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Start from the voltage divider equation (since it is Vo you are balancing), and see where you get to...

If bridge is balanced (without \$ R_L \$): \$ \frac {R_2}{R_1+R_2} = \frac {R_3}{R_g+R_3} \$

In the specific case that R3=RG (and R1=R2), then RL cancels out.

A Wheatstone bridge does not work by measuring Vo. It works by adjusting the known components, to balance the bridge (Vo=0).

In this case we will adjust R3 (whilst keeping R1=R2 constant), until Vo=0. The bridge will be balanced. Rg will = R3, and we know R3, as it is calibrated. For all values of RL we can measure Rg.

It is a common misuse to call a bridge arrangement where we measure the differential error voltage an use that to calculate the unknown, "Wheatstone" it is not.

But even if you do use the Vo measurement, this arrangement still works. If you had a two wire sensor, the RL causes a direct and immediate offset error. With the 3 wire arrangement, there is 0 error at balance, and the eror term grows ratiometrically with the change in Rg. For small changes in Rg (strain gauges, small temperature range RTD), this error term will be small.

A modern measurement system would also measure VoR3, and thus could calculate and remove RL, by brute arithmetic.

Start from the voltage divider equation (since it is Vo you are balancing), and see where you get to...

If bridge is balanced (without \$ R_L \$): \$ \frac {R_2}{R_1+R_2} = \frac {R_3}{R_g+R_3} \$

In the specific case that R3=RG (and R1=R2), then RL cancels out.

A Wheatstone bridge does not work by measuring Vo. It works by adjusting the known components, to balance the bridge (Vo=0).

In this case we will adjust R3 (whilst keeping R1=R2 constant), until Vo=0. The bridge will be balanced. Rg will = R3, and we know R3, as it is calibrated. For all values of RL we can measure Rg.

It is a common misuse to call a differential arrangement where we measure the differential voltage and use that to calculate the unknown a "bridge", "Wheatstone" or other. The true bridge arrangements have identities that are true at balance, and not true away from balance. Cancellation was able to eliminate computation, and often electrical precision.

But even if you do use the Vo measurement, this arrangement still works. If you had a two wire sensor, the RL causes a direct and immediate offset error. With the 3 wire arrangement, there is 0 error at balance, and the eror term grows ratiometrically with the change in Rg. For small changes in Rg (strain gauges, small temperature range RTD), this error term will be small.

A modern measurement system would also measure VoR3, and thus could calculate and remove RL, by brute arithmetic.

Start from the voltage divider equation (since it is Vo you are balancing), and see where you get to...

If bridge is balanced (without \$ R_L \$): \$ \frac {R_2}{R_1+R_2} = \frac {R_3}{R_g+R_3} \$

In the specific case that R3=RG (and R1=R2), then RL cancels out.

A Wheatstone bridge does not work by measuring Vo. It works by adjusting the known components, to balance the bridge (Vo=0).

In this case we will adjust R3 (whilst keeping R1=R2 constant), until Vo=0. The bridge will be balanced. Rg will = R3, and we know R3, as it is calibrated. For all values of RL we can measure Rg.

It is a common misuse to call a bridge arrangement where we measure the differential error voltage an use that to calculate the unknown, "Wheatstone" it is not.

Start from the voltage divider equation (since it is Vo you are balancing), and see where you get to...

If bridge is balanced (without \$ R_L \$): \$ \frac {R_2}{R_1+R_2} = \frac {R_3}{R_g+R_3} \$

In the specific case that R3=RG (and R1=R2), then RL cancels out.

A Wheatstone bridge does not work by measuring Vo. It works by adjusting the known components, to balance the bridge (Vo=0).

In this case we will adjust R3 (whilst keeping R1=R2 constant), until Vo=0. The bridge will be balanced. Rg will = R3, and we know R3, as it is calibrated. For all values of RL we can measure Rg.

It is a common misuse to call a bridge arrangement where we measure the differential error voltage an use that to calculate the unknown, "Wheatstone" it is not.

But even if you do use the Vo measurement, this arrangement still works. If you had a two wire sensor, the RL causes a direct and immediate offset error. With the 3 wire arrangement, there is 0 error at balance, and the eror term grows ratiometrically with the change in Rg. For small changes in Rg (strain gauges, small temperature range RTD), this error term will be small.

A modern measurement system would also measure VoR3, and thus could calculate and remove RL, by brute arithmetic.

Start from the voltage divider equation (since it is Vo you are balancing), and see where you get to...

If bridge is balanced (without \$ R_L \$): \$ \frac {R_2}{R_1+R_2} = \frac {R_3}{R_g+R_3} \$

Start from the voltage divider equation (since it is Vo you are balancing), and see where you get to...

If bridge is balanced (without \$ R_L \$): \$ \frac {R_2}{R_1+R_2} = \frac {R_3}{R_g+R_3} \$

In the specific case that R3=RG (and R1=R2), then RL cancels out.

A Wheatstone bridge does not work by measuring Vo. It works by adjusting the known components, to balance the bridge (Vo=0).

In this case we will adjust R3 (whilst keeping R1=R2 constant), until Vo=0. The bridge will be balanced. Rg will = R3, and we know R3, as it is calibrated. For all values of RL we can measure Rg.

It is a common misuse to call a bridge arrangement where we measure the differential error voltage an use that to calculate the unknown, "Wheatstone" it is not.