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Question
Derive the equation of the balanced state in a Wheatstone bridge using Kirchhoff’s laws.
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Solution
In this case, Kirchhoff’s junction rule applied to junctions D and B (see the figure) immediately gives us the relations I1 = I3 and I2 = I4. Next, we apply Kirchhoff’s loop rule to closed loops ADBA and CBDC.
The first loop gives
–I1R1 + 0 + I2R1 = 0 ...(Ig = 0)
And the second loop gives, upon using I3 = I1, I4 = I2
I2R4 + 0 – I1R3 = 0
From the equation, we obtain,
`I_1/I_2 = R_2/R_1`
Whereas from the equation, we obtain,
`I_1/I_2 = R_4/R_3`
Hence, we obtain the condition
`R_2/R_1 = R_4/R_3`
This last equation relating the four resistors is called the balance condition for the galvanometer to give zero or null deflection.
The Wheatstone bridge and its balance condition provide a practical method for the determination of unknown resistance. Let us suppose we have an unknown resistance, which we insert in the fourth arm; R4 is thus not known. Keeping known resistances R1 and R2 in the first and second arm of the bridge, we go on varying R3 till the galvanometer shows a null deflection. The bridge then is balanced, and from the balance condition the value of the unknown resistance R4 is given by, R4 = `R_3 R_2/R_1`
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