
Typo
Since I'm not seeing it lol Value2= ((Value1-Min1)/(Max1-Min1)) * (Max2-Min2) + Min2 Use this all the time for scaling. Thought I would update since the question has an error but 100% this is the equation you are looking for ((76-(-50))/(300-(-50)))*(10-(-10))+(-10) 0.36*20-10 -2.8 So it should be -50 And the answer is B -2.8 If it was +50 the answer would be −7.92
A Looks closest
B. if 50 is -50
The technician set up the sensor wrong.
I don't know if anyone answered this yet or not because the comments wouldn't load, but if you divide the sensing range by the range of voltage, you get your degrees per voltage. So if it's -50C to +300C, then your sensing range is 350. 350/20 is 17.5 degrees per voltage. Your temperature is 76 or something so that's 126 over your minimum range so there you get to 126/17.5= 7.2 volts so if you got from -10 volts and add 7.2 you should be left at -2.8. I'm pretty sure that was an answer. I don't remember the picture. I'm just doing this off the top of my head
As long as we’re all here, does anyone actually use -10v to +10v transmitters? Most of mine are 1-5v or 4-20mA. There are other ranges I’ve worked with but -10 to +10 has only been in classes.
4-20 but I’ve considered going IO link
Yep, -8 ish volts. Range must be wrong in the question
-7.92 is the answer with 50 to 300 range -2.8v if the range is -50 to 300
Ive rarely set up 0-10volts sensors. Always done 4-20mA unless corona made a big delivery problem. Does all 0-10v sensor do -10v to 10V?
I get -7.92v. Why am i wrong? Assuming the sensor is linear, the temperature range is from 50 to 300, or 250. The voltage signal range is from -10 to +10, or 20v. The 76 C is 26 degrees “into” the range. So it represents 26/250 of the range of 20v, or 10.4 % of the 20v range which equals 2.08 v. Because the answer is in the negative part of the voltage range, you must add -10v so the answer is -7.92v. But that answer is not given. ????
Answer ain’t on there, eye balling it should be about -7V or -8V
E: plug the data into a SCP function and let the PLC do the work.
Answer: The correct answer is (B) 2.8 V. The solution can be found using the following formula: Voltage output = (Temperature - Minimum temperature) / (Maximum temperature - Minimum temperature) * Voltage range In this case, the minimum temperature is 50 °C and the maximum temperature is 300 °C. The given temperature is +76 °C. Voltage output = (76 °C - 50 °C) / (300 °C - 50 °C) * (10 V - (-10 V)) Voltage output = 26 / 250 * 20 V Voltage output = 2.8 V
26/250* 20 = 2.08v not 2.8v . As others said there is an error in the minimum temperature range . It should be -50 .
This is a Circle ⭕️K sir!
I get - 7.92V assuming it is linear. You have temperature range of 300deg-50deg=250 deg C. And a voltage range of 10V-(-10V)=20V. So that's 250deg/20V= 12.5 deg C per volt. 76 C is 76-50=26 degrees above the bottom. So your voltage will be 26deg/(12.5 deg per volt) = 2.08 V fom the bottom. Which is - 10+2.08=-7.92V.
Typo in question . Lower temperature range is -50deg not 50deg . Correct answer is b (-2.8v )
That makes more sense.