# Wheatstone bridge

if r1 has value of 600 and r2=r3=r4=400 ohms. supply voltage is 10v.
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?

Your Wheatstone bridge is not balanced. You cannot have r1=600 and r2=r3=r4=400 ohms. Check all your resistances.

Since you supplied us with all four resistance values I'm assuming that you're looking for the voltage across the bridge. If that is the case then your formula is incorrect. Check Wikipedia.

vk6kro
I think you have the right answer but the wrong units. Assuming you are looking for the voltage between the junctions at the centre of the bridge.

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.

Wheatstone bridges are designed to use a sensitive galvanometer across the two junctions at the center of the bridge, and to zero the current thru the galvanometer using the bridge. If you are using a 120 ohm strain gauge, then you want (for example) three 400 ohm resistors, and a 280 ohm setting on the bridge in series with the strain gauge. Even better if you can use three 120 ohm resistors, because then you can get a direct reading of the strain gauge resistor on the bridge setting..

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

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