Obtain necessary condition for balancing state of wheat Stone Bridge by using Kirchoff's law. draw necessary circuit diagram

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Solution

I = 0 in galvanometer branch
Applying Kirchoff's law for loop ABCD A
$- I_{1} R_{1} + 0 + I_{2} R_{2} = 0$
$I_{1} R_{1} = 1_{2} R_{2}$
$\frac{I_{1}}{I_{2}} = \frac{R_{4}}{R_{1}}$ ......(1)
Applying Kirchoff's law for loop BCDB
$R_{3} I_{1} + I_{2} R_{4} + 0 = 4$
$I_{1} R_{3} = I_{2} R_{4}$
$\frac{I_{1}}{I_{2}} = \frac{R_{4}}{R_{3}}$ ........(2)
From equal (i) & (2)
$\frac{I_{1}}{I_{2}} = \frac{R_{2}}{R_{1}} = \frac{R_{4}}{R_{3}}$
$\frac{R_{2}}{R_{1}} = \frac{R_{4}}{R_{3}}$
$\frac{R_{1}}{R_{2}} = \frac{R_{3}}{R_{4}}$ Balancing condition for wheat stone bridge

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