Would a digital multimeter contain an A/D system?

AI Thread Summary
A digital multimeter does contain an analog-to-digital converter (ADC) to measure and display analog inputs like voltage or current on a digital screen. The term "A/D system" in the context of the Vernier device refers to this conversion process, which is essential for data collection in experiments. Additionally, some multimeters may feature data acquisition capabilities, allowing them to transmit digitized data to a computer via USB or serial connections. Understanding these components is crucial for effective use in laboratory settings. Overall, both devices utilize A/D systems for accurate data analysis and representation.
nargle
Messages
5
Reaction score
0
In the lab for my physics course we're using a Vernier "A/D system" to collect/transmit data from the experiment sensor to the computer for analysis. It's my first time using one, and we weren't told anything about it. In this sense does A/D mean "analog to digital" converter? Would a digital multimeter then also contain an A/D system? Just curious...

Thanks!
 
Physics news on Phys.org
Probably means Analog-Digital could also mean Acquisition+Data - it's just a model name.
A multimeter contains an ADC (analog digital converter) because it must measure an analog input (voltage or current) and display it on a digital display - numbers on the LCD
It might also have a data acquisition feature where it sends the digitised data to a PC with a USB or serial port.
 
Hmmm...I see. Thanks for the explanation.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top