Wheatstone Bridge with strain gauge explained

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By using an instrument called Wheatstone's resistance measuring bridge, unknown resistance values can be determined very precisely. This can then be used to infer, for example, temperature, light intensity, or mechanical deformation due to a specific mass. From the content:
0:00 Introduction Wheatstone bridge
0:27 History Wheatstone bridge
1:19 Calculation the Wheatstone bridge in the balanced and unbalanced state
3:07 Strain gauges - how they work
4:48 Wheatstone bridge with strain gauge - Quarter- half- and full bridge
5:34 Example - Wheatstone bridge without complicated mathematics
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1 июл 2024

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Комментарии : 10

@haydn995 2 года назад

That bass line hit

@davidschnapper1384 3 года назад

Great explanation!

@socialworker8605 3 года назад

sir can you explain about synchronous motor over excitation,..(magnetic neutralisation)

@viralvideosreacttv5523 3 года назад

Great info

@learnchannel-TV 3 года назад

Thanks, watch also our next videos. The vid RFID you would like.

@ram.4152 2 года назад

Given a pressure of 2kpa, nominal resistance of 120 ohm, diameter of strain gauge 0.1mm, a GF of 2. What will be the change in resistance value of the strain gauge?

@julesnogueira5101 5 месяцев назад

can the 4 gauges of a complete bridge be placed on the same face? None in compression

@Dampfhammer96 4 месяца назад

I hope that i understood it correctly: The bridge uses the ratio between the upper and the lower resistors (R1R2 and R3R4). In the formular is somewhere "R1/R2" and "R3/R4". If you increase both resistors at the same time, the ratio between them won't change much. That's also why the strain gauge 3 and 4 are "swapped" at the full bridge. The resistors that'll change have to be kind of unsymmetrical or "crossed".

@julesnogueira5101 4 месяца назад

@@Dampfhammer96 thank you for your answer. I think I understand the importance of having gauges on both sides. However, in my experience, I can't have gauges on the 2 sides, can I at least make a half bridge by having 2 gauges on the same side or is that not possible? If so, is the formula right: Vin/Vout = (K/4)*(ε1-ε2)?