- #1

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im getting the wrong answer and do not know how.

im using [(R4R2 - R3R1) / (R1+R4)(R2+R3)] * VS

i get an answer of [160000 - 240000 / (1000)(800)] * 10 = [-80000 / 80000] * 10 = -1ohm

but this is the incorrect answer. where am i going wrong?

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- Thread starter orla22
- Start date

- #1

- 22

- 0

im getting the wrong answer and do not know how.

im using [(R4R2 - R3R1) / (R1+R4)(R2+R3)] * VS

i get an answer of [160000 - 240000 / (1000)(800)] * 10 = [-80000 / 80000] * 10 = -1ohm

but this is the incorrect answer. where am i going wrong?

- #2

- 4,662

- 6

- #3

- 895

- 98

- #4

vk6kro

Science Advisor

- 4,081

- 40

One side of the bridge must have two equal resistors (both 400 ohms) so the voltage must be half the supply voltage. 5 v

The other side must have a 600 and a 400 ohm resistor. So, it must have 4/10 of the supply voltage at the junction. 4 v

So there must be 1 volt difference between the two junctions. Polarity depends on how you measure it.

- #5

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https://www.physicsforums.com/showthread.php?p=2199864#post2199864

- #6

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- #7

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hi all, could you possibly help me with this problem?

R1 and R2 are 2 guages with resistance of 100 ohm.

A steel bar has diameter 45mm and modulus of steel bar is 170GN/m^2 and guage factor of 2. How would I calculate Vout when a load of 35kN is applied?

i know what the Vout formula is and stress = tensile strain x YM.

thanks

R1 and R2 are 2 guages with resistance of 100 ohm.

A steel bar has diameter 45mm and modulus of steel bar is 170GN/m^2 and guage factor of 2. How would I calculate Vout when a load of 35kN is applied?

i know what the Vout formula is and stress = tensile strain x YM.

thanks

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