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Question
Use Kirchhoff's rules to obtain conditions for the balance condition in a Wheatstone bridge.
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Solution
Let us consider a Wheatstone bridge arrangement as shown below.
Applying Kirchhoff's loop law to the closed loop ADBA, we get
I1R−IgG−I2P=0 .....(1)
Here, G is the resistance of the galvanometer.
Applying Kirchhoff's loop law in the closed loop BDCB, we get
IgG+(I1+Ig)S−(I2−Ig)Q=0.....(2)
When the Wheatstone bridge is balanced, no current flows through the galvanometer, i.e. Ig=0
∴ From (1), we get
I1R−I2P=0
⇒I1R=I2P
`=>I_1/I_2=P/R" ......(3)"`
Similarly, from (2), we get
I1S−I2Q=0
⇒I1S=I2Q
`=>I_1/I_2=Q/S" .....(4)"`
From (3) and (4), we get
`P/R=Q/S`
or `P/Q=R/S`
This is the required balance condition in a Wheatstone bridge arrangement.
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