Wheatstone bridge formula derivation?
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The Wheatstone bridge formula can be derived by considering the four resistors and the current flowing through them.
Let’s assume that the resistance of each resistor is R1, R2, R3 and R4 respectively.
The Kirchhoff’s Voltage Law states that the sum of voltages around a closed loop must be zero, which implies:
V1 + V2 + V3 + V4 = 0
Now, let’s assume a current I flowing through the circuit in such a way that it enters the bottom left node and exits from the top right node. This current will divide into two paths, namely I1 and I2, where I1 flows through R1 and R3, and I2 flows through R2 and R4, as shown in the figure below.
By applying Ohm’s law to these two paths, we get:
V1/R1 + V2/R3 = I1 and V3/R2 + V4/R4 = I2
Adding these two equations and using the previously stated Kirchhoff’s Voltage Law, we get the following equation:
(V1/R1 + V3/R2) – (V2/R3 + V4/R4) = 0
Using the definition of the Wheatstone bridge from its name, we have:
V1/R1 = V2/R2 and V3/R3 = V4/R4
Substituting these values in the previous equation and rearranging, we get the final Wheatstone bridge formula:
R1*R4 – R2*R3 = 0