Derive the condition for balance of a Wheatstone's bridge using Kirchhoff's rules.

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Solution

Based on the circuit given in the figure , we have three known resistance's $R_{1}, R_{2}, R_{3}$ and an unknown variable resistor $R_{X}$ , a source of voltage, and a sensitive ammeter.
Kirchhoff's first rule is applied to find the currents in junctions B and D:
$I_{3} - I_{x} + I_{g} = 0$
$I_{1} - I_{2} - I_{g} = 0$
Now Kirchoff's second rule is used to find the voltage in the loops ABD and BCD:
$I_{3} . R_{3} - I_{g} . R_{g} - I_{1} . R_{1} = 0$
$I_{x} . R_{X} - I_{2} . R_{2} + I_{g} . R_{g} = 0$
The bridge is balanced and $I_{g}$ = 0, so the second set of equations can be rewritten as:
$I_{3} . R_{3} = I_{1} . R_{1}$
$I_{x} . R_{X} = I_{2} . R_{2}$
From first rule of Kirchoff.
$I_{3} = I_{x}$
$I_{1} = I_{2}$
$\frac{I_{3} . R_{3}}{I_{x} . R_{X}} = \frac{I_{1} . R_{1}}{I_{2} . R_{2}}$
$\frac{R_{1}}{R_{2}} = \frac{R_{3}}{R_{x}}$

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