Wheatstone Bridge Derivation
Suppose, on pressing the cell key K1, a current I flows through the cell, which splits into two parts at the end A. One part I1 flows through the resistance P in arm AB and the other part I2, through the resistance R in arm AD. The current I1 again comes to end B and gets divided into two parts. One part of it Ig flows through the galvanometer in arm BD and the remaining part (I1– Ig) flows through resistance Q in arm BC. At the end D, the current I2 from arm AD and the current Ig from arm BD, so the current flowing through the resistance S in arm DC will be (I2 + Ig).
So according to Kirchhoff’s law, in closed path ABDA,
I1P + IgG – I2R = 0 . . . (1)
And in closed path BCDB,
(I1 – Ig)Q – (I2 + Ig)S – IgG = 0 . . . (2)
The values of the resistors P, Q, R, and S are taken in such a way that no current flows through the galvanometer G when the key K2, is pressed. This is called the equilibrium state of the bridge, that is, in the equilibrium state of the bridge, the deflection in the galvanometer is zero (Ig = 0).
Putting Ig = 0, in the above equations,
I1P = I2R and I1Q = I2S
or
I1P / I1Q = I2R / I2S
or
P / Q = R / S
This is the necessary condition for the balance of the Wheatstone Bridge. With the help of the above formula, knowing the values of three resistors P, Q, and R, the value of the fourth resistance S can be found.
Wheatstone Bridge
Wheatstone bridge is a device that is used to find the resistance of a conductor, in 1842, scientist Wheatstone proposed a theory, which is called the principle of Wheatstone bridge after his name. we can prove or establish the formula for Wheatstone by using Kirchhoff laws. Wheatstone bridge is simply an electric circuit used to measure an unknown electric resistance by balancing two-point of a bridge. Let’s get started!
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