History of Kelvin Bridge
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In the nineteenth century, Baron Kelvin lectured in Glasgow as a Professor of Theoretical Physics. He is also the creator of the Kelvin contact and some resistance measurement equipment. William Thomson was his true name, and he is best known for inventing the Kelvin bridge (Thomson bridge), which is used to measure small resistances.
Definition of Kelvin Bridge
Kelvin bridge is used for measuring the unknown resistances having a value less than 1 Ω with high degree of accuracy.
A specialized version of the Wheatstone bridge network developed to minimise or considerably diminish the effect of lead and contact resistance, allowing for precise low resistance measurement.
Modification of Basic Kelvin Bridge
Figure 1 : Kelvin Bridge Basic Circuit (Linear)
The Wheatstone bridge's lead and contact resistance are represented by the resistor, RIC.
This is compensated for relatively low by the second set of Ra and Rb.
At balance ratio of Ra and Rb must equal to the ratio of R1 and R3 .
The galvanometer can be connected between these points A and B so as to obtain the null point.
It can demonstrate that when the null exists, Rx has the same value as the Wheatstone bridge, which is :
Rx = (R2R3)/R1
Therefore, a Kelvin bridge is balanced with :
Rx/R2 = R3/R1 = Rb/Ra
Derivation of Kelvin Bridge's equation
Figure 2 : Kelvin Bridge Circuit (Linear)
Same as Wheatstone Bridge, assume galvanometer has no current flowing through, the bridge is in balanced condition.
Start with, R1(Rb+Rx) = R3(Ra+R2)
Ra+R2 = (R1/R3)(Rb+Rx) ----- (1)
Since, Ra/Rb = R1/R3
Let, (Ra/Rb)+(Rb/Rb) = (R1/R3)+(R3/R3)
(Ra+Rb)/Rb = (R1+R3)/R3
Rb/(Ra+Rb) = R3/(R1+R3)
Rb = [R3/(R1+R3)][Ra+Rb] ----- (Note that : Ra+Rb = R)
Hence, Rb = [R3/(R1+R3)][R]
Since, Ra/Rb = R1/R3
also, Rb/Ra = R3/R1
Let, (Rb/Ra)+(Ra/Ra) = (R3/R1)+(R1/R1)
(Rb+Ra)/Ra = (R3+R1)/R1
Ra/(Rb+Ra) = R1/(R3+R1)
Ra = [R1/(R3+R1)][Rb+Ra] ----- (Note that : Ra+Rb = R)
Hence, Ra = [R1/(R3+R1)][R]
Substitute Ra and Rb into Equation (1) :
[R1/(R3+R1)][R]+R2 = (R1/R3) {[R3/(R1+R3)](R)+Rx}
R2+[(R1R)/(R3+R1)] = {(R1R3R)/[R3(R1+R3)]}+[(R1Rx)/R3]
R2 = (R1Rx)/R3
Therefore, Rx = (R2R3)/R1
Example with Solution : Kelvin Bridge
Question : Based on figure below, given that R1 = 20 Ω, R3 = 2R1 and ratio of Rb to Ra = 2000. Evaluate the value of Rx.
Figure 3 : Kelvin Bridge Circuit
Since, Rx/R3 = Ra/Rb
Given, R1 = 20 Ω and R3 = 2R1
R3 = 2(20 Ω)
R3 = 40 Ω
and, Rb/Ra = 2000
Ra/Rb = 1/2000
Hence, Rx = (Ra/Rb)(R3)
Rx = (1/2000)(40 Ω)
Rx = 0.02 Ω
Learning Video on Youtube : Kelvin Bridge
For those who are visual or kinesthetic learners, kindly check out this video which may give you a concept on how the kelvin double bridge device looks like in industrial visually.
Video Link : https://youtu.be/_10OnqWos9c
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