In a typical Wheatstone network, the resistances in cyclic order are $A = 10 \,\Omega $, $B = 5 \,\Omega $, $C = 4 \,\Omega $ and $D = 4 \,\Omega $ for the bridge to be balanced
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For a balance Wheatstone bridge.
$\frac{A}{B} = \frac{D}{C} \Rightarrow \frac{{10}}{5} \ne \frac{4}{4}$ (Unbalanced)
$\frac{{A'}}{B} = \frac{D}{C} \Rightarrow \frac{{A'}}{5} = \frac{4}{4}$$ \Rightarrow $ $A' = 5\,\Omega $
$A'$ $(5\,\Omega )$ is obtained by connecting a $10\,\Omega $ resistance in parallel with $A$.
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